worked exp mechanics 2.pdf

8/18/2019 Worked exp Mechanics 2.pdf

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College of EngineeringMech. Eng. Dept.Subject: Strength of MaterialsSecond Class Lecturer: Sadiq Muhsin Almosawy

S trength of materials is a branch of applied mechanics that deals with the behavior of solid bodies subjected to various types of loading. The solid bodies include axially-loadedbars, shafts, beams, and columns. The objective of analysis will be the determination of thestresses, strains, and deformations produced by the loads.Simple Stress ( σ ):

If a cylindrical bar is subjected to a direct pull or push along its axis, then it issaid to be subjected to tension or compression .

P P P P

Tension Compression

In SI systems of units load is measured in Newton (N) or Kiloewton (KN) or Meganewton (MN).

Normal stress ( ) : is the intensity of normal force per unit area

Stress = Area Load

AP

σ

stress may thus be compressive or tensile depending on the nature of the load and will bemeasured in units of Newton per square meter (N/m 2 ). This unit, called Pascal

1 Pa=1 N/m 2

1 KPa=1000 Pa=10 3 Pa1 MPa=10 6 Pa1 GPa=10 9 Pa

In the U.S. customary or foot-pound-second system of units, express stress inpounds per square inch (Psi) or kilopound per square inch (Ksi)


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Normal Strain ( ε):

If a bar is subjected to a direct load, and hence a stress, the bar will change inlength. If the bar has an original length (L) and changes in length by an amount ( L), the

strain produced is defined as follows:

Strain( )=lengthoriginallengthinchange

L Lδ

ε

P P

L L

Strain is thus a measure of the deformation of the material and is non-dimensional,i.e. it has no units. Tensile stresses and strains are considered positive sense. Compressivestresses and strains are considered negative in sense.Shear Stress ( τ) and Bearing Stress ( b ):

Shearing stress differs from both tensile and compressive stress in that it iscaused by forces acting along or parallel to the area resisting the forces, whereas tensileand compressive stresses are caused by forces perpendicular to the areas on which they act.For this reason, tensile and compressive stresses are called normal stresses, whereas ashearing stress may be called a tangential stress.

A shearing stress is produced whenever the applied loads cause one section of abody to tend to slide past its adjacent section.

Shear stress=shear resisting Area

load Shear

Q

Q Q

Q

A

Qτ


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Area resisting shear is the shaded area as shown above.

P P

P AP

Single shear stress

P P

P/2P

A

P/2

Double shear stress

Bearing stress is a normal stress that is produced by the compression of onesurface against another. The bearing area is defined as the projected area of the curvedbearing surface.

1P P 2

3

b

bb A

F σ

Consider the bolted connection shown above, this connection consists of a flat bar

A, a clevis C, and a bolt B that passes through holes in the bar and clevis. Consider thebearing stresses labeled 1, the projected area A b on which they act is rectangle having a

AP

τ

AP 2 /

τ

A C

B


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height equal to the thickness of the clevis and a width equal to the diameter of the bolt, thebearing force F b represented by the stresses labeled 1 is equal to P/2. The same area and thesame force apply to the stresses labeled 3. For the bearing stresses labeled 2, the bearingarea A b is a rectangle with height equal to the thickness of the flat bar and width equal to

the bolt diameter. The corresponding bearing force F b is equal to the load P.

Shear Strain ( γ ):

Shear strain is a measure of the distortion of the element due to shear. Shearstrain is measured in radians and hence is non-dimensional, i.e. it has no units.

Elastic Materials-Hook's Law:

A material is said to be elastic if it returns to its original, when load isremoved. In elastic material, stress is proportional to strain. Hook's law therefore states

that:Stress ( ) strain ( )

strainstress constant

Within the elastic limit, i.e. within the limits in which Hook's law applies, it hasbeen shown that:

E εσ

This constant is given the symbol E and termed the modulus of elasticity orYoung's modulus.

Poisson's Ratio ( υ ):

Consider the rectangular bar shown below subjected to a tensile load. Underthe action of this load the bar will increase in length by an amount L giving a longitudinalstrain in the bar of:

L L

Lδε

τ

τ

γ


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The bar will also exhibit a reduction in dimensions laterally i.e. its breadth anddepth will both reduce.

The associated lateral strains will both be equal, will be of opposite sense to thelongitudinal strain, and will be given by:

bb

d d

lat

δδε

Poisson's ratio is the ratio of the lateral and longitudinal strains and alwaysconstant

Poisson's ratio=Strainal Longitudin

Strain Lateral

L Ld d

/ /

δδ

Longitudinal Strain= E σ

Lateral Strain= E σ

υ

Modulus of Rigidity ( G ):

For materials within the elastic range the shear strain is proportional to the shearstress producing it.

