chapter-7

7Analysis of Stressand Strain

571

Plane Stress

Problem 7.2-1 An element in plane stress is subjected to stresses sx � 4750 psi,sy � 1200 psi, and txy � 950 psi, as shown in the figure.

Determine the stresses acting on an element oriented at an angle u � 60˚ from thex axis, where the angle u is positive when counterclockwise. Show these stresses on asketch of an element oriented at the angle u.

1200 psi

950 psi

4750 psi

Solution 7.2-1

sx1 � 2910 psi ;

sx1 �sx + sy

2+

sx � sy

2 cos(2u) + txy sin(2u)

u � 60°

sx � 4750 psi sy � 1200 psi txy � 950 psi

sy1 � 3040 psi ;sy1 � sx + sy � sx1

tx1y1 � �2012 psi ;

tx1y1 � �sx � sy

2 sin(2u) + txy cos(2u)

Problem 7.2-2 Solve the preceding problem for an element in plane stresssubjected to stresses sx � 100 MPa, sy � 80 MPa, and txy � 28 MPa, as shown inthe figure.

Determine the stresses acting on an element oriented at an angle u � 30˚ fromthe x axis, where the angle u is positive when counterclockwise. Show these stresseson a sketch of an element oriented at the angle u.

80 MPa

28 MPa

100 MPa

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572 CHAPTER 7 Analysis of Stress and Strain

Problem 7.2-3 Solve Problem 7.2-1 for an element in plane stress subjected to stresses sx � �5700 psi, sy � �2300 psi, and txy � 2500 psi, as shown in the figure.

Determine the stresses acting on an element oriented at an angle u � 50˚ from the xaxis, where the angle u is positive when counterclockwise. Show these stresses on a sketchof an element oriented at the angle u.

Solution 7.2-2

sx1 � 119.2 MPa ;

sx1 �sx + sy

2+

sx � sy

2 cos (2u) + txy sin (2u)

u � 30°

sx � 100 MPa sy � 80 MPa txy � 28MPa

sy1 � 60.8 MPa ;sy1 � sx + sy � sx1

tx1y1 � 5.30 MPa ;


2 sin(2u) + txy cos(2u)

2300 psi

2500 psi

5700 psi

Solution 7.2-3

sx1 � �1243 psi ;

sx1 �sx + sy

2+

sx � sy


u � 50°

sx � �5700 psi sy � �2300 psi txy � 2500 psi

sy1 � �6757 psi ;sy1 � sx + sy � sx1

tx1y1 � 1240 psi ;


2 sin (2u) + txy cos (2u)

Problem 7.2-4 The stresses acting on element A in the web of a train rail are found to be 40 MPa tension in the horizontaldirection and 160 MPa compression in the vertical direction(see figure). Also, shear stresses of magnitude 54 MPa act in thedirections shown.

Determine the stresses acting on an element oriented at acounterclockwise angle of 52˚ from the horizontal. Show thesestresses on a sketch of an element oriented at this angle.

A

54 MPa

160 MPa

SideView

CrossSection

40 MPa

A

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SECTION 7.2 Plane Stress 573

Problem 7.2-5 Solve the preceding problem if the normal and shear stresses acting on element A are 6500 psi, 18,500 psi,and 3800 psi (in the directions shown in the figure).

Determine the stresses acting on an element oriented at acounterclockwise angle of 30˚ from the horizontal. Show thesestresses on a sketch of an element oriented at this angle.

Solution 7.2-4

sx1 � �136.6 MPa ;

sx1 �sx + sy

2+

sx � sy


u � 52°

sx � 40 MPa sy � �160 MPa txy � �54 MPa

sy1 � 16.6 MPa ;sy1 � sx + sy � sx1

tx1y1 � �84.0 MPa ;



A

3800 psi

18,500 psi

SideView

CrossSection

6500 psi

A

Solution 7.2-5

sx1 � �3041 psi ;

sx1 �sx + sy

2+

sx � sy


u � 30°

sx � 6500 psi sy ��18500 psi txy � �3800 psi

sy1 � �8959 psi ;sy1 � sx + sy � sx1

tx1y1 � �12725 psi ;



Problem 7.2-6 An element in plane stress from the fuselage of an airplane is subjected to compressive stresses of magnitude 27 MPa in the horizontal direction and tensilestresses of magnitude 5.5 MPa in the vertical direction (see figure). Also, shear stressesof magnitude 10.5 MPa act in the directions shown.

Determine the stresses acting on an element oriented at a clockwise angle of 35˚from the horizontal. Show these stresses on a sketch of an element oriented at thisangle. 10.5 MPa

5.5 MPa

27 MPa

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Problem 7.2-7 The stresses acting on element B in the web of a wide-flange beam are found to be 14,000 psi com-pression in the horizontal direction and 2600 psi compres-sion in the vertical direction (see figure). Also, shear stressesof magnitude 3800 psi act in the directions shown.

Determine the stresses acting on an element oriented at acounterclockwise angle of 40˚ from the horizontal. Showthese stresses on a sketch of an element oriented at this angle.

Solution 7.2-6

sx1 � �6.4 MPa ;

sx1 �sx + sy

2+

sx � sy


u � �35°

sx � �27 MPa sy � 5.5 MPa txy � �10.5 MPa

sy1 � �15.1 MPa ;sy1 � sx�sy � sx1

tx1y1 � �18.9 MPa ;



B

3800 psi

2600 psi

14,000 psi

SideView

CrossSection

B

Solution 7.2-7

sx1 � �13032 psi ;

sx1 �sx + sy

2+

sx � sy


u � 40°

txy � �3800 psi sx � �14000 psi sy � �2600 psi

sy1 � �3568 psi ; sy1 � sx + sy � sx1

tx1y1 � 4954 psi ;



Problem 7.2-8 Solve the preceding problem if the normal and shear stresses acting on element B are 46 MPa, 13 MPa, and 21 MPa (in the directions shown in the figure) andthe angle is 42.5˚ (clockwise).

B46 MPa

13 MPa

21 MPa

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Problem 7.2-9 The polyethylene liner of a settling pond is subjected to stresses and , as shown by the plane-stresselement in the first part of the figure.

Determine the normal and shear stresses acting on aseam oriented at an angle of to the element, asshown in the second part of the figure. Show thesestresses on a sketch of an element having its sidesparallel and perpendicular to the seam.

30°

txy � �120 psisx � 350 psi, sy �112 psi,

Solution 7.2-8

sx1 � �51.9 MPa ;

sx1 �sx + sy

2+

sx � sy


u � �42.5°

sx � �46 MPa sy � �13 MPa txy � 21 MPa

sy1 � �7.1 MPa ; sy1 � sx + sy � sx1

tx1y1 � �14.6 MPa ;



y

xO

120 psi

112 psi

350 psi

Seam

30°

Solution 7.2-9 Plane stress (angle � )

sy1� sx + sy � sx1

� 275 psi ;� �163 psi ;

tx1y1� �

sx � sy

2 sin 2u + txy cos 2u

� 187 psi ;

sx1�

sx + sy

2+

sx � sy

2 cos 2u + txy sin 2u

u � 30°

sx � 350 psi sy �112 psi txy � �120 psi The normal stress on the seam equals tension.

The shear stress on the seam equals , actingclockwise against the seam. ;

163 psi

;187 psi

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Problem 7.2-10 Solve the preceding problem if the normaland shear stresses acting on the element are

, and , and the seam is orientedat an angle of to the element (see figure).22.5°

txy � �560 kPasy � 300 kPasx � 2100 kPa,

y

xO

560 kPa

300 kPa

2100 kPa

Seam

22.5°

Solution 7.2-10 Plane stress (angle � )

u � 22.5°

sx � 2100 kPa sy � 300 kPa txy ��560 kPaThe normal stress on the seam equals tension.The shear stress on the seam equals , actingclockwise against the seam. ;

1030 kPa ;

1440 kPa


� 960 kPa ;� �1030 kPa ;

tx1y1� �

sx � sy


� 1440 kPa ;

sx1�

sx + sy

2+

sx�sy


Problem 7.2-11 A rectangular plate of dimensions is formed by welding two triangular plates (see figure). The plate is subjected to a tensilestress of in the long direction and a compressive stress of in theshort direction.

Determine the normal stress acting perpendicular to the line of the weldand the shear acting parallel to the weld. (Assume that the normal stress is positive when it acts in tension against the weld and the shear stress ispositive when it acts counterclockwise against the weld.)

tw

swtw

sw

350 psi500 psi

3.0 in. * 5.0 in. 350 psi

Weld500 psi3 in.

5 in.

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Problem 7.2-12 Solve the preceding problem for a plate of dimensions subjected to a compressive stress of in the

long direction and a tensile stress of 12.0 MPa in the short direction (see figure).

2.5 MPa100 mm * 250 mm

Solution 7.2-11 Biaxial stress (welded joint)

� 275 psi

sx1�

sx + sy

2+

sx � sy


u � arctan 3 in.

5 in.� arctan 0.6 � 30.96°

sx � 500 psi sy � �350 psi txy � 0

STRESSES ACTING ON THE WELD


� �125 psi

tx1y1� �

sx � sy

2 sin 2u + txy cos 2u � �375 psi

tw � 375 psi ;sw � �125 psi ;

12.0 MPa

Weld 2.5 MPa100 mm250 mm

Solution 7.2-12 Biaxial stress (welded joint)

� �0.5 MPa

sx1�

sx + sy

2+

sx � sy


u � arctan 100 mm

250 mm� arctan 0.4 � 21.80°

sx � �2.5 MPa sy � 12.0 MPa txy � 0

STRESSES ACTING ON THE WELD


� 10.0 MPa

tx1y1� �

sx � sy

2 sin 2u + txy cos 2u � 5.0 MPa

tw � �5.0 MPa ;sw � 10.0 MPa ;

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Problem 7.2-13 At a point on the surface of a machine the materialis in biaxial stress with , and asshown in the first part of the figure. The second part of the figureshows an inclined plane cut through the same point in the materialbut oriented at an angle .

Determine the value of the angle between zero and suchthat no normal stress acts on plane . Sketch a stress elementhaving plane as one of its sides and show all stresses acting onthe element.

aaaa

90°u

u

aa

sy � �1600 psi,sx � 3600 psiy

x

a

a

O

1600 psi

3600 psiu

Solution 7.2-13 Biaxial stress

Find angle for .

� normal stress on plane a-a

For , we obtain

‹ 2u �112.62° and u � 56.31°

cos 2u � �1000

2600sx1

� 0

� 1000 + 2600 cos 2u(psi)

sx1�

sx + sy

2+

sx � sy


s

s � 0u

txy � 0

sy � �1600 psi

sx � 3600 psi

STRESS ELEMENT

� �2400 psi tx1y1

� �sx � sy



� 2000 psi ;sx1

� 0 u � 56.31°

Problem 7.2-14 Solve the preceding problem for and (see figure).sy � �50 MPa

sx � 32 MPa y

x

a

a

O

50 MPa

32 MPau

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Problem 7.2-15 An element in plane stress from the frame of a racing car is oriented at a known angle u (see figure). On this inclined element, the normal andshear stresses have the magnitudes and directions shown in the figure.

Determine the normal and shear stresses acting on an element whose sides are parallel to the xy axes, that is, determine sx, sy, and txy. Show the results on a sketch of an element oriented at u � 0˚.


Find angles for .

� normal stress on plane a-a

For , we obtain

‹ 2u � 77.32° and u � 38.66° ;

cos 2u �9

41sx1

� 0

� �9 + 41 cos 2u ( MPa)

sx1�

sx + sy

2+

sx � sy


s

s � 0u

txy � 0

sy � �50 MPa

sx � 32 MPa

STRESS ELEMENT

� �40 MPa ;

tx1y1� �

sx � sy



� �18 MPa ;sx1

� 0 u � 38.66°

2475 psi3950 psi

14,900 psi

y

xO

u = 40°

Solution 7.2-15

Transform from

sx1 � �12813 psi ;

sx1 �sx + sy

2+

sx � sy


u � �40°

txy � 2475 psisx � �14900 psi sy � �3950 psi

u � 40° to u � 0°

sy1 � �6037 psi ;sy1 � sx + sy � sx1

tx1y1 � �4962 psi ;



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Problem 7.2-16 Solve the preceding problem for the element shown in thefigure. 24.3 MPa

62.5 MPa

24.0 MPa

y

xO

u = 55°

Solution 7.2-16

Transform from

sx1 � 56.5 MPa ;

sx1 �sx + sy

2+

sx � sy


u � �55°

txy � �24 MPasx � �24.3 MPa sy � 62.5 MPa

u � 55° to u � 0°

sy1 � �18.3 MPa ;sy1 � sx + sy � sx1

tx1y1 � �32.6 MPa ;



Problem 7.2-17 A plate in plane stress is subjected to normal stresses sx and sy and shear stress txy, as shown in the figure. At counterclockwise angles u � 35˚ and u � 75˚ from the x axis, the normal stress is 4800 psi tension.

If the stress sx equals 2200 psi tension, what are the stresses sy and txy?

y

xO

txy

sy

sx = 2200 psi

Solution 7.2-17

Find and

sx1 �sx + sy

2+

sx � sy


txysy

At u �35° and u � 75°, sx1 � 4800 psi

sx � 2200 psi sy unknown txy unknown For

(1)or 0.32899 sy + 0.93969 txy � 3323.8 psi

* cos (70°) + txy sin(70°)

4800 psi �2200 psi + sy

2+

2200 psi � sy

2

sx1 � 4800 psi

u � 35°

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For

* cos (150°) + txy sin(150°)

4800 psi �2200 psi + sy

2+

2200 psi � sy

2

sx1 � 4800 psi

u � 75°: (2)

Solve Eqs. (1) and (2):

sy � 3805 psi txy � 2205 psi ;

or 0.93301sy + 0.50000 txy � 4652.6 psi

Problem 7.2-18 The surface of an airplane wing is subjected to plane stresswith normal stresses sx and sy and shear stress txy, as shown in the figure. At acounterclockwise angle u � 32˚ from the x axis, the normal stress is 37 MPatension, and at an angle u � 48˚, it is 12 MPa compression.

If the stress sx equals 110 MPa tension, what are the stresses sy and txy?

y

xO

txy

sy

sx = 110 MPa

Solution 7.2-18

Find and

For

* cos (64°) + txy sin (64°)

37 MPa �110 MPa + sy

2+

110 MPa � sy

2

sx1 � 37 MPa

u � 32°

sx1 �sx + sy

2+

sx �sy


txysy

At u � 48°, sx1 � �12 MPa (compression)

At u � 32°, sx1 � 37 MPa (tension)

sx � 110 MPa sy unknown sxy unknown (1)

For

(2)

Solve Eqs. (1) and (2):

sy � �60.7 MPa txy � �27.9 MPa ;

or 0.55226sy + 0.99452txy � �61.25093 MPa

* cos (96°) + txy sin (96°)

�12 MPa �110 MPa + sy

2+

110 MPa � sy

2

sx1 � �12 MPa

u � 48°:

or 0.28081sy + 0.89879txy � �42.11041 MPa

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Problem 7.2-19 At a point in a structure subjected to plane stress, the stresses are sx � �4100 psi, sy � 2200 psi, and txy � 2900 psi (the sign convention for thesestresses is shown in Fig. 7-1). A stress element located at the same point in the structure(but oriented at a counterclockwise angle u1 with respect to the x axis) is subjected tothe stresses shown in the figure (sb, tb, and 1800 psi).

Assuming that the angle u1 is between zero and 90˚, calculate the normal stress sb, theshear stress tb, and the angle u1

1800 psi

O x

y

tbsb

u1

Solution 7.2-19

For

Find

Stress

Angle

sx1 �sx + sy

2+

sx � sy


u1

sb � sx + sy �1800 psi sb � �3700 psi ;sb

sb, tb, and u1

sx1 � 1800 psi sy1 � sb tx1y1 � tb

u � u1:

txy � 2900 psisx � �4100 psi sy � 2200 psi

SOLVE NUMERICALLY:

Shear Stress

tb � 3282 psi ;

tb � �sx � sy

2 sin 12u12 + txy cos 12u12

tb

2u1 � 87.32° u1 � 43.7° ;

+ 2900 psi sin 12u121800 psi � �950 psi � 3150 psi cos12u12

Principal Stresses and Maximum Shear Stresses

When solving the problems for Section 7.3, consider only the in-plane stresses (the stresses in the xy plane).

Problem 7.3-1 An element in plane stress is subjected to stresses sx � 4750 psi, sy � 1200 psi, and txy � 950 psi (see thefigure for Problem 7.2-1).

Determine the principal stresses and show them on a sketch of a properly oriented element.

Solution 7.3-1

PRINCIPAL STRESSES

up1 � 14.08°

up1 �

atana 2 txy

sx � syb

2


s2 � 962 psi ;s1 � 4988 psi ;

s2 �sx + sy

2+

sx � sy

2 cos 12up22 + txy sin 12up22

s1 �sx + sy

2+

sx � sy


up2 � up1 + 90° up2 � 104.08°

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SECTION 7.3 Principal Stresses and Maximum Shear Stresses 583

Problem 7.3-2 An element in plane stress is subjected to stresses sx � 100 MPa, sy � 80 MPa, and txy � 28 MPa (see thefigure for Problem 7.2-2).


Solution 7.3-2

PRINCIPAL STRESSES

up2 � up1 + 90° up2 � 125.17°

up1 � 35.2°

up1 �

atana 2 txy

sx � syb

2

sx � 100 MPa sy � 80 MPa txy � 28 MPa

s2 � 60 MPa ;s1 � 120 MPa ;

s2 �sx + sy

2+

sx � sy

2 cos12up22 + txy sin12up22

s1 �sx + sy

2+

sx � sy


Problem 7.3-3 An element in plane stress is subjected to stresses sx � �5700 psi, sy � �2300 psi, and txy � 2500 psi(see the figure for Problem 7.2-3).


Solution 7.3-3

PRINCIPAL STRESSES

up1 � up2 + 90° up1 � 62.1°

up2 � �27.89°

up2 �

atana 2txy

sx � syb

2


s2 � �7023 psi ;s1 � �977 psi ;

s2 �sx + sy

2+

sx � sy


s1 �sx + sy

2+

sx � sy


Problem 7.3-4 The stresses acting on element A in the web of a train rail are found to be 40 MPa tension in the horizontaldirection and 160 MPa compression in the vertical direction (see figure). Also, shear stresses of magnitude 54 MPa act in thedirections shown (see the figure for Problem 7.2-4).


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Problem 7.3-5 The normal and shear stresses acting on element A are 6500 psi, 18,500 psi, and 3800 psi (in the directionsshown in the figure) (see the figure for Problem 7.2-5).

Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly orientedelement.

Solution 7.3-4

PRINCIPAL STRESSES

up2 � up1 + 90° up2 � 75.8°

up1 � �14.2°

up1 �

atana 2txy

sx � syb

2

sx � 40 MPa sy � �160 MPa txy � �54 MPa

s2 � �173.6 MPa ;s1 � 53.6 MPa ;

s2 �sx + sy

2+

sx � sy


s1 �sx + sy

2+

sx � sy

2 cos 12up1

2 + txy sin 12up12

Solution 7.3-5

PRINCIPAL ANGLES

s2 �sx + sy

2+

sx � sy


s1 �sx + sy

2+

sx � sy

2 cos 12up1

2 + txy sin 12up12

up2 � up1 + 90° up2 � 81.55°

up1 � �8.45°

up1 �

atana 2txy

sx � syb

2

sx � 6500 psi sy � �18500 psi txy � �3800 psi

MAXIMUM SHEAR STRESSES

saver �sx + sy

2 saver � �6000 psi ;

us1 � up1 � 45° us1 � �53.4° ;

tmax � Aasx � sy

2b2

+ txy2 tmax � 13065 psi ;

s2 � �19065 psi

s1 � 7065 psi

Problem 7.3-6 An element in plane stress from the fuselage of an airplane is subjected to compressive stresses of magnitude 27 MPa in the horizontal direction and tensile stresses of magnitude 5.5 MPa in the vertical direction. Also, shear stresses of magnitude 10.5 MPa act in the directions shown (see the figure for Problem 7.2-6).


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Problem 7.3-7 The stresses acting on element B in the web of a wide-flange beam are found to be 14,000 psi compressionin the horizontal direction and 2600 psi compression in the vertical direction. Also, shear stresses of magnitude 3800 psi actin the directions shown (see the figure for Problem 7.2-7).


Solution 7.3-6

PRINCIPAL ANGLES

s2 �sx + sy

2+

sx � sy


s1 �sx + sy

2+

sx � sy

2 cos 12up1

2 + txy sin 12up12

up1 � up2 + 90° up1 � 106.43°

up2 � 16.43°

up2 �

atana 2txy

sx � syb

2

sx � �27 MPa sy � 5.5 MPa txy � �10.5 MPa


saver �sx + sy

2 saver � �10.8 MPa ;

us1 � up1 � 45° us1 � 61.4°

tmax � 19.3 MPa ;

tmax � A asx � sy

2b2

+ txy2

s2 � �30.1 MPa

s1 � 8.6 MPa

Solution 7.3-7

PRINCIPAL ANGLES

s1 �sx + sy

2+

sx � sy

2 cos12up1

2 + txy sin12up12

up1 � up2 + 90° up1 � 106.85°

up2 � 16.85°

up2 �

atana 2txy

sx � syb

2

txy � �3800 psisx � �14000 psi sy � �2600 psi


saver �sx + sy

2 saver � �8300 psi ;

us1 � up1 � 45° us1 � 61.8° ;


2b2

+ txy2 tmax � 6851 psi ;

s2 � �15151 psi

s1 � �1449 psi

s2 �sx + sy

2+

sx � sy


Problem 7.3-8 The normal and shear stresses acting on element B are sx � �46 MPa, sy � �13 MPa, and txy � 21 MPa(see figure for Problem 7.2-8).


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Problem 7.3-9 A shear wall in a reinforced concrete building is subjected to a vertical uniform load of intensity q and a horizontalforce H, as shown in the first part of the figure. (The force Hrepresents the effects of wind and earthquake loads.) As aconsequence of these loads, the stresses at point A on the surfaceof the wall have the values shown in the second part of the figure (compressive stress equal to 1100 psi and shear stress equal to 480 psi).

(a) Determine the principal stresses and show them on a sketchof a properly oriented element.

(b) Determine the maximum shear stresses and associatednormal stresses and show them on a sketch of a properly oriented element.

Solution 7.3-8

PRINCIPAL ANGLES

s2 �sx + sy

2+

sx � sy


s1 �sx + sy

2+

sx � sy

2 cos 12up1

2 + txy sin 12up12

up1 � up2 + 90° up1 � 64.08°

up2 � �25.92°

up2 �

atana 2txy

sx � syb

2

sx � �46 MPa sy � �13 MPa txy � 21 MPa


saver �sx � sy

2 saver � �29.5 MPa ;

us1 � up1 � 45° us1 � 19.08° ;

tmax � Aasx � sy

2b2

+ txy2 tmax � 26.7 MPa ;

s2 � �56.2 MPa

s1 � �2.8 MPa

1100 psi

480 psi

A

A

q

H

Solution 7.3-9 Shear wall

(a) PRINCIPAL STRESSES

For

For sx1� �1280 psi2up � 138.89°:

2up � �41.11°: sx1� 180 psi

sx1�

sx + sy

2+

sx � sy


2up � 138.89° and up � 69.44°

2up � �41.11° and up � �20.56°

tan 2up �2txy

sx � sy� �0.87273

sx � 0 sy � �1100 psi txy � �480 psi

s2 � �1280 psi and up2� 69.44°

Therefore, s1 �180 psi and up1� �20.56° f ;

07Ch07.qxd 9/27/08 1:18 PM Page 586


(b) MAXIMUM SHEAR STRESSES

and

saver �sx + sy

2� �550 psi ;

t � �730 psius2� up1

+ 45° � 24.44°

us1� up1

� 45° � �65.56° and t � 730 psi


2b2 + t2

xy � 730 psi

f ;

Problem 7.3-10 A propeller shaft subjected to combined torsion and axial thrust is designed to resist a shear stress of 56 MPa and a compressive stress of 85 MPa(see figure).

(a) Determine the principal stresses and show them on a sketch of a properlyoriented element.

(b) Determine the maximum shear stresses and associated normal stressesand show them on a sketch of a properly oriented element.

85 MPa

56 MPa

Solution 7.3-10


s2 �sx + sy

2+

sx � sy


s1 �sx + sy

2+

sx � sy

2 cos 12up1

2 + txy sin 12up12

up1 � up2 + 90° up1 � 116.4° ;up2 � 26.4°

up2 �

atana 2txy

sx � syb

2

sx � �85 MPa sy � 0 MPa Txy � �56 MPa


saver �sx + sy

2 saver � �42.5 MPa ;

us1 � up1 � 45° us1 � 71.4° ;tmax � 70.3 MPa ;


2b2 + txy

2

;s2 � �112.8 MPa

;s1 � 27.8 MPa

07Ch07.qxd 9/27/08 1:18 PM Page 587


Problems 7.3-11 sx � 2500 psi, sy � 1020 psi, txy � �900 psi

(a) Determine the principal stresses and show them on a sketch of a properlyoriented element.

(b) Determine the maximum shear stresses and associated normal stresses andshow them on a sketch of a properly oriented element.

y

xO

txy

sx

sy

Solution 7.3-11


s2 �sx + sy

2+

sx � sy


s1 �sx + sy

2+

sx � sy

2 cos 12up1

2 + txy sin 12up12

up2 � 90 ° + up1 up2 � 64.71°

up1 � 25.29°

tan(2up) �2txy

sx � sy up1 �

atana 2txy

sx � syb

2

sx � 2500 psi sy � 1020 psi txy � �900 psi Therefore,


saver �sx + sy

2 saver � 1760 psi ;

t2 � �1165 psi ;us2 � up1 � 45° us2 � 19.71° and

t1 � 1165 psi ;us1 � up1 � 45° us1 � �70.3° and

tmax � 1165 psi


2b2

+ txy2

s2 � 595 psi ;For up2 � 64.7°:

For up1 � �25.3°: s1 � 2925 psi ;

Problems 7.3-12 sx � 2150 kPa, sy � 375 kPa, txy � �460 kPa

(a) Determine the principal stresses and show them on a sketch of a properly oriented element. (b) Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly

oriented element.

Solution 7.3-12


tan(2up) �2txy

sx � sy up1 �

atana 2txy

sx�syb

2

sx � 2150 kPa sy � 375 kPa txy � �460 kPa

s2 �sx + sy

2+

sx � sy


s1 �sx + sy

2+

sx � sy

2 cos 12up1

2 + txy sin 12up12

up2 � 90° + up1 up2 � 76.30°

up1 � �13.70°

Probs. 7.3-11 through 7.3-16

07Ch07.qxd 9/27/08 1:18 PM Page 588


Therefore,



2b2

+ txy2

;s2 � 263 kPaFor up2 � 76.3°

;For up1 � �13.70° s1 � 2262 kPa

saver �sx + sy

2 saver � 1263 kPa

and t2 ��1000 kPa;us2 � up1 + 45° us2 � 31.3°

and t1 � 1000 kPa;us1 � up1 � 45° us1 � �58.7°

tmax � 145 psi

Problems 7.3-13 sx � 14,500 psi, sy � 1070 psi, txy � 1900 psi


oriented element.

Solution 7.3-13


s2 �sx + sy

2+

sx � sy


s1 �sx + sy

2+

sx � sy

2 cos 12up1

2 + txy sin 12up12

up2 � 90° + up1 up2 � 97.90°

up1 � 7.90°

tan(2up) �2txy

sx � sy up1 �

atana 2txy

sx�syb

2

sx � 14500 psi sy � 1070 psi txy � 1900 psi Therefore,


saver �sx + sy

2 saver � 7785 psi

and t2 ��6979 psi;us2 � up1 + 45° us2 � 52.9°

and t1 � 6979 psi;us1 � up1 � 45° us1 � �37.1°

tmax � 6979 psi


2b2

+ txy2

;For up2 � 97.9° s2 � 806 psi

;For up1 � 7.90° s1 � 14764 psi

Problems 7.3-14 sx � 16.5 MPa, sv � �91 MPa, txy � �39 MPa


oriented element.

07Ch07.qxd 9/27/08 1:18 PM Page 589


Solution 7.3-14


Therefore,

For

For s2 � �103.7 MPaup2 � 72.0°

up1 � �17.98° s1 � 29.2 MPa

s2 �sx + sy

2+

sx � sy


s1 �sx + sy

2+

sx � sy

2 cos 12up1

2 + txy sin 12up12

up2 � 90° + up1 up2 �72.02°

up1 � �17.98°

tan(2up) �2txy

sx � sy up1 �

atana 2txy

sx�syb

2

sx � 16.5 MPa sy � �91 MPa txy � �39 MPa (b) MAXIMUM SHEAR STRESSES

saver �sx + sy

2 saver � �37.3 MPa

and t2 � �66.4 MPa;us2 � up1 + 45° us2 � 27.0°

and t1 � 66.4 MPa;us1 � up1 � 45° us1 � �63.0°

tmax � 9631.7 psi


2b2

+ txy2

Problems 7.3-15 sx � �3300 psi, sy � �11,000 psi, txy � 4500 psi


oriented element.

Solution 7.3-15


s2 �sx + sy

2+

sx � sy


s1 �sx + sy

2+

sx � sy


up2 � 90° + up1 up2 � 114.73°

up1 � 24.73°

tan(2up) �2txy

sx � sy up1 �

atana 2txy

sx�syb

2

s � �3300 psi sy ��11000 psi txy � 4500 psi Therefore,For

For


saver �sx + sy

2 saver � �7150 psi

and t2 � �5922 psi;us2 � up1 + 45° us2 � 69.7°

and t1 � 5922 psi;us1 � up1 � 45° us1 � �20.3°

tmax � 5922 psi


2b2

+ txy2

up2 � 114.7° s2 � �13072 psi

up1 � 24.7° s1 � �1228 psi

07Ch07.qxd 9/27/08 1:18 PM Page 590


Problems 7.3-16 sx � �108 MPa, sy � 58 MPa, txy � �58 MPa


oriented element.

Solution 7.3-16


Therefore,

For

For ;s2 � �126.3 MPaup2 � 17.47°

;up1 � 107.47° s1 � 76.3 MPa

s2 �sx + sy

2+

sx � sy


s1 �sx + sy

2+

sx � sy

2 cos 12up1

2 + txy sin 12up12

up1 � 90° + up2 up1 � 107.47°

up2 � 17.47°

tan(2up) �2txy

sx � sy up2 �

atana 2txy

sx�syb

2

sx � �108 MPa sy � 58 MPa txy � �58 MPa (b) MAXIMUM SHEAR STRESSES

saver �sx + sy


and t2 ��101.3 MPa;us2 � up1 + 45° us2 � 152.47°

and t1 � 101.3 MPa;us1 � up1 � 45° us1 � 62.47°

tmax � 14686.1 psi


2b2

+ txy2

Problem 7.3-17 At a point on the surface of a machine component, the stresses acting on the x face of a stress element are sx � 5900 psi and txy � 1950 psi(see figure).

What is the allowable range of values for the stress sy if the maximum shear stressis limited to t0 � 2500 psi?

y

xO

txy = 1950 psi

sy

sx = 5900 psi

07Ch07.qxd 9/27/08 1:18 PM Page 591


Problem 7.3-18 At a point on the surface of a machine component the stresses acting on the x face of a stress element are sx � 42 MPa and txy � 33 MPa (see figure).

What is the allowable range of values for the stress sy if the maximum shearstress is limited to t0 � 35 MPa?

Solution 7.3-17

Find the allowable range of values for if the

maximum allowable shear stresses is

(1)

Solve for

sy � a9029

2771b psisy � sx� J

22tmax2�txy

2

� a22tmax2 �txy

2b Ksy


2b2

+ txy2

tmax � 2500 psi

sy

sx � 5900 psi sy unknown txy � 1950 psi Therefore,

From Eq. (1):

tmax(sy1) � A asx � sy1

2b2

+ txy2

2771 psi … sy … 9029 psi

2.5 ksi

tmax(σy1)

σy1

2.771 ksi 9.029 ksi

y

xO

txy = 33 MPa

sy

sx = 42 MPa

Solution 7.3-18

Find the allowable range of values for if themaximum allowable shear stresses is tmax � 35 MPa

sy

sx � 42 MPa sy unknown txy � 33 MPa(1)tmax � A a

sx � sy

2b2

+ txy2

07Ch07.qxd 9/27/08 1:18 PM Page 592


35 MPa

18.7 MPa 65.3 MPa

tmax (σy1)

σy1

Problem 7.3-19 An element in plane stress is subjected to stresses sx � 5700 psi and txy � �2300 psi (see figure). It is known that one of the principal stresses equals6700 psi in tension.