γ τ

StrainShear StressShear =Constant

γ τ =G

The constant G is termed the modulus of rigidity.

P P

b

d

2

bδ

2

d δ


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Example 1: A 25 mm square cross-section bar of length 300 mm carries an axialcompressive load of 50 KN . Determine the stress set up in the bar and its change of lengthwhen the load is applied. For the bar material E=200 GN/m 2.

Cross-section area of the bar(A)=25×10 -3×25×10 -3=625×10 -6 m2

AP

σ

= 63

106251050 =80000000 N/m 2

=80 MN/ m 2

E σ

ε

=9

6

10200

1080 =0.0004

L L εδ

L=0.0004×300×10 -3=0.12×10 -3m

L=0.12 mm

KN 50

mm300

mm25


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Example 2: Two circular bars, one of brass and the other of steel, are to be loaded by ashear load of 30 KN . Determine the necessary diameter of the bars a) in single shear b) indouble shear, if the shear stress in the two materials must not exceed 50 MN/m 2 and 100

MN/m 2 respectively.

a) Single Shear

AF

ττF

A

For brass material

A= 63

10501030 =0.0006 m 2

A= r2

π

Ar

π

0006.0r

r=13.8197×10 -3 mthe diameter of the bar (d)=27.639×10 -3 m

For steel material

A= 63

101001030 =0.0003 m 2

π0003.0

r =9.772×10 -3 m

the diameter of the bar (d)=19.544×10 -3 m

b) Double Shear

AF

2τ

τ2F

A

For brass material

A= 63

105021030 =0.0003 m 2

π

0003.0r =9.772×10 -3 m

the diameter of the bar (d)=19.544×10 -3 m For steel material

A= 63

1010021030 =0.00015 m 2

π00015.0

r =6.909×10 -3 m

the diameter of the bar (d)=13.819×10 -3 m


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Example 3: The 80 kg lamp is supported by two rods AB and BC as shown. If AB has adiameter of 10 mm and BC has a diameter of 8 mm , determine the average normal stress ineach rod.

FBC=0.625F BA ……………..(1)

FBC=1308-1.44337F BA ……….(2)

1308-1.44337F BA=0.625F BA

FBA=632.38 N

FBC=395.2375 N

BA

BA BA A

F σ = 23 )105(

38.632π

BA=8.051877×106 Pa

BA=8.051877 MPa

BC

BC BC A

F σ = 23 )104(

2375.395π

BC=7.863149×106 Pa

BC=7.863149 MPa

A

B

C

5

43

60

BC F BAF

N 8.78481.980

060cos54

0

BA BC

x

F F

F

08.78460sin53

0

BA BC

y

F F

F


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Example 4: Shafts and pulleys are usually fastened together by means of a key, as shown.Consider a pulley subjected to a turning moment T of 1 KN.m keyed by a 10 mm×10

mm×75 mm key to the shaft. The shaft is 50 mm in diameter. Determine the shear stress ona horizontal plane through the key.

0o

M

0025.0101 3 F F=40000 NF=40 KN

AF

τ F

FA is the shaded area

= 333

10751010

1040

=53.333×10 6 N/m 2

=53.333 MN/m 2

Example 5: Consider a steel bolt 10 mm in diameter and subjected to an axial tensile loadof 10 KN as shown. Determine the average shearing stress in the bolt head, assumingshearing on a cylindrical surface of the same diameter as the bolt.A= dtA= ×10×10 -3×8×10 -3=0.000251327 m 2

AF

τ

=000251327.0

1010 3

=39.7888 ×10 6 N/m 2

=39.7888 MN/m 2

mm50

mm10

mm10

T

F

o

KN 1

mm75

mm10

mm8

KN 10


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Example 6: The bar shown has a square cross section for which the depth and thickness are 40 mm . If an axial force of 800 N is applied along the centroidal axis of the bar's crosssectional area, determine the average normal stress and average shear stress acting on thematerial along a) section plane a-a and b) section plane b-b .

a) section plane a-a

AP

σ

= 33 10401040800

=500 KN/m 2

AF

τ

F=0=0

b) section plane b-b

d=60sin

40 =46.188 mm

AF 2σ

= 33 104010188.4660sin800 =375 KN/m 2

A

F 1τ

= 33 104010188.4660cos800 =216.50645 KN/m 2

a

a b

b

60 N 800

N 800 N 800

N 800 N 80060

1F

2F

d


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Example 7: Determine the total increase of length of a bar of constant cross sectionhanging vertically and subject to its own weight as the only load. The bar is initiallystraight.