(a) Determine the stress sy.(b) Determine the other principal stress and the orientation of the principal planes,

then show the principal stresses on a sketch of a properly oriented element.

y

xO5700 psi

sy

2300 psi

Solve for

Therefore, 18.7 MPa … sy … 65.3 MPa

sy � a65.3

18.7b MPasy � sx� J

22tmax2�txy

2

� a22tmax2 �txy

2b Ksy From Eq. (1):

tmax(sy1) � A asx � sy1

2b2

+ txy2

Solution 7.3-19

(a) STRESS

Because is smaller than a given principal stress,we know that the given stress is the larger principalstress.

s1 �sx + sy

2+ A a

sx � sy

2b2

+ txy2

s1 � 6700 psi

sy

sy

sx � 5700 psi sy unknown txy � �2300 psi Solve for

(b) PRINCIPAL STRESSES

up2 � 90° + up1 up2 � 66.50°

up1 � �23.50°

tan (2up) �2txy

sx � sy up1 �

atana 2txy

sx �syb

2

sy sy � 1410 psi ;

07Ch07.qxd 9/27/08 1:18 PM Page 593


Problem 7.3-20 An element in plane stress is subjected to stresses sx � �50 MPa and txy � 42 MPa (see figure). It is known that one of the principal stresses equals33 MPa in tension.

(a) Determine the stress sy.(b) Determine the other principal stress and the orientation of the principal planes,

then show the principal stresses on a sketch of a properly oriented element.

y

xO

42 MPa

sy

50 MPa

Solution 7.3-20

(a) STRESS

Because is smaller than a given principal stress,we know that the given stress is the larger principalstress.

Solve for

(b) PRINCIPAL STRESSES

tan(2up) �2txy

sx � sy up2 �

atana 2txy

sx �syb

2

sy sy � 11.7 MPa ;

s1 �sx + sy

2+ A a

sx � sy

2b2

+ txy2

s1 � 33 MPa

sy

sy

sx � �50 MPa sy unknown txy � 42 MPa

Therefore,

For

For up2 � �26.8° : s2 � �71.3 MPa ;up1 � 63.2° : s1 � 33.0 MPa ;

+ txy sin 12up22s2 �

sx + sy

2+

sx � sy

2 cos 12up22

+ txy sin 12up12

s1 �sx + sy

2+

sx � sy

2 cos 12up1

2up1 � 90° + up2 up1 � 63.15°

up2 � �26.85°

+ txy sin 12up22s2 �

sx + sy

2+

sx � sy

2 cos 12up22

+ txy sin12up12

s1 �sx + sy

2+

sx � sy

2 cos 12up1

2 Therefore,

For

For up2 � 66.5° : s2 � 410 psi ;up1 � �23.5° : s1 � 6700 psi ;

07Ch07.qxd 9/27/08 1:18 PM Page 594

SECTION 7.4 Mohr’s Circle 595

Mohr’s CircleThe problems for Section 7.4 are to be solved using Mohr’s circle. Consider only thein-plane stresses (the stresses in the xy plane).

Problem 7.4-1 An element in uniaxial stress is subjected to tensile stresses sx � 11,375 psi, as shown in the figure. Using Mohr’s circle, determine:

(a) The stresses acting on an element oriented at a counterclockwise angle u � 24°from the x axis.

(b) The maximum shear stresses and associated normal stresses.Show all results on sketches of properly oriented elements.

y

xO11,375 psi

Solution 7.4-1

(a) ELEMENT AT

PointDœ: sy1 � R � R cos (2u)

tx1y1 � �4227 psi ;tx1y1 � �R sin (2u)

sx1 � 9493 psi ;Point D: sx1 � R + R cos(2u)

Point C: sc � R sc � 5688 psi

2u � 48° R �sx

2 R � 5688 psi

; u � 24°



saver � R saver � 5688 psi ;tmax � �R tmax � �5688 psi ;

Point S2: us2 �90°

2 us2 � 45° ;

tmax � R tmax � 5688 psi ;

Point S1: us1 ��90°

2 us1 ��45° ;

sy1 � 1882 psi ;

Problem 7.4-2 An element in uniaxial stress is subjected to tensile stresses sx � 49 MPa, as shown in the figure Using Mohr’s circle, determine:

(a) The stresses acting on an element oriented at an angle u � �27° from the xaxis (minus means clockwise).


y

xO49 MPa

Solution 7.4-2

(a) ELEMENT AT

Point C: sc � R sc � 24.5 MPa

2u � �54.0° R �sx

2 R � 24.5 MPa

u � �27°

sx � 49 MPa sy � 0 MPa txy � 0 MPa Point D:

Point

sy1 � 10.1 MPa ;D

œ

sy1 � R � R cos ( |2u| )

tx1y1 � 19.8 MPa ;tx1y1 � �R sin (2u)

sx1 � 38.9 MPa ; sx1 � R + R cos (|2u|)

07Ch07.qxd 9/27/08 1:19 PM Page 595



Point S1:

tmax � R tmax � 24.5 MPa ; ;us1 � �45.0°

us1 ��90°

2

Point S2:

saver � R saver � 24.5 MPa ;tmax � �R tmax � �24.5 MPa ;

us2 �90°

2 us2 � 45.0° ;

Problem 7.4-3 An element in uniaxial stress is subjected to compressive stresses of magnitude 6100 psi, as shown in the figure. Using Mohr’s circle, determine:

(a) The stresses acting on an element oriented at a slope of 1 on 2 (see figure).(b) The maximum shear stresses and associated normal stresses.Show all results on sketches of properly oriented elements.

y

xO

6100 psi

12

Solution 7.4-3

(a) ELEMENT AT A SLOPE OF 1 ON 2

Point C:

Point D:

tx1y1 � �R sin (2u) tx1y1 � 2440 psi ;sx1 � �4880 psi ;

sx1 � R + R cos (2u)

sc � R sc � �3050 psi

2u � 53.130° R �sx

2 R � �3050 psi

;u � atana1

2b u � 26.565°

sx � �6100 psi sy � 0 psi txy � 0 psi Point


Point S1:

Point S2:

saver � R saver � �3050 psi ;tmax � R tmax � �3050 psi ;

us2 � �45° ; us2 ��90°

2

tmax � �R tmax � 3050 psi ;

us1 �90°

2 us1 � 45° ;

sy1 � �1220 psi

Dœ

: sy1 � R � R cos (2u)

Problem 7.4-4 An element in biaxial stress is subjected to stresses sx � �48 MPa and sy � 19 MPa, as shown in the figure. Using Mohr’s circle, determine:

(a) The stresses acting on an element oriented at a counterclockwise angle u � 25°from the x axis.

(b) The maximum shear stresses and associated normal stresses.

Show all results on sketches of properly oriented elements.

y

xO

19 MPa

48 MPa

07Ch07.qxd 9/27/08 1:19 PM Page 596


Solution 7.4-4

(a) ELEMENT AT

sy1 � 7.0 MPa

Point Dœ: sy1 � sc + R cos(2u)

tx1y1 � 25.7 MPa ;tx1y1 � �R sin(2u)

sx1 � �36.0 MPa ;Point D: sx1 � sc � R cos(2u)

sc � �14.5 MPaPoint C: sx � sx + R

2u � 50.0 deg R �|sx| + |sy|

2 R � 33.5 MPa

; u � 25°

sx � �48 MPa sy � 19 MPa txy � 0 MPa (b) MAXIMUM SHEAR STRESSES

saver � sc saver � �14.5 MPa ;tmax � �R tmax � �33.5 MPa ;

us2 � �45.0° ;


2

tmax � R tmax � 33.5 MPa ;us1 � 45.0° ;


2

Problem 7.4-5 An element in biaxial stress is subjected to stresses sx � 6250 psi and sy � �1750 psi, as shown in the figure. Using Mohr’s circle, determine:

(a) The stresses acting on an element oriented at a counterclockwise angle u � 55° from the x axis.


y

xO

1750 psi

6250 psi

Solution 7.4-5

(a) ELEMENT AT

sy1 � 4250 psiPoint Dœ: sy1 � sc� R cos(2u)

tx1y1 � �3464 psi ;tx1y1 � �R sin (2u)

sx1 � 250 psi ;Point D: sx1 � sc + R cos(2u)

Point C: sc � sx � R sc � 2250 psi

2u � 120° R �|sx| + |sy|

2 R � 4000 psi

u � 60°

sx � 6250 psi sy � �1750 psi txy � 0 psi (b) MAXIMUM SHEAR STRESSES

saver � sc saver � 2250 psi ;tmax � R tmax � 4000 psi ;


2 us2 � 45° ;

tmax � R tmax � 4000 psi ;us1 � �45° ;


2

07Ch07.qxd 9/27/08 1:19 PM Page 597


Problem 7.4-6 An element in biaxial stress is subjected to stresses sx � �29 MPa and sy � 57 MPa, as shown in the figure. Using Mohr’s circle, determine:

(a) The stresses acting on an element oriented at a slope of 1 on 2.5 (see figure).(b) The maximum shear stresses and associated normal stresses.Show all results on sketches of properly oriented elements.

y

xO

57 MPa

29 MPa

1

2.5

Solution 7.4-6

(a) ELEMENT AT A SLOPE OF 1 ON 2.5

sy1 � 45.1 MPa

Point Dœ: sy1 � sc + R cos (2u)

;tx1y1 � R sin (2u) tx1y1 � 29.7 MPa

sx1 � �17.1 MPa ;Point D: sx1 � sc � R cos (2u)

Point C: sc � sx + R sc � 14.0 MPa

2u � 43.603° R �|sx| + |sy|

2 R � 43.0 MPa

;u � atana 1

2.5b u � 21.801°

sx � �29 MPa sy � 57 MPa txy � 0 MPa (b) MAXIMUM SHEAR STRESSES

saver � sc saver � 14.0 MPa ;tmax � �R tmax � �43.0 MPa ;

us2 � �45.0° ;


2

tmax � R tmax � 43.0 MPa ;


2 us1 � 45.0° ;

Problem 7.4-7 An element in pure shear is subjected to stresses txy � 2700 psi, as shown in the figure. Using Mohr’s circle, determine:

(a) The stresses acting on an element oriented at a counterclockwise angle u � 52° fromthe x axis.

(b) The principal stresses.Show all results on sketches of properly oriented elements.

y

xO

2700 psi

07Ch07.qxd 9/27/08 1:19 PM Page 598


Solution 7.4-7

(a) ELEMENT AT

sy1 � �2620 psi ;

Point Dœ: sy1 � �R cos (2u � 90°)

tx1y1 � �653 psi ;tx1y1 � �R sin (2u � 90°)

sx1 � 2620 psi ;Point D: sx1 � R cos (2u � 90°)

2u � 104.0° R � txy R � 2700 psi

u � 52°

sx � 0 psi sy � 0 psi txy � 2700 psi (b) PRINCIPAL STRESSES

s2 � �R s2 � �2700 psi ;

;up2 � �45°

Point P2: up2 ��90°

2

s1 � R s1 � 2700 psi ;

Point P1: up1 �90°

2 up1 � 45° ;

Problem 7.4-8 An element in pure shear is subjected to stresses txy � �14.5 MPa, asshown in the figure. Using Mohr’s circle, determine:

(a) The stresses acting on an element oriented at a counterclockwise angle u � 22.5°from the x axis

(b) The principal stresses.Show all results on sketches of properly oriented elements.

y

xO

14.5 MPa

Solution 7.4-8

(a) ELEMENT AT

sy1 � 10.25 MPa ;

Point Dœ: sy1 � R cos (2u � 90°)

tx1y1 � �10.25 MPa ;tx1y1 � R sin (2u � 90°)

sx1 � �10.25 MPa ;Point D: sx1 � �R cos (2u � 90°)

R � |txy| R � 14.50 MPa

2u � 45.00°

u � 22.5°

sx � 0 MPa sy � 0 MPa txy � �14.5 MPa (b) PRINCIPAL STRESSES

s2 � �14.50 MPa ;

s2 � �R

up2 � �135.0° ;


2

s1 � R s1 � 14.50 MPa ;

;up1 � 135.0°Point P1: up1 �270°

2

07Ch07.qxd 9/27/08 1:19 PM Page 599

Problem 7.4-10 sx � 27 MPa, sy � 14 MPa, txy � 6 MPa, u � 40°Using Mohr’s circle, determine the stresses acting on an element oriented at an

angle u from the x axis. Show these stresses on a sketch of an element oriented at theangle u. (Note: The angle u is positive when counterclockwise and negative whenclockwise.)

y

xO

txy

sx

sy


Problem 7.4-9 An element in pure shear is subjected to stresses txy � 3750 psi, as shown in the figure. Using Mohr’s circle, determine:

(a) The stresses acting on an element oriented at a slope of 3 on 4 (see figure).(b) The principal stresses.

Show all results on sketches of properly oriented elements.O

3

4

y

x

3750 psi

Solution 7.4-9

(a) ELEMENT AT A SLOPE OF 3 ON 4

sy1 � �3600 psi ;Point Dœ: sy1 � �R cos (2u � 90°)

tx1y1 � 1050 psi ;tx1y1 � �R sin (2u � 90°)

sx1 � 3600 psi ;Point D: sx1 � R cos (2u � 90°)

2u � 73.740° R � txy R � 3750 psi

u � atana3

4b u � 36.870°

sx � 0 psi sy � 0 psi txy � 3750 psi (b) PRINCIPAL STRESSES

s2 � �R s2 � �3750 psi ;up2 � �45° ;


2

s1 � R s1 � 3750 psi ;

Poin P1: up1 �90°

2 up1 � 45° ;


07Ch07.qxd 9/27/08 1:19 PM Page 600


Solution 7.4-10

a � atana txy

sx � saverb a � 42.71°

R � 2(sx � saver)2

+ txy

2 R � 8.8459 MPa

saver �sx + sy

2 saver � 20.50 MPa

u � 40°

sx � 27 MPa sy � 14 MPa txy � 6 MPa

sy1 � 13.46 MPa ;

Point Dœ: sy1 � saver � R cos (b)

tx1y1 � �5.36 MPa ;tx1y1 � �R sin (b)

;sx1 � 27.5 MPa

Point D: sx1 � saver + R cos (b)

b � 2u � a b � 37.29°

Problem 7.4-11 sx � 3500 psi, sy � 12,200 psi, txy � �3300 psi, u � �51°Using Mohr’s circle, determine the stresses acting on an element oriented at an angle u from the x axis. Show these stresses on a

sketch of an element oriented at the angle u. (Note: The angle u is positive when counterclockwise and negative when clockwise.)

Solution 7.4-11

a � atana txy

sx � saverb a � 37.18°


+ txy

2 R � 5460 psi

saver �sx + sy


;u � �51°

sx � 3500 psi sy � 12200 psi txy � �3300 psi

;sy1 � 3718 psi

Point Dœ: sy1 � saver � R cos (b)

tx1y1 � �3569 psi ;tx1y1 � �R sin (b)

sx1 � 11982 psi ;


b � 180° + 2u � a b � 40.82°

Problem 7.4-12 sx � �47 MPa, sy � �186 MPa, txy � �29 MPa, u � �33°Using Mohr’s circle, determine the stresses acting on an element oriented at an angle u from the x axis. Show these stresses on a

sketch of an element oriented at the angle u. (Note: The angle u is positive when counterclockwise and negative when clockwise.)

Solution 7.4-12

b � �2u � a b � 43.35°

a � atana ` txy

sx � saver` b a � 22.65°


+ txy

2 R � 75.3077 MPa

saver �sx + sy

2 saver � �116.50 MPa

u � �33°

sx ��47MPa sy ��186MPa txy ��29MPa

sy1 � �171.3 MPa ;Point Dœ: sy1 � saver � R cos (b)

tx1y1 ��51.7 MPa ;tx1y1 ��R sin (b)

sx1 � �61.7 MPa ;


07Ch07.qxd 9/27/08 1:19 PM Page 601


Problem 7.4-13 sx � �1720 psi, sy � �680 psi, txy � 320 psi, u � 14°Using Mohr’s circle, determine the stresses acting on an element oriented at an angle u from the x axis. Show these

stresses on a sketch of an element oriented at the angle u. (Note: The angle u is positive when counterclockwise and negativewhen clockwise.)