: is the specific weight ( weight/unit volume )A: is the cross-sectional area

AE Aydy

d γ

δ L

d 0

δδ

= L

AE Aydy

0

γ

= L

ydy AE

A

0

γ

= L

y AE

A

0

2

21γ

= 22

L AE Aγ

= AE L AL2 .γ

W= AL

= AE

LW 2

.

dy

yAγ

dy

y

L


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Example 8: A member is made from a material that has a specific weight γ and modulus of elasticity E . If its formed into a cone having the dimensions shown, determine how far itsend is displaced due to gravity when its suspended in the vertical position.

r x =

L y

L y

r x

v= y x 23π

w(y)= v= y x 23π

= y L

yr 2

22

3γ

π

w(y)= 322

3 y

L

r γ π

From equilibrium P(y)=w(y)

P(y)= 322

3 y

L

r γ π

A(y)= x2= 222

y L

r π

d = E y

L

r

dy y L

r

E y Ady yP

22

2

32

2

3)()(

π

γ π

d = ydy E 3γ

L L

E ydy

d 00 3

γ δδ

E L6

2γ δ

r

y y

L x

)( yw

)( yP


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Example 9: A solid truncated conical bar of circular cross section tapers uniformly from adiameter d at its small end to D at the large end. The length of the bar is L. Determine theelongation due to an axial force P applied at each end as shown.

yd r 2

L

d D

x y 22

L xd D

y )22

(

L xd Dd

r )22

(2

A(x)= r2

A(x)=2

)22

(2 L

xd Dd π

d = E x A

Pdx)(

= E

L xd Dd

Pdx2

)22

(2

π

= L

E L xd Dd

Pdx

02

)22

(2

π

= L

L xd Dd

d D L

E

P

0

1

)22

(2)

22(

π

= L

L

xd Dd d D E

PL

0)22

(2

)22

(π=

2)

22(

222)

22(

d d D E

PLd Dd d D

E

PL

ππ

=)

44()

44(

22 d Dd E

PLdD D

E

PL

ππ= 22

114d Dd dD D E

PLπ

=dDE PL

π4

d D

L

P P

r y

dx x

P P

22d D


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Example 10: Determine the smallest dimensions of the circular shaft and circular end copif the load it is required to support is 150 KN . The allowable tensile stress, bearing stress,and shear stress is ( t ) allow =175 MPa , is ( b ) allow =275 MPa , and τ allow =115 MPa .

( b)allow =b

b

AF

275×10 6=b A

310150

Ab=0.0005454 m2

Ab= 224d

π

d2= ππ0005454.044 b A

d2=0.026353 m=26.353 mm( t)allow =

AP

175×10 6= A

310150

A=0.0008571 m 2

A= ])1030([4

2321d

π =0.0008571

d1=0.04462 m=44.62 mm

allow = AF

115×10 6= A

310150

A=0.0013043 m 2

1. A=t d0.0013043= t× ×30×10 -3

t=0.013839 m=13.839 mm2. A=t d2

0.0013043= t× ×26.353×10-3

t=0.01575 m=15.75 mm

KN P 150

2d

1d

mm30

t


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Statically Indeterminate Members:

If the values of all the external forces which act on a body can bedetermined by the equations of static equilibrium alone, then the force system is staticallydeterminate.

In many cases the forces acting on a body cannot be determined by the equations of static alone because there are more unknown forces than the equations of equilibrium. Insuch case the force system is said to be statically indeterminate.

P

3 R2 R

1 R

1 R

2 R 3 R

1P2P

P

3 R2 R

1 R 4 R

3 R2 R

1 R

4 R

P

1 M


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Example 11: A square bar 50 mm on a side is held rigidly between the walls and loadedby an axial force of 150 KN as shown. Determine the reactions at the end of the bar and theextension of the right portion. Take E=200 GPa .

R1+R

2=150×10 3 ………………(1)

1= 2

933

32

933

31

102001050105010150

102001050105010100 R R

0.1R 1=0.15R 2

R1=1.5R 2 ………………….. (2)

From equations (1) and (2)

1.5R 2 + R 2=150×103

R2 =60000 NR1=90000 N

2= 9333

933

32

10200105010501015060000

102001050105010150 R

2=0.000018 m2=0.018 mm

mm100 mm150

KN 150KN 150

2 R1 R


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Example 12: A steel bar of cross section 500 mm 2 is acted upon by the forces shown.Determine the total elongation of the bar. For steel, E=200 GPa .