Solution 7.4-13

b � 2u + a b � 64.16°

a � atana txy

|sx � saver|b a � 36.16°


+ txy

2 R � 644.0 psi

saver �sx + sy


u � 14°


sy1 � �919 psi ;Point Dœ: sy1 � saver + R cos (b)

tx1y1 � R sin (b) tx1y1 � 580 psi ;sx1 � �1481 psi ;

Point D: sx1 � saver � R cos(b)

Problem 7.4-14 sx � 33 MPa, sy � �9 MPa, txy � 29 MPa, u � 35°Using Mohr’s circle, determine the stresses acting on an element oriented at an angle u from the x axis. Show these


Solution 7.4-14

b � 2u � a b � 15.91°

a � atana ` txy

sx � saver` b a � 54.09°


+ txy2 R � 35.8050 MPa

saver �sx + sy

2 saver � 12.00 MPa

u � 35°

sx � 33 MPa sy � �9 MPa txy � 29 MPa

sy1 � �22.4 MPa ;Point Dœ : sy1 � saver � R cos (b)

tx1y1 � �9.81 MPa ;tx1y1 � �R sin (b)

sx1 � 46.4 MPa ;


Problem 7.4-15 sx � �5700 psi, sy � 950 psi, txy � �2100 psi, u � 65°Using Mohr’s circle, determine the stresses acting on an element oriented at an angle u from the x axis. Show these


07Ch07.qxd 9/27/08 1:19 PM Page 602


Solution 7.4-15

b � 180° � 2u + a b � 82.28°

a � atana |txy|

|sx � saver|b a � 32.28°


+ txy

2 R � 3933 psi

saver �sx + sy


u � 65°

sx � �5700 psi sy � 950 psi txy � �2100 psi

sy1 � �2904 psi ;Point Dœ: sy1 � saver � R cos (b)

tx1y1 � R sin (b) tx1y1 � 3897 psi ;sx1 � �1846 psi ;


Problems 7.4-16 sx � �29.5 MPa, sy � 29.5 MPa, txy � 27 MPaUsing Mohr’s circle, determine (a) the principal stresses and (b) the maximum

shear stresses and associated normal stresses. Show all results on sketches of properlyoriented elements.

y

xO

txy

sx

sy

Solution 7.4-16


up2 � up1 � 90° up2 � �21.2° ;

up1 �180° � a

2 up1 � 68.8° ;

a � atana ` txy

sx � saver` b a � 42.47°


+ txy2 R � 39.9906 MPa

saver �sx + sy

2 saver � 0 MPa

sx � �29.5 MPa sy � 29.5 MPa txy � 27 MPa


tmax � R tmax � 40.0 MPa ;

Point S1: saver � 0 MPa ;

us2 � 90° + us1 us2 � 113.8° ;

us1 �90°�a

2 us1 � 23.8° ;

Point P2: s2 � �R s2 � �40.0 MPa ;

Point P1: s1 � R s1 � 40.0 MPa ;


07Ch07.qxd 9/27/08 1:19 PM Page 603


Problems 7.4-17 sx � 7300 psi, sy � 0 psi, txy � 1300 psiUsing Mohr’s circle, determine (a) the principal stresses and (b) the maximum shear stresses and associated normal

stresses. Show all results on sketches of properly oriented elements.

Solution 7.4-17


up2 �a + 180°

2 up2 � 99.8°

;up1 �a

2 up1 � 9.80°

a � atana ` txy

sx � saver` b a � 19.60°


+ txy2 R � 3875 psi

saver �sx + sy





Point S1: saver � 3650 psi ;

us2 � 90° + us1 us2 � 54.8°

us1 ��90° + a

2 us1 � �35.2°

s2 � �225 psi

Point P2: s2 � �R + saver

s1 � 7525 psi ;

Point P1: s1 � R + saver

Problems 7.4-18 sx � 0 MPa, sy � �23.4 MPa, txy � �9.6 MPaUsing Mohr’s circle, determine (a) the principal stresses and (b) the maximum shear stresses and associated normal


Solution 7.4-18


up2 � up1 + 90° up2 � 70.32° ;

up1 ��a

2 up1 � �19.68° ;

a � atana ` txy

sx � saver` b a � 39.37°


+ txy2 R � 15.1344 MPa

saver �sx + sy


sx � 0 MPa sy � �23.4 MPa txy � �9.6 MPa


tmax � R tmax � 15.13 MPa ;Point S1: saver � �11.70 MPa

us2 � 90 ° + us1 us2 � 25.3°

us1 ��90° � a

2 us1 � �64.7° ;

s2 � �26.8 MPa ;Point P2: s2 � �R + saver

s1 � 3.43 MPa ;Point P1: s1 � R + saver

07Ch07.qxd 9/27/08 1:19 PM Page 604


Problems 7.4-19 sx � 2050 psi, sy � 6100 psi, txy � 2750 psiUsing Mohr’s circle, determine (a) the principal stresses and (b) the maximum shear stresses and associated normal


Solution 7.4-19


;up2 ��a

2 up2 � �26.8°

;up1 �180° � a

2 up1 � 63.2°

a � atana ` txy

sx � saver` b a � 53.63°


+ txy2 R � 3415 psi

saver �sx + sy





Point S1: saver � 4075 psi ;

us2 � 90° + us1 us2 � 71.8°

us1 ��90° + a

2 us1 � �18.2° ;

s2 � 660 psi ;


s1 � 7490 psi ;


Problems 7.4-20 sx � 2900 kPa, sy � 9100 kPa, txy � �3750 kPaUsing Mohr’s circle, determine (a) the principal stresses and (b) the maximum shear stresses and associated normal


Solution 7.4-20


up2 �a

2 up2 � 25.2° ;

;up1 �a + 180°

2 up1 � 115.2°

a � atana ` txy

sx � saver` b a � 50.42°


+ txy2 R � 4865.4393 kPa

saver �sx + sy

2 saver � 6000 kPa

sx � 2900 kPa sy � 9100 kPa txy � �3750 kPa


tmax � R tmax � 4865 kPa ;Point S1: saver � 6000 kPa ;

us2 � 90° + us1 us2 � 160.2° ;

us1 �90° + a

2 us1 � 70.2° ;

s2 � 1135 kPa ;Point P2: s2 � �R + saver

s1 � 10865 KPa ;


07Ch07.qxd 9/27/08 1:19 PM Page 605


Problems 7.4-21 sx � �11,500 psi, sy � �18,250 psi, txy � �7200 psiUsing Mohr’s circle, determine (a) the principal stresses and (b) the maximum shear stresses and associated normal


Solution 7.4-21


up2 �180° � a

2 up2 � 57.6° ;

up1 ��a

2 up1 � �32.4° ;

a � atana ` txy

sx � saver` b a � 64.89°


+ txy2 R � 7952 psi

saver �sx + sy


txy � �7200 psi

sx � �11500 psi sy � �18250 psi


tmax � R tmax � 7952 psi ;Point S1: saver � �14875 psi ;

us2 � 90° + us1 us2 � 192.6°

;us1 �270° � a

2 us1 � 102.6°

s2 � �22827 psi ;Point P2: s2 � �R + saver

s1 � �6923 psi ;


Problems 7.4-22 sx � �3.3 MPa, sy � 8.9 MPa, txy � �14.1 MPaUsing Mohr’s circle, determine (a) the principal stresses and (b) the maximum shear stresses and associated normal


Solution 7.4-22

a � atana ` txy

sx � saver` b a � 66.6°


+ txy2 R � 15.4 MPa

saver �sx + sy

2 saver � 2.8 MPa

txy ��14.1 MPa

sx ��3.3 MPa sy � 8.9 MPa (a) PRINCIPAL STRESSES


s1 � 18.2 MPa ;


up2 �a

2 up2 � 33.3°

;up1 �a + 180°

2 up1 � 123.3°

07Ch07.qxd 9/27/08 1:19 PM Page 606


Problems 7.4-23 sx � 800 psi, sy � �2200 psi, txy � 2900 psiUsing Mohr’s circle, determine (a) the principal stresses and (b) the maximum shear stresses and associated normal


Solution 7.4-23


;up2 �180° + a

2 up2 � 121.3°

;up1 �a

2 up1 � 31.3°

a � atana ` txy

sx � saver` b a � 62.65°


+ txy2 R � 3265 psi

saver �sx + sy


sx � 800 psi sy � �2200 psi txy � 2900 psi


tmax � R tmax � 3265 psi

Point S1: saver � �700 psi ;

us2 � 90° + us1 us2 � 76.3° ;

us1 ��90° + a

2 us1 � �13.7° ;

s2 � �3965 psi ;


s1 � 2565 psi ;



us1 �90° + a

2 us1 � 78.3°

s2 � �12.6 MPa ;

tmax � R tmax � 15.4 MPa ;Point S1: saver � 2.8 MPa ;

us2 � 90° + us1 us2 � 168.3°

07Ch07.qxd 9/27/08 1:19 PM Page 607


Hooke’s Law for Plane StressWhen solving the problems for Section 7.5, assume that the material is linearly elastic with modulus of elasticity E and Poisson’s ratio n.

Problem 7.5-1 A rectangular steel plate with thickness t � 0.25 in. is subjected to uniform normal stresses �x and �y, as shown in the figure. Strain gages A and B, oriented in the x and y directions, respectively, are attached to the plate. The gage readings give normal strains ex � 0.0010 (elongation) and ey � �0.0007 (shortening).

Knowing that E � 30 � 106 psi and � � 0.3, determine the stresses �x and �y and the change �t in the thickness of the plate.

sy

sx

y

xOB A

Solution 7.5-1 Rectangular plate in biaxial stress

SUBSTITUTE NUMERICAL VALUES:

Eq. (7-40a):

Eq. (7-40b):

sy �E

(1 � �)2(ây + �âx) � �13,190 psi ;

sx �E

(1 � �)2(âx + �ây) � 26,040 psi ;

E � 30 * 106 psi � � 0.3

t � 0.25 in. âx � 0.0010 ây � �0.0007 Eq. (7-39c):

(Decrease in thickness)

¢t � âzt � �32.1 * 10�6 in. ;

âz � ��

E (sx + sy) � �128.5 * 10�6

Problem 7.5-2 Solve the preceding problem if the thickness of the steel plate is t � 10 mm, the gage readings are ex � 480 � 10�6 (elongation) and ey � 130 � 10�6 (elongation), the modulus is E � 200 GPa, and Poisson’s ratio is � � 0.30.

Solution 7.5-2 Rectangular plate in biaxial stress

SUBSTITUTE NUMERICAL VALUES:

Eq. (7-40a):

sx �E

(1 � �)2 (âx + �ây) � 114.1 MPa ;

E � 200 GPa � � 0.3

ây � 130 * 10�6

t � 10 mm âx � 480 * 10�6 Eq. (7-40b):

Eq. (7-39c):


¢t � âz t � �2610 * 10�6 mm ;

âz � ��

E (sx + sy) � �261.4 * 10�6

sy �E

(1 � �)2 (ây + �âx) � 60.2 MPa ;

Probs. 7.5-1 and 7.5-2

07Ch07.qxd 9/27/08 1:20 PM Page 608

Problem 7.5-3 Assume that the normal strains and for an element in plane stress (see figure) are measured with strain gages.

(a) Obtain a formula for the normal strain in the direction in terms of , and Poisson’s ratio .

(b) Obtain a formula for the dilatation in terms of , and Poisson’s ratio .�

Px, Pye�Px, Py

zPz

PyPx y

x

z

Osx

txy

sy

Solution 7.5-3 Plane stressGiven:

(a) NORMAL STRAIN

Eq. (7-34c):

Eq. (7-36a):

Eq. (7-36b):

Substitute and into the first equation and simplify:

âz � ��

1 � � (âx + ây) ;

sysx

sy �E

(1 � �2) (ây + �âx)

sx �E

(1 � �2) (âx + �ây)

âz � ��

E (sx + sy)

âz

âx, ây, � (b) DILATATION

Eq. (7-47):

Substitute and from above and simplify:

e �1 � 2�

1 � � (âx + ây) ;

sysx

e �1 � 2�

E (sx + sy)

Problem 7.5-4 A magnesium plate in biaxial stress is subjected to tensile stresses sx� 24 MPa and sy� 12 MPa (see figure). The corresponding strains in the plate are �x � 440 � 10�6 and �y � 80 � 10�6.

Determine Poisson’s ratio and the modulus of elasticity Efor the material.

�

sy

sx

y

xO


POISSON’S RATION AND MODULUS OF ELASTICITY

Eq. (7-39a):

Eq. (7-39b): ây �1

E (sy � �sx)

âx �1

E (sx � �sy)

âx � 440 * 10�6 ây � 80 * 10�6

sx � 24 MPa sy � 12 MPa Substitute numerical values:

Solve simultaneously:

� � 0.35 E � 45 GPa ;

E (80 * 10�6) � 12 MPa � � (24 MPa)

E (440 * 10�6) � 24 MPa � � (12 MPa)

SECTION 7.5 Hooke’s Law for Plane Stress 609


07Ch07.qxd 9/27/08 1:20 PM Page 609


Problem 7.5-5 Solve the preceding problem for a steel plate with sx � 10,800 psi (tension), sy� �5400 psi (compres-sion), ex � 420 � 10�6 (elongation), and ey � �300 � 10�6 (shortening).


POISSON’S RATIO AND MODULUS OF ELASTICITY

Eq. (7-39a):

Eq. (7-39b): ây �1

E (sy � �sx)

âx �1

E (sx � �sy)

âx � 420 * 10�6 ây � �300 * 10�6

sx � 10,800 psi sy � �5400 psi Substitute numerical values:

Solve simultaneously:

� � 1/3 E � 30 * 106 psi ;

E (�300 * 10�6) � �5400 psi � � (10,800 psi)

E (420 * 10�6) � 10,800 psi � � (�5400 psi)

Problem 7.5-6 A rectangular plate in biaxial stress (see figure) is subjected to normal stresses �x� 90 MPa (tension) and�y� �20 MPa (compression). The plate has dimensions and is made of steel with and

.

(a) Determine the maximum in-plane shear strain in the plate.(b) Determine the change in the thickness of the plate.(c) Determine the change in the volume of the plate.¢V

¢tgmax

� � 0.30E � 200 GPa400 * 800 * 20 mm


Dimensions of Plate: Shear Modulus (Eq. 7-38):

(a) MAXIMUM IN-PLANE SHEAR STRAIN

Principal stresses:

Eq. (7-26):

Eq. (7-35): gmax �tmax

G� 715 * 10�6 ;

tmax �s1 � s2

2� 55.0 MPa

s1 � 90 MPa s2 � �20 MPa

G �E

2(1 + �)� 76.923 GPa

400 mm * 800 mm * 20 mm

E � 200 GPa � � 0.30

sx � 90 MPa sy � �20 MPa (b) CHANGE IN THICKNESS

Eq. (7-39c):


(c) CHANGE IN VOLUME

From Eq. (7-47):

Also,

(Increase in volume)

� 896 mm3 ;‹ ¢V � (6.4 * 106 mm3)(140 * 10�6)

a1 � 2�

Eb(sx + sy) � 140 * 10�6

V0 � (400)(800)(20) � 6.4 * 106 mm3

¢V � V0a1 � 2�

Eb(sx + sy)

¢t � âz t � �2100 * 10�6 mm ;

âz � ��

E (sx + sy) � �105 * 10�6

07Ch07.qxd 9/27/08 1:20 PM Page 610


Problem 7.5-7 Solve the preceding problem for an aluminum plate with sx� 12,000 psi (tension), sy� �3,000 psi (compression), dimensions 20 � 30 � 0.5 in., E � 10.5 � 106 psi, and .� � 0.33


Dimensions of Plate: .Shear Modulus (Eq. 7-38):

(a) MAXIMUM IN-PLANE SHEAR STRAIN

Principal stresses:

Eq. (7-26):

Eq. (7-35): gmax �tmax

G� 1,900 * 10�6 ;

tmax �s1 � s2

2� 7,500 psi

s2 � �3,000 psi

s1 � 12,000 psi

G �E

2(1 + �)� 3.9474 * 106 psi

20 in. * 30 in. * 0.5 in

E � 10.5 * 106 psi � � 0.33

sx � 12,000 psi sy � �3,000 psi (b) CHANGE IN THICKNESS

Eq. (7-39c):



From Eq. (7-47):

Also,


� 0.0874 in.3 ;‹ ¢V � (300 in.3)(291.4 * 10�6)

a1 � 2�

Eb(sx + sy) � 291.4 * 10�6

V0 � (20)(30)(0.5) � 300 in.3

¢V � V0 a1 � 2�

Eb(sx + sy)

¢t � âz t � �141 * 10�6 in. ;� �282.9 * 10�6

âz � ��

E (sx + sy)

Problem 7.5-8 A brass cube 50 mm on each edge is compressed in two perpendicular directions by forces P � 175 kN (see figure).

Calculate the change �V in the volume of the cube and the strain energy U stored in the cube, assuming E � 100 GPa and � � 0.34.

P = 175 kN

P = 175 kN

Solution 7.5-8 Biaxial stress-cube

Side

E � 100 GPa � � 0.34 ( Brass)

b � 50 mm P � 175 kN

CHANGE IN VOLUME

Eq. (7-47):

(Decrease in volume)

;¢V � eV0 � �56 mm3

V0 � b3 � (50 mm)3 � 125 * 103mm3

e �1 � 2�

E (sx + sy) � �448 * 10�6

sx � sy � �P

b2� �

(175 kN)

(50 mm)2� �70.0 MPa

07Ch07.qxd 9/27/08 1:20 PM Page 611


STRAIN ENERGY

Eq. (7-50):

� 0.03234 MPa

u �1

2E (sx

2+ sy

2 � 2�sxsy)� 4.04 J ;

U � uV0 � (0.03234 MPa)(125 * 103 mm3)

Problem 7.5-9 A 4.0-inch cube of concrete is compressed in biaxial stress by means of a framework that is loaded as shown in the figure.

Assuming that each load equals 20 k, determine the change in the volume of the cube and the strain energy stored in the cube.U

¢VF

(E � 3.0 * 106 psi, � � 0.1)

F

F

Solution 7.5-9 Biaxial stress – concrete cube

Joint :

sx � sy � �P

b2 � �1768 psi

� 28.28 kips

P � F12

A

CHANGE IN VOLUME

Eq. (7-47):

(Decrease in volume)

STRAIN ENERGY

Eq. (7-50):

U � uV0 � 60.0 in.-lb ;� 0.9377 psi

u �1

2E (sx

2+ sy

2 � 2�sxsy)

¢V � eV0 � �0.0603 in.3 ;V0 � b3 � (4 in.)3 � 64 in.3

e �1 � 2�

E (sx + sy) � �0.0009429.

F � 20 kips

� � 0.1

E � 3.0 * 106 psi

b � 4 in

Problem 7.5-10 A square plate of width b and thickness t is loaded by normal forces and , and by shear forces V, as shown in the figure. These forces produce uniformly distributed stresses acting on the side faces of the place.

Calculate the change �V in the volume of the plate and the strain energy U stored in the plate if the dimensions are b � 600 mm and t � 40 mm, the plate is made of magnesium with E � 45 GPa and v � 0.35, and the forces are Px � 480 kN, Py � 180 kN, and V � 120 kN.

Probs. 7.5-10 and 7.5-11

PyPx

Py

Py

PxPx

y

t

b

b

V

V

V

V

xO

07Ch07.qxd 9/27/08 1:20 PM Page 612


Solution 7.5-10 Square plate in plane stress

CHANGE IN VOLUME

Eq. (7-47): e �1 � 2�

E (sx + sy) � 183.33 * 10�6

txy �V

bt� 5.0 MPaV � 120 kN

sy �Py

bt� 7.5 MPaPy � 180 kN

sx �Px

bt� 20.0 MPa Px � 480 kN

v � 0.35 (magnesium)E � 45 GPa

t � 40 mmb � 600 mm


STRAIN ENERGY

Eq. (7-50):

Substitute numerical values:

U � uV0 � 67.0 N # m � 67.0 J ;u � 4653 Pa

G �E

2(1 + �)� 16.667 GPa

u �1

2E(sx

2+ sy

2 � 2�sxsy) +

txy2

2G

¢V � eV0 � 2640 mm3 ;V0 � b2t � 14.4 * 106 mm3

Problem 7.5-11 Solve the preceding problem for an aluminum plate with , , , ,, , and .V � 15 kPy � 20 kPx � 90 k

� � 0.33E � 10,600 ksit � 1.0 in.b � 12 in.

Solution 7.5-11 Square plate in plane stress

CHANGE IN VOLUME

Eq. (7-47):


¢V � eV0 � 0.0423 in.3 ;V0 � b2t � 144 in.3

e �1 � 2�

E(sx + sy) � 294 * 10�6

V � 15 k txy �V

bt� 1250 psi

Py � 20 k sy �Py

bt� 1667 psi

Px � 90 k sx �Px

bt� 7500 psi

E � 10,600 ksi � � 0.33 ( aluminum)

b � 12.0 in. t � 1.0 in. STRAIN ENERGY

Eq. (7-50):


U � uV0 � 373 in.-lb ;u � 2.591 psi

G �E

2(1 + �)� 3985 ksi

u �1

2E (sx

2+ sy

2 � 2�sxsy) +

txy2

2G

07Ch07.qxd 9/27/08 1:20 PM Page 613


Problem 7.5-12 A circle of diameter d � 200 mm is etched on a brass plate (see figure). The plate has dimensions 400 � 400 � 20 mm. Forces are applied to the plate, producing uniformly distributed normal stresses �x � 42 MPa and �y � 14 MPa.

Calculate the following quantities: (a) the change in length �acof diameter ac; (b) the change in length �bd of diameter bd; (c) the change �t in the thickness of the plate; (d) the change �V in the volume of the plate, and (e) the strain energy U stored in the plate. (Assume E � 100 GPa and v � 0.34.)

sy

sx

sx

sy

yz

b

d

ca

x

Solution 7.5-12 Plate in biaxial stress

Dimensions:Diameter of circle:

(a) CHANGE IN LENGTH OF DIAMETER IN DIRECTION

Eq. (7-39a):

(increase)

(b) CHANGE IN LENGTH OF DIAMETER IN DIRECTION

Eq. (7-39b):

(decrease)� �560 * 10�6 mm ;

¢bd � ây d

ây �1

E(sy � �sx) � �2.80 * 10�6

y

¢ac � âx d � 0.0745 mm ;

âx �1

E(sx � �sy) � 372.4 * 10�6

x

E � 100 GPa � � 0.34 (Brass)

d � 200 mm400 * 400 * 20 (mm)

sx � 42 MPa sy � 14 MPa (c) CHANGE IN THICKNESS

Eq. (7-39c):

(decrease)

(d) CHANGE IN VOLUME

Eq. (7-47):

(increase)

(e) STRAIN ENERGY

Eq. (7-50):

U � uV0 � 25.0 N # m � 25.0 J ;� 7.801 * 10�3 MPa

u �1

2E(sx

2+ sy

2 � 2�sx sy)

¢V � eV0 � 573 mm3 ;V0 � (400)(400)(20) � 3.2 * 106 mm3

e �1 � 2�

E(sx + sy) � 179.2 * 10�6

¢t � âz t � �0.00381 mm ;� �190.4 * 10�6

âz � ��

E (sx + sy)

07Ch07.qxd 9/27/08 1:20 PM Page 614

SECTION 7.6 Triaxial Stress 615

Triaxial StressWhen solving the problems for Section 7.6, assume that the material is linearly elastic with modulus of elasticity E and Poisson’s ratio n.

Problem 7.6-1 An element of aluminum in the form of a rectangularparallelepiped (see figure) of dimensions ., , and

. is subjected to triaxial stresses ,, and acting on the , and faces,

respectively.Determine the following quantities: (a) the maximum shear stress

in the material; (b) the changes , and in the dimensions of theelement; (c) the change in the volume; and (d) the strain energy stored in the element. (Assume and .)

Probs. 7.6-1 and 7.6-2� � 0.33E � 10,400 ksi

U¢V¢c¢a, ¢b

tmax

zx, ysz � �1,000 psisy � �4,000 psisx � 12,000 psic � 3.0 in

b � 4.0 ina � 6.0 in

y

x

z

a

b

c

O

Solution 7.6-1 Triaxial stress

(a) MAXIMUM SHEAR STRESS

(b) CHANGES IN DIMENSIONS

��350.0 * 10�6

Eq. (7-53 c): âz �sz

E�

�

E (sx + sy)

��733.7 * 10�6

Eq. (7- 53 b): ây �sy

E�

�

E (sz + sx)

�1312.5 * 10�6

Eq. (7-53 a): âx �sx

E�

�

E (sy + sz)

tmax �s1 � s3

2� 8,000 psi ;

s3 � �4,000 psi

s1 � 12,000 psi s2 � �1,000 psi

E � 10,400 ksi � � 0.33 ( aluminum)

a � 6.0 in. b � 4.0 in. c � 3.0 in.

sz � �1,000 psi

sx � 12,000 psi sy � �4,000 psi


Eq. (7-56):

(d) STRAIN ENERGY

U � u (abc) � 685 in.-lb ;

� 9.517 psi

Eq. (7-57a): u �1

2 (sx âx + sy ây + sz âz)

¢V � e (abc) � 0.0165 in.3 ( increase) ;V � abc

e �1 � 2�

E(sx + sy + sz) � 228.8 * 10�6

¢c � câz � �0.0011 in. ( decrease)

¢b � bây � �0.0029 in. ( decrease)

¢a � aâx � 0.0079 in. ( increase)

M ;

07Ch07.qxd 9/27/08 1:22 PM Page 615


Problem 7.6-2 Solve the preceding problem if the element is steel ( = 200 GPA, ) with dimensions = 300 mm,= 150 mm, and = 150 mm and the stresses are , , and .sz � �40 MPasy � �40 MPasx � �60 MPacb

a� � 0.30E

Solution 7.6-2 Triaxial stress

(a) MAXIMUM SHEAR STRESS

(b) CHANGES IN DIMENSIONS

Eq. (7-53 c): âz �sz

E�

�

E (sx + sy) � �50.0 * 10�6

Eq. (7-53 b): ây �sy

E�

�

E (sz + sx) � �50.0 * 10�6

Eq. (7-53 a): âx �sx

E�

�

E (sy + sz) � �180.0 * 10�6

tmax �s1 � s3

2� 10.0 MPa ;

s3 � �60 MPa

s1 � �40 MPa s2 � �40 MPa

E � 200 GPa � � 0.30 (steel)

a � 300 mm b � 150 mm c � 150 mm

sz � �40 MPa

sx � �60 MPa sy � �40 MPa


Eq. (7-56):

(d) STRAIN ENERGY

U � u (abc) � 50.0 N # m � 50.0 J ;

� 0.00740 MPa

Eq. (7-57 a): u �1


¢V � e(abc) � �1890 mm3 (decrease) ;V � abc

e �1 � 2�

E (sx + sy + sz) � �280.0 * 10�6

¢c � câz � �0.0075 mm. (decrease)

¢b � bây � �0.0075 mm (decrease)

¢a � aâx � �0.0540 mm (decrease)

M ;

Problem 7.6-3 A cube of cast iron with sides of length = 4.0 in. (see figure) is tested in a laboratory under triaxial stress. Gages mountedon the testing machine show that the compressive strains in the materialare and

Determine the following quantities: (a) the normal stresses , andacting on the , and faces of the cube; (b) the maximum shear

stress in the material; (c) the change in the volume of thecube; and (d) the strain energy stored in the cube. (Assume = 14,000ksi and .)

Probs. 7.6-3 and 7.6-4

� � 0.25EU

¢Vtmax

zx, ysz

sx, sy

Py � Pz � �37.5 * 10�6.Px � �225 * 10�6

a y

x

z

a

a

a

O

07Ch07.qxd 9/27/08 1:22 PM Page 616


Solution 7.6-3 Triaxial stress (cube)

(a) NORMAL STRESSES

Eq. (7-54a):

In a similar manner, Eqs. (7-54 b and c) give

(b) MAXIMUM SHEAR STRESS

tmax �s1 � s3

2� 1050 psi ;

s3 � �4200 psi

s1 � �2100 psi s2 � �2100 psi

sy � �2100 psi sz � �2100 psi ;

� �4200 psi ;

sx �E

(1 + �)(1 � 2�)[(1 � �)âx + �(ây + âz)]

E � 14,000 ksi � � 0.25 (cast iron)

âz � �37.5 * 10�6 a � 4.0 in.

âx � �225 * 10�6 ây � �37.5 * 10�6 (c) CHANGE IN VOLUME

(d) STRAIN ENERGY

U � ua3 � 35.3 in.-lb ;� 0.55125 psi

Eq. (7-57a): u �1


¢V � ea3 � �0.0192 in.3 ( decrease) ;V � a3

Eq. (7-55): e � âx + ây + âz � �0.000300

Problem 7.6-4 Solve the preceding problem if the cube is granite ( ) with dimensions = 75 mm andcompressive strains and Py � Pz � �270 * 10�6.Px � �720 * 10�6

aE � 60 GPa, � � 0.25

Solution 7.6-4 Triaxial stress (cube)

(a) NORMAL STRESSES

Eq.(7-54a):

In a similar manner, Eqs. (7-54 b and c) give

(b) MAXIMUM SHEAR STESS

s3 � �64.8 MPa

s1 � �43.2 MPa s2 � �43.2 MPa

sy � �43.2 MPa sz � �43.2 MPa ;

� �64.8 MPa ;

sx �E

(1 + �)(1 � 2�)[(1 � �)âx + �(âx + âz)]

� � 0.25 (Granite)

âz � �270 * 10�6 a � 75 mm E � 60 GPa

âx � �720 * 10�6 ây � �270 * 10�6


(d) STRAIN ENERGY

U � ua3 � 14.8 N # m � 14.8 J ;� 0.03499 MPa � 34.99 kPa

Eq. (7-57 a): u �1

2(sxâx + syây + szâz)

¢V � ea3 � �532 mm3 ( decrease) ;V � a3

Eq. (7-55): e � âx + ây + âz � �1260 * 10�6

tmax �s1 � s3

2� 10.8 MPa ;

07Ch07.qxd 9/27/08 1:22 PM Page 617


Problem 7.6-5 An element of aluminum in triaxial stress (see figure) is subjected to stresses (tension), (compression), and (com-pression). It is also known that the normal strains in the x and y directionsare (elongation) and (short-ening).

What is the bulk modulus for the aluminum?

Probs. 7.6-5 and 7.6-6

K

Py � �502.3 * 10�6Px � 7138.8 * 10�6

sz � �3090 psisy � �4750 psisx � 5200 psi

Solution 7.6-5 Triaxial stress (bulk modulus)

Find .

Eq. (7-53 b): ây �sy

E�

�

E (sx + sy)

Eq. (7-53 a): âx �sx

E�

�

E (sy + sz)

K

ây � �502.3 * 10�6

sz � �3090 psi âx � 713.8 * 10�6

sx � 5200 psi sy � �4750 psi Substitute numerical values and rearrange:

(1)

(2)

Units: = psi

Solve simultaneously Eqs. (1) and (2):

Eq. (7-16): K �E

3(1 � 2�)� 10.0 * 10�6 psi ;

E � 10.801 * 106 psi � � 0.3202

E

(�502.3 * 10�6) E � �4750 � 2110 �

(713.8 * 10�6) E � 5200 + 7840 �

y

x

z

Osxsx

sz

sy

sy

sz

Problem 7.6-6 Solve the preceding problem if the material is nylon subjected to compressive stresses

, , and , and the normal strains are and

(shortenings).Py � �320 * 10�6

Px � �740 * 10�6sz � �2.1 MPasy � �3.6 MPasx � �4.5 MPa

Solution 7.6-6 Triaxial stress (bulk modulus)

Find .

Eq. (7-53 b): ây �sy

E�

�

E (sz + sx)

Eq. (7-53 a): âx �sx

E�

�

E (sy + sz)

K

ây � �320 * 10�6

sz � �2.1 MPa âx � �740 * 10�6

sx � �4.5 MPa sy � �3.6 MPa Substitute numerical values and rearrange:

(1)

(2)

Units: = MPa

Solve simultaneously Eqs. (1) and (2):

Eq. (7-16): K �E

3(1 � 2�)� 5.0 GPa ;

E � 3,000 MPa � 3.0 GPa � � 0.40

E

(�320 * 10�6) E � �3.6 + 6.6 �

(�740 * 10�6) E � �4.5 + 5.7 �

07Ch07.qxd 9/27/08 1:22 PM Page 618


Problem 7.6-7 A rubber cylinder of length and cross-sectionalarea is compressed inside a steel cylinder by a force thatapplies a uniformly distributed pressure to the rubber (see figure).

(a) Derive a formula for the lateral pressure between therubber and the steel. (Disregard friction between the rubberand the steel, and assume that the steel cylinder is rigid whencompared to the rubber.)

(b) Derive a formula for the shortening of the rubber cylinder.d

p

FSALR

LS

R

F

S

F

Solution 7.6-7 Rubber cylinder

(a) LATERAL PRESSURE

or 0 � �p � �a�F

A� pb

Eq. (7-53 a): âx �sx

E�

�

E (sy + sz)

âx � âz � 0

sz � �p

sx � �p sy � �F

A

(b) SHORTENING

Substitute for and simplify:

(Positive represents an increase in strain, that is,elongation.)

(Positive represents a shortening of the rubbercylinder.)

d

d �(1 + �)(1 � 2�)

(1 � �)aFL

EAb ;

d � �âyL

ây

ây �F

EA

(1 + �)(�1 + 2�)

1 � �

p

� �F

EA�

�

E (�2p)

Eq. (7-53 b): ây �sy

E�

�

E (sz + sx)

Solve for p: p �v

1 � v aF

Ab ;

Problem 7.6-8 A block of rubber is confined between plane parallel walls of a steel block (see figure). A uniformly dis-tributed pressure is applied to the top of the rubber block bya force .

(a) Derive a formula for the lateral pressure between therubber and the steel. (Disregard friction between therubber and the steel, and assume that the steel block isrigid when compared to the rubber.)

(b) Derive a formula for the dilatation of the rubber.(c) Derive a formula for the strain-energy density of the

rubber.u

e

p

Fp0

SR

FF

SR

S

07Ch07.qxd 9/27/08 1:22 PM Page 619


Solution 7.6-8 Block of rubber

(a) LATERAL PRESSURE

OR 0 � �p � � (�p0) ‹ p � �p0 ;

Eq. (7-53 a): âx �sx

E�

�

E (sy + sz)

âx � 0 ây Z 0 âz Z 0

sy � �p0 sz � 0

sx � �p

(b) DILATATION

Substitute for :

(c) STRAIN ENERGY DENSITY

Eq. (7-57b):

Substitute for and :

u �(1 � �2)p0

2

2E ;

psx, sy, sz,

u �1

2E (sx

2+ sy

2+ sz

2) �v

E (sx sy + sx sz + sy sz)

e � �(1 + �)(1 � 2�)p0

E ;

p

�1 � 2�

E (�p � p0)

Eq. (7-56): e �1 � 2�

E (sx + sy + sz)

Problem 7.6-9 A solid spherical ball of brass ( ) is lowered into the ocean to a depth of 10,000 ft.The diameter of the ball is 11.0 in.

Determine the decrease in diameter, the decrease in volume, and the strain energy of the ball.U¢V¢d

� � 0.34E � 15 * 106 psi

Solution 7.6-9 Brass sphere

Lowered in the ocean to depth

Diameter

Sea water:

Pressure:

DECREASE IN DIAMETER

(decrease) ¢d � â0d � 1.04 * 10�3 in. ;

Eq. (7-59): â0 �s0

E(1 � 2�) � 94.53 * 10�6

s0 � gh � 638,000 lb/ft2 � 4431 psi

g � 63.8 lb/ft3d � 11.0 in.

h � 10,000 ft

E � 15 * 10�6 psi � � 0.34 DECREASE IN VOLUME

(decrease)

STRAIN ENERGY

U � uV0 � 438 in.-lb ;

u �3(1 � 2�)s0

2

2E� 0.6283 psi

Use Eq. (7-57 b) with sx � sy � sz � s0:

¢V � eV0 � 0.198 in.3 ;

V0 �4

3pr 3 �

4

3(p)a11.0 in.

2b3

� 696.9 in.3

Eq. (7-60): e � 3â0 � 283.6 * 10�6

07Ch07.qxd 9/27/08 1:22 PM Page 620


Problem 7.6-10 A solid steel sphere ( , = 0.3) is subjected to hydrostatic pressure such that its volume isreduced by 0.4%.

(a) Calculate the pressure .(b) Calculate the volume modulus of elasticity for the steel.(c) Calculate the strain energy stored in the sphere if its diameter is d � 150 mmU

Kp

p�E � 210 GPa

Solution 7.6-10 Steel sphere

Hydrostatic Pressure. = Initial volume

(a) PRESSURE

Pressure ;p � s0 � 700 MPa

or s0 �Ee

3(1 � 2�)� 700 MPa

Eq.(7-60): e �3s0(1 � 2�)

E

Dilatation: e �¢V

V0� 0.004

¢V � 0.004 V0

V0

E � 210 GPa � � 0.3 (b) VOLUME MODULUS OF ELASTICITY

(c) STRAIN ENERGY ( = diameter)= 150 mm r = 75 mm

From Eq. (7-57b) with

U � uV0 � 2470 N # m � 2470 J ;

V0 �4pr 3

3� 1767 * 10�6 m3

u �3(1 � 2�)s0

2

2E� 1.40 MPa

sx � sy � sz � s0:d

d

Eq. (7-63): K �s0

E�

700 MPa

0.004� 175 GPa ;

Problem 7.6-11 A solid bronze sphere (volume modulus of elasticity K � 14.5 � 106 psi) is suddenly heated around itsouter surface. The tendency of the heated part of the sphere to expand produces uniform tension in all directions at the centerof the sphere.

If the stress at the center is 12,000 psi, what is the strain? Also, calculate the unit volume change e and the strain-energydensity u at the center.

Solution 7.6-11 Bronze sphere (heated)

(tension at the center)

STRAIN AT THE CENTER OF THE SPHERE

Combine the two equations:

â0 �s0

3K� 276 * 10�6 ;

Eq. (7-61): K �E

3(1 � 2�)

Eq. (7-59): â0 �s0

E(1 � 2�)

s0 � 12,000 psiK � 14.5 * 106 psi UNIT VOLUME CHANGE AT THE CENTER

STRAIN ENERGY DENSITY AT THE CENTER

u � 4.97 psi ;

u �3(1 � 2�)s0

2

2E�

s20

2K

Eq. (7-57b) with sx � sy � sz � s0:

Eq. (7-62): e �s0

K� 828 * 10�6 ;

07Ch07.qxd 9/27/08 1:22 PM Page 621


Plane Strain

When solving the problems for Section 7.7, consider only the in-plane strains (thestrains in the xy plane) unless stated otherwise. Use the transformation equationsof plane strain except when Mohr’s circle is specified (Problems 7.7-23 through7.7-28).

Problem 7.7-1 A thin rectangular plate in is subjected tostresses sx and sy, as shown in part (a) of the figure on the next page. Thewidth and height of the plate are and , respectively.Measurements show that the normal strains in the and directions are

and , respectively. With reference to part (b) of the figure, which shows a two-dimensional

view of the plate, determine the following quantities: (a) the increase in thelength of diagonal ; (b) the change in the angle between diagonal and the axis; and (c) the change in the angle between diagonal andthe axis.y

Odc¢cxOdf¢fOd

¢d

Py � �125 * 10�6Px � 195 * 10�6

yxh � 4.0 in.b � 8.0 in.

biaxial stress

sy

sx

y

b

h

xO

(a)

y

x

z

b

h

(b)

f

c

d


(a) INCREASE IN LENGTH OF DIAGONAL

âx1�

âx + ây

2+

âx � ây

2 cos 2u +

gxy

2 sin 2u

Ld � 1b2+ h2 � 8.944 in.

f � arctan h

b� 26.57°

ây � �125 * 10�6 gxy � 0

b � 8.0 in. h � 4.0 in. âx � 195 * 10�6

(b) CHANGE IN ANGLE

Minus sign means line rotates clockwise (angle decreases).

(c) CHANGE IN ANGLE

Angle increases the same amount that decreases.

¢c � 128 * 10�6 rad (increase) ;

fcc

¢f � 128 * 10�6 rad (decrease) ;f

Od

For u � f � 26.57°: a � �128.0 * 10�6 rad

Eq. (7-68): a � �(âx � ây) sinu cosu � gxy sin2u

f

¢d � âx1Ld � 0.00117 in. ;

For u � f � 26.57°, âx1� 130.98 * 10�6

Probs. 7.7-1 and 7.7-2

07Ch07.qxd 9/27/08 1:24 PM Page 622



âx1�

âx + ây

2+

âx � ây

2 cos 2u +

gxy

2 sin 2u

Ld � 1b2+ h2 � 170.88 mm

f � arctan h

b� 20.56°

ây � �320 * 10�6 gxy � 0

b � 160 mm h � 60 mm âx � 410 * 10�6

(b) CHANGE IN ANGLE

Minus sign means line rotates clockwise (angle decreases.)

(c) CHANGE IN ANGLE

Angle increases the same amount that decreases.

¢c � 240 * 10�6 rad (increase) ;

fcc

¢f � 240 * 10�6 rad (decrease) ;

fOd

For u � f � 20.56°: a � �240.0 * 10�6 rad

Eq. (7-68): a � �(âx�ây) sin u cos u �gxy sin2u

f

¢d � âx1Ld � 0.0547 mm ;

For u � f � 20.56°: âx1 � 319.97 * 10�6

Problem 7.7-3 A thin square plate in is subjected to stresses and , as shown in part (a) of the figure . The width of the plate is Measurements show that thenormal strains in the and directions are and , respectively.

With reference to part (b) of the figure, which shows atwo-dimensional view of the plate, determine the followingquantities: (a) the increase in the length of diagonal ;(b) the change in the angle between diagonal andthe axis; and (c) the shear strain associated with diagonals

and (that is, find the decrease in angle ).cedcfOdgx

Odf¢f

Od¢d

Py � 113 * 10�6Px � 427 * 10�6yx

b � 12.0 in.sysx

biaxial stresssy

sx

y

b

b

xO

(a)

y

x

z b

b e

c d

f

(b)

f

Problem 7.7-2 Solve the preceding problem if , and Py � �320 * 10�6.b � 160 mm, h � 60 mm, Px � 410 * 10�6

SECTION 7.7 Plane Strain 623

PROBS. 7.7-3 and 7.7-4

07Ch07.qxd 9/27/08 1:24 PM Page 623


Solution 7.7-3 Square plate in biaxial stress


¢d � âx1Ld � 0.00458 in. ;

For u � f � 45°: âx1� 270 * 10�6

âx1�

âx + ây

2+

âx � ây

2 cos 2u +

gxy

2 sin 2u

Ld � b12 � 16.97 in.

f � 45° gxy � 0

ây � 113 * 10�6

b � 12.0 in. âx � 427 * 10�6

(b) CHANGE IN ANGLE


(c) SHEAR STRAIN BETWEEN DIAGONALS

(Negative strain means angle increases)

g � �314 * 10�6 rad ;ced

For u � f � 45°: gx1y1� �314 * 10�6 rad

Eq. (7-71b): gx1y1

2� �

âx �ây

2 sin 2u +

gxy

2 cos 2u

¢f � 157 * 10�6 rad (decrease) ;

fOd

For u � f � 45°: a � �157 * 10�6 rad

Eq. (7-68): a � �(âx � ây) sinu cosu � gxy sin2u

f

Problem 7.7-4 Solve the preceding problem if , and Py � 211 * 10�6.b � 225 mm, Px � 845 * 10�6

Solution 7.7-4 Square plate in biaxial stress

Ld � b12 � 318.2 mm

ây � 211 * 10�6 f � 45° gxy � 0

b � 225 mm âx � 845 * 10�6


(b) CHANGE IN ANGLE


¢f � 317 * 10�6 rad (decrease) ;

fOd

For u � f � 45°: a � �317 * 10�6 rad

Eq. (7-68): a � �(âx � ây) sin u cos u � gxysin2u

f

¢d � âx1Ld � 0.168 mm ;

For u � f � 45°: âx1� 528 * 10�6

âx1�

âx + ây

2+

âx � ây

2 cos 2u +

gxy

2 sin 2u

07Ch07.qxd 9/27/08 1:24 PM Page 624


Solution 7.7-5 Element in plane strain

ây1� 239 * 10�6

âx1� 461 * 10�6 gx1y1

� 225 * 10�6

For u � 50°:

ây1� âx + ây � âx1

gx1y1

2� �

âx � ây

2 sin 2u +

gxy

2 cos 2u

âx1�

âx + ây

2+

âx �ây

2 cos 2u +

gxy

2 sin 2u

gxy � 180 * 10�6

âx � 220 * 10�6 ây � 480 * 10�6

Problem 7.7-6 Solve the preceding problem for the following data: , , and .u � 37.5°

gxy �310 * 10�6,Py � �170 * 10�6Px � 420 * 10�6

(c) SHEAR STRAIN BETWEEN DIAGONALS

Eq. (7- 71b): gx1y1

2� �

âx �ây

2 sin 2u +

gxy

2 cos 2u

(Negative strain means angle increases)

g � �634 * 10�6 rad ced

For u � f � 45°: gx1y1� �634 * 10�6 rad

Problem 7.7-5 An element of material subjected to (see figure) has strains as follows: , and

.Calculate the strains for an element oriented at an angle and show these

strains on a sketch of a properly oriented element.u � 50°

gxy � 180 * 10�6Px � 220 * 10�6, Py � 480 * 10�6

plane strain

ey

ex

gxy

y

xO 1

1


07Ch07.qxd 9/27/08 1:24 PM Page 625


Problem 7.7-7 The strains for an element of material in (see figure) are as follows: , and

Determine the principal strains and maximum shear strains, and show these strains on sketches of properly orientedelements.

gxy � �350 * 10�6.Py � 140 * 10�6Px � 480 * 10�6, plane strain


ây1� �101 * 10�6

âx1� 351 * 10�6 gx1y1

� �490 * 10�6

For u � 37.5°:


gx1y1

2� �

âx � ây

2 sin 2u +

gxy

2 cos 2u

âx1�

âx + ây

2+

âx � ây

2 cos 2u +

gxy

2 sin 2u

gxy � 310 * 10�6

âx � 420 * 10�6 ây � �170 * 10�6


PRINCIPAL STRAINS

up2� 67.1° â2 � 66 * 10�6 ;

‹ up1� �22.9° â1 � 554 * 10�6 ;� 554 * 10�6

âx1�

âx + ây

2+

âx � ây

2 cos 2u +

gxy

2 sin 2u

For up � �22.9°:

up � �22.9° and 67.1°

2up � �45.8° and 134.2°

tan 2up �gxy

âx � ây� �1.0294

â1 � 554 * 10�6 â2 � 66 * 10�6

� 310 * 10�6; 244 * 10�6

â1, 2 �âx + ây

2; A a

âx � ây

2b2

+ agxy

2b2

gxy � �350 * 10�6

âx � 480 * 10�6 ây � 140 * 10�6

MAXIMUM SHEAR STRAINS

âaver �âx + ây

2� 310 * 10�6

gmin � �488 * 10�6 ;us2

� us1+ 90° � 22.1°

gmax � 488 * 10�6 ;us1

� up1� 45° � �67.9° or 112.1°

gmax � 488 * 10�6

� 244 * 10�6

gmax

2� A a

âx �ây

2b2

+ agxy

2b2

07Ch07.qxd 9/27/08 1:24 PM Page 626


Problem 7.7-8 Solve the preceding problem for the following strains: , , and.gxy � �360 * 10�6

Py � �450 * 10�6Px � 120 * 10�6


PRINCIPAL STRAINS

â2 � �502 * 10�6 ;up2� 73.9°

‹ up1� 163.9° â1 � 172 * 10�6 ;� 172 * 10�6

âx1�

âx + ây

2+

âx � ây

2 cos 2u +

gxy

2 sin 2u

For up � 163.9°:

up � 163.9° and 73.9°

2up � 327.7° and 147.7°

tan 2up �gxy

âx � ây� �0.6316

â1 � 172 * 10�6 â2 � �502 * 10�6

� �165 * 10�6; 377 * 10�6

â1,2 �âx + ây

2; A a

âx � ây

2b2

+ agxy

2b2

gxy � �360 * 10�6

âx � 120 * 10�6 ây � �450 * 10�6 MAXIMUM SHEAR STRAINS

âaver �âx + ây

2� �165 * 10�6

gmin � �674 * 10�6 ;us2

� us1� 90° � 28.9°

gmax � 674 * 10�6 ;us1

� up1� 45° � 118.9°

gmax � 674 * 10�6

� 337 * 10�6

gmax

2� A a

âx � ây

2b2

+ agxy

2b2

07Ch07.qxd 9/27/08 1:24 PM Page 627


Problem 7.7-9 A element of material in (see figure) is subjected to strains ,and

Determine the following quantities: (a) the strains for an element oriented at an angle , (b) the principal strains,and (c) the maximum shear strains. Show the results on sketches of properly oriented element.

u � 75°gxy � 420 * 10�6.

Px � 480 * 10�6, Py � 70 * 10�6plane strain


ây1� 348 * 10�6

âx1� 202 * 10�6 gx1y1

� �569 * 10�6

For u � 75°:


gx1y1

2� �

âx � ây

2 sin 2u +

gxy

2 cos 2u

âx1�

âx + ây

2+

âx �ây

2 cos 2u +

gxy

2 sin 2u

gxy � 420 * 10�6

âx � 480 * 10�6 ây � 70 * 10�6

up2� 112.8° â2 � �18 * 10�6 ;

â1 � 568 * 10�6 ;‹ up1� 22.8°

� 568 * 10�6

âx1�

âx + ây

2+

âx � ây

2 cos 2u +

gxy

2 sin 2u

For up � 22.85°:

PRINCIPAL STRAINS

up � 22.85° and 112.85°

2up � 45.69° and 225.69°

tan 2up �gxy

âx � ây�1.0244

â1 � 568 * 10�6 â2 � �18 * 10�6

� 275 * 10�6; 293 * 10�6

â1,2 �âx + ây

2; A a

âx � ây

2b2

+ agxy

2b2


âaver �âx + ây

2� 275 * 10�6

gmin � �587 * 10�6 ;us2

� us1+ 90° � 67.8°

gmax � 587 * 10�6 ;us1

� up1� 45° � �22.2° or 157.8°

gmax � 587 * 10�6

gmax

2� A a

âx � ây

2b2

+ agxy

2b2

� 293 * 10�6

07Ch07.qxd 9/27/08 1:24 PM Page 628


Problem 7.7-10 Solve the preceding problem for the following data: , ,, and .u � 45°gxy � 780 * 10�6

Py � �430 * 10�6Px � �1120 * 10�6


ây1� �1165 * 10�6

âx1� �385 * 10�6 gx1y1

� 690 * 10�6

For u � 45°:


gx1y1

2� �

âx � ây

2 sin 2u +

gxy

2 cos 2u

âx1�

âx + ây

2+

âx � ây

2 cos 2u +

gxy

2 sin 2u

gxy � 780 * 10�6

âx � �1120 * 10�6 ây � �430 * 10�6

up2� 155.7° â2 � �1296 * 10�6 ;

‹ up1� 65.7° â1 � �254 * 10�6 ;

PRINCIPAL STRAINS

� �254 * 10�6

âx1�

âx + ây

2+

âx � ây

2 cos 2u +

gxy

2 sin 2u

For up � 65.7°:

up � 65.7° and 155.7°

2up � 131.5° and 311.5°

tan 2up �gxy

âx � ây� �1.1304

â1 � �254 * 10�6 â2 � �1296 * 10�6

� �775 * 10�6; 521 * 10�6

â1, 2 �âx + ây

2; A a

âx � ây

2b2

+ agxy

2b2


âaver �âx + ây

2� �775 * 10�6

gmin � �1041 * 10�6 ;us2

� us1+ 90° � 110.7°

gmax � 1041 * 10�6 ;us1

� up1� 45° � 20.7°

gmax � 1041 * 10�6

� 521 * 10�6

gmax

2� A a

âx � ây

2b2

+ agxy

2b2

07Ch07.qxd 9/27/08 1:24 PM Page 629


Solution 7.7-11 Steel plate in biaxial stress

UNITS: All stresses in psi.

STRAIN IN BIAXIAL STRESS (EQS. 7-39)

(1)

(2)

(3)

STRAINS AT ANGLE (EQ. 7-71a)

(4)Solve for sy: sy � 2400 psi

(23,400�1.3sy) cos 60°+ a1

2b a 1

30 * 106b

407 * 10�6 � a1

2b a 1

30 * 106b (12,600 + 0.7sy)

âx1�

âx + ây

2+

âx � ây

2 cos 2u +

gxy

2 sin 2u

f � 30°

âz � ��

E (sx�sy) � �

0.3

30 * 106 (18,000�sy)

ây �1

E (sy � �sx ) �

1

30 * 106 (sy � 5400)

âx �1

E (sx � �sy) �

1

30 * 106 (18,000 � 0.3sy)

Strain gage: f � 30° â � 407 * 10�6

E � 30 * 106 psi � � 0.30

sx � 18,000 psi gxy � 0 sy � ? MAXIMUM IN-PLANE SHEAR STRESS

STRAINS FROM EQS. (1), (2), AND (3)

MAXIMUM SHEAR STRAINS (EQ. 7-75)

� 104 * 10�6 ;gyz � 0 (gmax)yz

yz plane: (gmax) yz

2� A a

ây � âz

2b2

+ agyz

2b2

� 780 * 10�6 ;gxz � 0 (gmax) xz

xz plane: (gmax)xz

2� A a

âx � âz

2b2

+ agxz

2b2

� 676 * 10�6 ;gxy � 0 (gmax) xy

xy plane: (gmax) xy

2� A a

âx � ây

2b2

+ agxy

2b2

âz � �204 * 10�6

âx � 576 * 10�6 ây � �100 * 10�6

(tmax)xy �sx � sy

2� 7800 psi ;

Problem 7.7-11 A steel plate with modulus of elasticity and Poisson’s ratio is loaded in by normal stresses and (see figure). A strain gage is bonded to the plate at an angle .

If the stress is 18,000 psi and the strain measured by the gage is ,what is the maximum in-plane shear stress and shear strain ? What is themaximum shear strain in the plane? What is the maximum shear strain in the plane?yz(gmax)yz

xz(gmax)xz

(gmax)xy(tmax)xy

P � 407 * 10�6sx

f � 30°sysxbiaxial stress� � 0.30

E � 30 * 106 psi

sy

sx

y

x

z

f

Probs. 7.7-11 and 7.7-12

07Ch07.qxd 9/27/08 1:24 PM Page 630


Problem 7.7-12 Solve the preceding problem if the plate is made of aluminum with and , the stressis , the angle is , and the strain is 946 * 10�6.P21°f86.4 MPasx

� � 1/3E � 72 GPa

Solution 7.7-12 Aluminum plate in biaxial stress

UNITS: All stresses in MPa.

STRAIN IN BIAXIAL STRESS (EQS. 7-39)

(1)

(2)

(3)

STRAINS AT ANGLE (EQ. 7-71a)

Solve for sy: sy � 21.55 MPa (4)

+ a1

2b a 1

72,000b a115.2 �

4

3 syb

cos 42°

946 * 10�6 � a1

2b a 1

72,000b a57.6 +

2

3 syb

âx1�

âx + ây

2+

âx � ây

2 cos 2u +

gxy

2 sin 2u

f � 21°

âz � ��

E (sx�sy) � �

1/3

72,000 (86.4�sy)

ây �1

E (sy � �sx) �

1

72,000 (sy � 28.8)

âx �1

E (sx � �sy) �

1

72,000 (86.4 �

1

3 sy)

Strain gage: f � 21° â � 946 * 10�6

� � 1/3E � 72 GPa

sx � 86.4 MPa gxy � 0 sy � ? MAXIMUM IN-PLANE SHEAR STRESS

STRAINS FROM EQS. (1), (2), AND (3)

MAXIMUM SHEAR STRAINS (EQ. 7-75)

� 399 * 10�6 ;gyz � 0 (gmax)yz

yz plane: (gmax) yz

2� A a

ây � âz

2b2

+ agyz

2b2

� 1600 * 10�6 ;gxz � 0 (gmax) xz

xz plane: (gmax)xz

2� A a

âx � âz

2b2

+ agxz

2b2

� 1200 * 10�6 ;gxy � 0 (gmax) xy

xy plane: (gmax) xy

2� A a

âx � ây

2b2

+ agxy

2b2

âz � �500 * 10�6

âx � 1100 * 10�6 ây � �101 * 10�6

(tmax)xy �sx � sy

2� 32.4 MPa ;

07Ch07.qxd 9/27/08 1:24 PM Page 631


Problem 7.7-13 An element in plane stress is subjected to stresses sx � �8400 psi, sy � 1100 psi, and txy � �1700 psi (see figure). The material is aluminum with modulusof elasticity E � 10,000 ksi and Poisson’s ratio .

Determine the following quantities: (a) the strains for an element oriented at an angle, (b) the principal strains, and (c) the maximum shear strains. Show the results on

sketches of properly oriented elements.

Probs. 7.7-13 and 7.7-14

u � 30°

� � 0.33y

xO

txy

sy

sx


HOOKE’S LAW (EQS. 7-34 AND 7-35)

FOR


� 267 * 10�6

gx1y1� 868 * 10�6

� 434 * 10�6

gx1y1

2� �

âx � ây

2 sin 2u +

gxy

2 cos 2u

� �756 * 10�6

âx1�

âx + ây

2+

âx � ây

2 cos 2u +

gxy

2 sin 2u

u � 30°:

gxy �txy

G�

2txy(1 + �)

E� �452.2 * 10�6

ây �1

E(sy � �sx) � 387.2 * 10�6

âx �1

E(sx � �sy) � �876.3 * 10�6

txy � �1700 psi E � 10,000 ksi � � 0.33

sx � �8400 psi sy � 1100 psi PRINCIPAL STRAINS

FOR

;up2� 9.8° â2 � �916 * 10�6

;‹ up1� 99.8° â1 � 426 * 10�6

� �916 * 10�6

âx1�

âx + ây

2+

âx � ây

2 cos 2u +

gxy

2 sin 2u

up � 9.8°:

up � 9.8° and 99.8°

2up � 19.7° and 199.7°

tan 2up �gxy

âx � ây� 0.3579

â1 � 426 * 10�6 â2 � �961 * 10�6

� �245 * 10�6; 671 * 10�6

â1,2 �âx + ây

2; A a

âx � ây

2b2

+ agxy

2b2

07Ch07.qxd 9/27/08 1:24 PM Page 632


Problem 7.7-14 Solve the preceding problem for the following data: sx � �150 MPa, sy � �210 MPa, txy � �16 MPa, and . The material is brass with and .� � 0.34E � 100 GPau � 50°


âaver �âx + ây

2� �245 * 10�6

;gmin � �1342 * 10�6

us2� us1

+ 90° � 144.8°

;gmax � 1342 * 10�6

us1� up1

� 45° � 54.8°

gmax � 1342 * 10�6

� 671 * 10�6

gmax

2� A a

âx � ây

2b2

+ agxy

2b2


HOOKE’S LAW (EQS. 7-34 AND 7-35)

FOR


� �907 * 10�6

gx1y1� �717 * 10�6

� �358.5 * 10�6

gx1y1

2� �

âx � ây

2 sin 2u +

gxy

2 cos 2u

� �1469 * 10�6

âx1�

âx + ây

2+

âx � ây

2 cos 2u +

gxy

2 sin 2u

u � 50°:

gxy �txy

G�

2txy(1 + �)

E� �429 * 10�6

ây �1

E(sy � �sx) � �1590 * 10�6

âx �1

E(sx � �sy) � �786 * 10�6

txy � �16 MPa E � 100 GPa � � 0.34

sx � �150 MPa sy � �210 MPa

PRINCIPAL STRAINS

up � 76.0° and 166.0°

2up � 151.9° and 331.9°

tan 2up �gxy

âx � ây� �0.5333

â1 � �732 * 10�6 â2 � �1644 * 10�6

� �1188 * 10�6; 456 * 10�6

â1,2 �âx + ây

2; A a

âx � ây

2b2

+ agxy

2b2

07Ch07.qxd 9/27/08 1:24 PM 3


Problem 7.7-15 During a test of an airplane wing, the strain gage readings from a rosette (see figure) are as follows: gage , ; gage , ; and gage , .

Determine the principal strains and maximum shear strains, and show them on sketches ofproperly oriented elements.

Probs. 7.7-15 and 7.7-16

� 80 * 10�6C360 * 10�6B520 * 10�6A45°

FOR

;up2� 76.0° â2 � �1644 * 10�6

; ‹ up1� 166.0° â1 � �732 * 10�6

� �1644 * 10�6

âx1�

âx + ây

2+

âx � ây

2 cos 2u +

gxy

2 sin 2u

up � 76.0°: MAXIMUM SHEAR STRAINS

âaver �âx + ây

2� �1190 * 10�6

;gmin � �911 * 10�6

us2� us1

� 90° � 31.0°

;gmax � 911 * 10�6

us1� up1

� 45° � 121.0°

gmax � 911 * 10�6

� 456 * 10�6

gmax

2� A a

âx � ây

2b2

+ agxy

2b2

y

CB

A xO

45°

45°

07Ch07.qxd 9/27/08 1:24 PM Page 634


Problem 7.7-16 A strain rosette (see figure) mounted on the surface of an automobile frame gives the following readings: gage , ; gage , ; and gage , .

Determine the principal strains and maximum shear strains, and show them on sketches of properly oriented elements.�160 * 10�6C180 * 10�6B310 * 10�6A

45°

Solution 7.7-15 strain rosette 45°

FROM EQS. (7-77) AND (7-78) OF EXAMPLE 7-8:

PRINCIPAL STRAINS

â1 � 551 * 10�6 â2 � �111 * 10�6

� 220 * 10�6; 331 * 10�6

â1,2 �âx + ây

2; A a

âx � ây

2b2

+ agxy

2b2

gxy � 2âB � âA � âC � 280 * 10�6

âx � âA � 520 * 10�6 ây � âC � �80 * 10�6

âC � �80 * 10�6

âA � 520 * 10�6 âB � 360 * 10�6


âaver �âx + ây

2� 220 * 10�6

;gmin � �662 * 10�6

us2� us1

+ 90° � 57.5°

;gmax � 662 * 10�6

us1� up1

� 45° � �32.5°or 147.5°

gmax � 662 * 10�6

� 331 * 10�6

gmax

2� A a

âx � ây

2b2

+ agxy

2b2

;up2� 102.5° â2 � �111 * 10�6

;‹ up1� 12.5° â1 � 551 * 10�6

For

� 551 * 10�6

âx1�

âx + ây

2+

âx � ây

2 cos 2u +

gxy

2 sin 2u

up � 12.5°:

up � 12.5° and 102.5°

2up � 25.0° and 205.0°

tan 2up �gxy

âx � ây� 0.4667

07Ch07.qxd 9/27/08 1:24 PM Page 635


Solution 7.7-16 strain rosette 45°

FROM EQS. (7-77) AND (7-78) OF EXAMPLE 7-8:

PRINCIPAL STRAINS

FOR

;up2 � 102.0° â2 � �182 * 10�6

;‹ up1 � 12.0° â1 � 332 * 10�6

� 332 * 10�6

âx1�

âx + ây

2+

âx � ây

2 cos 2u +

gxy

2 sin 2u

up � 12.0°:

up � 12.0° and 102.0°

2up � 24.1° and 204.1°

tan 2up �gxy

âx � ây� 0.4468

â1 � 332 * 10�6 â2 � �182 * 10�6

� 75 * 10�6; 257 * 10�6

â1,2 �âx + ây

2; A a

âx � ây

2b2

+ agxy

2b2

gxy � 2âB � âA � âC � 210 * 10�6

âx � âA � 310 * 10�6 ây � âC � �160 * 10�6

âC � �160 * 10�6

âA � 310 * 10�6 âB � 180 * 10�6 MAXIMUM SHEAR STRAINS

âaver �âx + ây

2� 75 * 10�6

;gmin � �515 * 10�6

us2� us1

+ 90° � 57.0°

;gmax � 515 * 10�6

us1� up1

� 45° � �33.0° or 147.0°

gmax � 515 * 10�6

� 257 * 10�6

gmax

2� A a

âx � ây

2b2

+ agxy

2b2

07Ch07.qxd 9/27/08 1:24 PM Page 636


Problem 7.7-17 A solid circular bar of diameter in. is subjected to an axial force and a torque (see figure). Strain gages

and mounted on the surface of the bar give readingand . The bar is made of steel

having psi and .

(a) Determine the axial force and the torque .(b) Determine the maximum shear strain and the maximum

shear stress in the bar.tmax

gmax

TP

� � 0.29E � 30 * 106Pb � �55 * 10�6

Pa � 100 * 10�6BA

TPd � 1.5 d

C

B

A

P

T

C

45∞

Solution 7.7-17 Circular bar (plane stress)

Bar is subjected to a torque and an axial force .

STRAIN GAGES

ELEMENT IN PLANE STRESS

AXIAL FORCE

SHEAR STRAIN

� �(0.1298 * 10�6)T (T � lb-in.)

gxy �txy

G�

2txy(1 + �)

E� �

32T(1 + �)

pd 3E

;âx �sx

E�

4P

pd 2E P �

pd 2Eâx

4� 5300 lb

P

âx � 100 * 10�6 ây � ��âx � �29 * 10�6

sx �P

A�

4P

pd 2 sy � 0 txy � �

16T

pd 3

At u � 45°: âB � �55 * 10�6

At u � 0°: âA � âx � 100 * 10�6

Diameter d � 1.5 in.

E � 30 * 106 psi � � 0.29

PT STRAIN AT

(1)

Substitute numerical values into Eq. (1):

Solve for :

MAXIMUM SHEAR STRAIN AND MAXIMUM SHEAR STRESS

; tmax � Ggmax � 2580 psi

; gmax � 222 * 10�6 rad

� 111 * 10�6 rad

Eq. (7-75): gmax

2� A a

âx � ây

2b2

+ agxy

2b2

gxy � �(0.1298 * 10�6)T � �180.4 * 10�6 rad

; T � 1390 lb-inT

�55 * 10�6 � 35.5 * 10�6 � (0.0649 * 10�6)T

âx1� âB � �55 * 10�6 2u � 90°

âx1�

âx + ây

2+

âx � ây

2 cos 2u +

gxy

2 sin 2u

u � 45°

07Ch07.qxd 9/27/08 1:24 PM Page 637


Problem 7.7-18 A cantilever beam of rectangular cross section (width , height ) is loaded by a force that acts at the midheight of the beam and isinclined at an angle to the vertical (see figure). Two straingages are placed at point , which also is at the midheight of the beam. Gage measures the strain in the horizontaldirection and gage measures the strain at an angle b = 60to the horizontal. The measured strains are and .

Determine the force and the angle , assuming thematerial is steel with and .� � 1/3E � 200 GPa

aPPb � �375 * 10�6

Pa � 125 * 10�6°B

AC

a

Ph � 100 mmb � 25 mm

h

bP

C

C

B

A

b

b

a

h

Solution 7.7-18 Cantilever beam (plane stress)

Beam loaded by a force acting at an angle .

Axial force

Shear force

(At the neutral axis, the bending moment produces nostresses.)

STRAIN GAGES


âx � 125 * 10�6 ây � ��âx � �41.67 * 10�6

txy � �3V

2A� �

3P cos a

2bh

sx �F

A�

P sin a

bh sy � 0

At u � 60°: âB � �375 * 10�6

At u � 0°: âA � âx � 125 * 10�6

V � P cos a

F � P sin a

h � 100 mm

E � 200 GPa � � 1/3 b � 25 mm

aP HOOKE’S LAW

(1)

(2)

FOR :

(3)

Substitute into Eq. (3):

(4)

SOLVE EQS. (1) AND (4):

P � 125 kN ;tan a � 0.5773 a � 30° ;

or P cos a � 108,260 N

� (3.464 * 10�9)P cos a

�375 * 10�6 � 41.67 * 10�6 � 41.67 * 10�6

âx1� âB � �375 * 10�6 2u � 120°

âx1�

âx + ây

2+

âx � ây

2 cos 2u +

gxy

2 sin 2u

u � 60°

� �(8.0 * 10�9) P cos a

gxy �txy

G� �

3P cos a

2bhG� �

3(1 + �) P cos a

bhE

P sin a � bhEâx � 62,500 N

âx �sx

E�

P sin a

bhE

Problem 7.7-19 Solve the preceding problem if the cross-sectional dimensions are and , the gageangle is b = 75 , the measure strains are and , and the material is a magnesium alloywith modulus and Poisson’s ratio � � 0.35.E � 6.0 * 106 psi

Pb � �266 * 10�6Pa � 171 * 10�6°

h � 3.0 in.b � 1.0 in.

Probs. 7.7-18 and 7.7-19

07Ch07.qxd 9/27/08 1:24 PM Page 638


Problem 7.7-20 A strain rosette, or delta rosette, consists of threeelectrical-resistance strain gages arranged as shown in the figure. Gage measuresthe normal strain in the direction of the axis. Gages and measure thestrains and in the inclined directions shown.

Obtain the equations for the strains , and associated with the axis.xygxyPx, Py

PcPb

CBxPa

A60°

Solution 7.7-19 Cantilever beam (plane stress)

Beam loaded by a force acting at an angle .

Axial force Shear foce (At the neutral axis, the bending moment producesno stresses.)

STRAIN GAGES


âx � 171 * 10�6 ây � ��âx � �59.85 * 10�6

txy � �3V

2A� �

3P cos a

2bh

sx �F

A�

P sin a

bh sy � 0

At u � 75°: âB � �266 * 10�6

At u � 0°: âA � âx � 171 * 10�6

V � P cos aF � P sin a

h � 3.0 in.

E � 6.0 * 106 psi � � 0.35 b � 1.0 in.

aP HOOKE’S LAW

(1)

(2)

FOR :

(3)

Substitute into Eq. (3):

(4)


P � 5000 lb ;tan a � 0.7813 a � 38° ;

or P cos a � 3939.8 lb

�(56.25 * 10�9)P cos a

�266 * 10�6 � 55.575 * 10�6 � 99.961 * 10�6

âx1� âB � �266 * 10�6 2u � 150°

âx1�

âx + ây

2+

âx � ây

2 cos 2u +

gxy

2 sin 2u

u � 75°

� �(225.0 * 10�9)P cos a

gxy �txy

G� �

3P cos a

2bhG� �

3(1 + v)P cos a

bhE

P sin a � bhEâx � 3078 lb

âx �sx

E�

P sin a

bhE

y

CB

A

xO

60°60°

60°

Solution 7.7-20 Delta rosette ( strain rosette)60°

STRAIN GAGES

FOR u � 0°: âx � âA ;

Gage C at u � 120° Strain � âC

Gage B at u � 60° Strain � âB

Gage A at u � 0° Strain � âA

FOR :

(1)âB �âA

4+

3ây

4+

gxy13

4

âB �âA + ây

2+

âA � ây

2 (cos 120°) +

gxy

2 (sin 120°)

âx1�

âx + ây

2+

âx � ây

2 cos 2u +

gxy

2 sin 2u

u � 60°

07Ch07.qxd 9/27/08 1:24 PM Page 639


Problem 7.7-21 On the surface of a structural component in a space vehicle, the strainsare monitored by means of three strain gages arranged as shown in the figure. During a certain maneuver, the following strains were recorded:

, and .Determine the principal strains and principal stresses in the material, which is a

magnesium alloy for which and . (Show the principal strains and principal stresses on sketches of properly oriented element.)

� � 0.35E � 6000 ksi

Pc � 200 * 10�6Pb � 200 * 10�6

Pa � 1100 * 10�6,

FOR :

(2)âC �âA

4+

3ây

4�

gxy13

4

âC �âA + ây

2+

âA � ây

2 (cos 240°)

gxy

2(sin 240°)

âx1�

âx + ây

2+

âx � ây

2 cos 2u +

gxy

2 sin 2u

u � 120° SOLVE EQS. (1) AND (2):

gxy �2

13 (âB � âC) ;

ây �1

3 (2âB + 2âC � âA) ;

y

CB

AxO

30°

Solution 7.7-21 strain rosette 30-60-90°

Magnesium alloy:

STRAIN GAGES

FOR

FOR

FOR

Solve for

PRINCIPAL STRAINS

â1 � 1550 * 10�6 â2 � �250 * 10�6

� 650 * 10�6; 900 * 10�6

â1,2 �âx + ây

2; A a

âx � ây

2b2

+ agxy

2b2

gxy: gxy � 1558.9 * 10�6

� 0.43301gxy

200 * 10�6 � 650 * 10�6+ 225 * 10�6

âx1� âC �

âx + ây

2+

âx � ây

2 cos 2u +

gxy

2 sin 2u

u � 150°:

u � 90°: ây � âB � 200 * 10�6

u � 0°: âx � âA � 1100 * 10�6

Gage C at u � 150° âC � 200 * 10�6

Gage B at u � 90° âB � 200 * 10�6

Gage A at u � 0° PA � 1100 * 10�6

E � 6000 ksi � � 0.35

FOR :

up2 � 120° â2 � �250 * 10�6 ; ‹ up1 � 30° â1 � 1550 * 10�6 ;

� 1550 * 10�6

âx1�

âx + ây

2+

âx � ây

2 cos 2u +

gxy

2 sin 2u

up � 30°

2up � 60° up � 30°

tan 2up �gxy

âx � ây� 13 � 1.7321

07Ch07.qxd 9/27/08 1:24 PM Page 640


PRINCIPAL STRESSES (see Eqs. 7-36)


s1 � 10,000 psi s2 � 2,000 psi ;

s1 �E

1 � �2 (â1 + �â2) s2 �

E

1 � �2 (â2 + �â1)

Problem 7.7-22 The strains on the surface of an experimental device made of pure aluminum ( , ) and tested in a space shuttle were measured by means of strain gages. The gages were oriented as shown in the figure, and the measured strains were , , and .

What is the stress in the x direction?sx

Pc � �39.44 * 10 *�6

Pb � 1496 * 10�6Pa � 1100 * 10�6

� � 0.33E � 70 Gpa

y

CB

AxO 40°40°

Solution 7.7-22 strain rosette 40-40-100°

Pure aluminum:

STRAIN GAGES

FOR

FOR :

(1)0.41318ây + 0.49240gxy � 850.49 * 10�6

âx � 1100 * 10�6; then simplify and rearrange:

Substitute âx1� âB � 1496 * 10�6 and

âx1�

âx + ây

2+

âx � ây

2 cos 2u +

gxy

2 sin 2u

u � 40°

u � 0°: âx � âA � 1100 * 10�6

Gage C at u � 140° âC � �39.44 * 10�6

Gage B at u � 40° âB � 1496 * 10�6

Gage A at u � 0° âA � 1100 * 10�6

E � 70 GPa � � 0.33 FOR

(2)


HOOKE’S LAW

sx �E

1 � �2 (âx + �ây) � 91.6 MPa ;

ây � 200.3 * 10�6 gxy � 1559.2 * 10�6

0.41318ây � 0.49240gxy � �684.95 * 10�6

âx � 1100 * 10�6; then simplify and rearrange:

Substitute âx1� âc � �39.44 * 10�6 and

âx1�

âx + ây

2+

âx � ây

2 cos 2u +

gxy

2 sin 2u

u � 140°:

07Ch07.qxd 9/27/08 1:24 PM Page 641


Problem 7.7-23 Solve Problem 7.7-5 by using Mohr’s circle for plane strain.


POINT C:

POINT D

POINT D

gx1y1� �225 * 10�6

gx1y1

2� �R sin b � �112.4 * 10�6

âx1� 350 * 10�6 � R cos b � 239 * 10�6

(u � 140°):¿

gx1y1� 225 * 10�6

gx1y1

2� R sin b � 112.4 * 10�6

âx1� 350 * 10�6

+ R cos b � 461 * 10�6

(u � 50°):

âx1� 350 * 10�6

b � 180° � a � 2u � 45.30°

a � arctan 90

130� 34.70°

� 158.11 * 10�6

R � 2(130 * 10�6)2+ (90 * 10�6)2

gxy � 180 * 10�6 gxy

2� 90 * 10�6 u � 50°

âx � 220 * 10�6 ây � 480 * 10�6

07Ch07.qxd 9/27/08 1:24 PM Page 642




gxy � 310 * 10�6 gxy

2� 155 * 10�6 u � 37.5°

âx � 420 * 10�6 ây � �170 * 10�6 POINT C:

POINT D

POINT D

gx1y1� 490 * 10�6

gx1y1

2� R sin b � 244.8 * 10�6

âx1� 125 * 10�6 � R cos b � �101 * 10�6

œ (u � 127.5°):

gx1y1� �490 * 10�6

gx1y1

2� �R sin b � �244.8 * 10�6

âx1� 125 * 10�6

+ R cos b � 351 * 10�6

(u � 37.5°):

âx1� 125 * 10�6

b � 2u � a � 47.28°

a � arctan 155

295� 27.72°

� 333.24 * 10�6

R � 2(295 * 10�6)2+ (155 * 10�6)2

07Ch07.qxd 9/27/08 1:24 PM Page 643




gxy � �350 * 10�6 gxy

2� �175 * 10�6

âx � 480 * 10�6 ây � 140 * 10�6


gmin � �488 * 10�6

Point S2: âaver � 310 * 10�6

gmax � 2R � 488 * 10�6

Point S1: âaver � 310 * 10�6

2us1� 2us2

+ 180° � 224.17° us1� 112.1°

2us2� 90°�a � 44.17° us2

� 22.1°

POINT C:

PRINCIPAL STRAINS

Point P2: â2 � 310 * 10�6 � R � 66 * 10�6

Point P1: â1 � 310 * 10�6+ R � 554 * 10�6

2up1� 2up2

+ 180° � 314.2° up1� 157.1°

2up2� 180° � a � 134.2° up2

� 67.1°

âx1� 310 * 10�6

a � arctan 175

170� 45.83°

� 243.98 * 10�6

R � 2(175 * 10�6)2+ (170 * 10�6)2

07Ch07.qxd 9/27/08 1:24 PM Page 644



gxy � �360 * 10�6 gxy

2� �180 * 10�6

âx � 120 * 10�6 ây � �450 * 10�6


gmin � �674 * 10�6

Point S2: âaver � �165 * 10�6

gmax � 2R � 674 * 10�6

Point S1: âaver � �165 * 10�6

2us1� 2us2

+ 180° � 237.72° us1� 118.9°

2us2� 90° � a � 57.72° us2

� 28.9°

Point C:

PRINCIPAL STRAINS

Point P2: â2 � �165 * 10�6 � R � �502 * 10�6

Point P1: â1 � R � 165 * 10�6 � 172 * 10�6

2up1� 2up2

+ 180° � 327.72° up1� 163.9°

2up2� 180° � a � 147.72° up2

� 73.9°

âx1� �165 * 10�6

a � arctan 180

285� 32.28°

� 337.08 * 10�6

R � 2(285 * 10�6)2+ (180 * 10�6)2


07Ch07.qxd 9/27/08 1:24 PM Page 645




Point C:

Point D

Point D

gx1y1� 569 * 10�6

gx1y1

2� R sin b � 284.36 * 10�6

âx1� 275 * 10�6

+ R cos b � 348 * 10�6

œ (u � 165°):

gx1y1� �569 * 10�6

gx1y1

2� �R sin b � �284.36 * 10�6

âx1� 275 * 10�6 � R cos b � 202 * 10�6

(u � 75°):

âx1� 275 * 10�6

b � a + 180° � 2u � 75.69°

a � arctan 210

205� 45.69°

� 293.47 * 10�6

R � 2(205 * 10�6)2+ (210 * 10�6)2

gxy � 420 * 10�6 gxy

2� 210 * 10�6 u � 75°

âx � 480 * 10�6 ây � 70 * 10�6

PRINCIPAL STRAINS

Point P2: â2 � 275 * 10�6 � R � �18 * 10�6

Point P1: â1 � 275 * 10�6+ R � 568 * 10�6

2up2� 2up1

+ 180° � 225.69° up2� 112.8°

2up1� a � 45.69° up1

� 22.8°


gmin � �587 * 10�6

Point S2: âaver � 275 * 10�6

gmax � 2R � 587 * 10�6

Point S1: âaver � 275 * 10�6

2us1� 2us2

+ 180° � 315.69° us1� 157.8°

2us2� 90° + a � 135.69° us2

� 67.8°

07Ch07.qxd 9/27/08 1:24 PM Page 646




gxy � 780 * 10�6 gxy

2� 390 * 10�6 u � 45°

âx � �1120 * 10�6 ây � �430 * 10�6

PRINCIPAL STRAINS

Point P2: â2 � �775 * 10�6 � R � �1296 * 10�6

Point P1: â1 � �775 * 10�6+ R � �254 * 10�6

2up2� 2up1

+ 180° � 311.50° up2� 155.7°

2up1� 180° � a � 131.50° up1

� 65.7°

Point C:

Point D:

Point D :

gx1y1� �690 * 10�6

gx1y1

2� �R sin b � �345 * 10�6

âx1� �775 * 10�6 � R cos b � �1165 * 10�6

(u � 135°)¿

gx1y1

2� R sin b � 345 * 10�6 gx1y1

� 690 * 10�6

âx1� �775 * 10�6

+ R cos b � �385 * 10�6

(u � 45°):

âx1� �775 * 10�6

b � 180° � a � 2u � 41.50°

a � arctan 390

345� 48.50°

� 520.70 * 10�6

R � 2(345 * 10�6)2+ (390 * 10�6)2


gmin � �1041 * 10�6

Point S2: âaver � �775 * 10�6

gmax � 2R � 1041 * 10�6

Point S1: âaver � �775 * 10�6

2us2� 2us1

+ 180° � 221.50° us2� 110.7°

2us1� 90° � a � 41.50° us1

� 20.7°

07Ch07.qxd 9/27/08 1:24 PM Page 647

07Ch07.qxd 9/27/08 1:24 PM Page 648

chapter-7

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