15 KN 10 KN

For portion AB

1= AE PL = 96

33

1020010500105001050 =0.00025 m=0.25 mm

For portion BC

2= AE PL = 96

3

102001050011035 =0.00035 m=0.35 mm

For portion CD

3= AE PL

= 963

10200105005.11045

=0.000675 m=0.675 mm

T= 1+ 2+ 3T=0.25+0.35+0.675=1.275 mm

KN 50 KN 45

mm500 m1 m5.1

A B C D

A BKN 50 KN 50 KN 50

KN 15KN 35

C B

KN 45KN 45 C D


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Example 13: Member AC shown is subjected to a vertical force of 3 KN . Determine theposition x of this force so that the average compressive stress at C is equal to the averagetensile stress in the tie rod AB . The rod has a cross-sectional area of 400 mm 2 and thecontact area at C is 650 mm 2.

0 yF

FAB+F C-3000=0FAB+F C=3000 ………………………(1)

AB= C

C

C

AB

AB

AF AF

66 1065010400 C AB F F

FAB=0.6153 F C …………………….(2)


FC=1857.24 NFAB=1142.759 N

0 A M

FC×200×10-3-3000×x=0

30002.024.1857

x =0.123816 m=123.816 mm

mm200

KN 3 x

A

B

C

mm200

KN 3 x

A

C

ABF

C F


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Example 14: The bar AB is considered to be absolutely rigid and is horizontal before theload of 200 KN is applied. The connection at A is a pin, and AB is supported by the steelrod EB and the copper rod CD . The length of CD is 1m , of EB is 2 m . The cross sectionalarea of CD is 500 mm 2, the area of EB is 250 mm 2. Determine the stress in each of the

vertical rods and the elongation of the steel rod. Neglect the weight of AB . For copper E=120 GPa , for steel E=200 GPa .

FCo×1+F s×2-200×103×1.5=0

FCo=300×103-2 F s …………..(1)

12Cos δδ

s=2 Co

CoCo

Co

ss

s

E A LF

E A LF

2

9696 10120105001

21020010250

2 Cos F F

FCo=1.2 F s …………………….(2)


Fs=93750 NFCo=112500 N

s

ss A

F σ = 610250

93750 =375000000 Pa

s=375 MPa

Co

CoCo A

F σ = 610500

112500 =225000000 Pa

Co=225 MPa

E

Lss

σδ = 9

6

10200210375 =0.00375 m=3.75 mm

A B

C

D

E

m1 mm500 mm500

KN 200

KN 200

sF CoF

y A

x A

A B

C

D

E

Coδsδ

0 A M


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Thermal Stresses:

A change in temperature can cause a material to change its dimensions. If thetemperature increases, generally a material expands, whereas if the temperature decreasesthe material will contract.

The deformation of a member having a length L can be calculated using theformula:

T= × T×L

T= AE FL = × T×L

T=E× × T

: Linear coefficient of thermal expansion. The units measure strain per degree of temperature. They are (1/ºF) in the foot-pound-second system and (1/ºC) or (1/ºK) in SIsystem.

T: Change in temperature of the member.L: The original length of the member.

T: The change in length of the member.

Example 15: The A-36 steel bar shown is constrained to just fit between two fixedsupports when T 1=60º F . If the temperature is raised to T 2=120º F determine the averagenormal thermal stress developed in the bar. For steel =6.6×10 -6 1/ºF , E=29×10 3 Ksi .

0 yF FA-FB=F

T- F=0T= × T×LT=E× × T

=29×10 3 ×6.6×10 -6 ×(120-60)=11.484 Ksi

in5.0

in5.0

in20

AF

BF

A

B

T δ F δ


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Example 16: A 2014-T6 aluminum tube having a cross sectional area of 600 mm 2 is usedas a sleeve for an A-36 steel bolt having a cross sectional area of 400 mm 2. When thetemperature is T 1=15º C , the nut hold the assembly in a snug position such that the axialforce in the bolt is negligible. If the temperature increases T 2=80º C , determine the average

normal stress in the bolt and sleeve. For aluminum =23×10-6

1/ºC , E=73.1 GPa, for steel =12×10 -6 1/ºC , E=200 GPa.

0 yF Fsl-Fb=0

Fsl=F b=F

F slT slF bT b )()()()( δδδδδ

[ × T×L+ AE FL ]b=[ × T×L-

AE FL ]sl

12×10 -6×0.15×(80-15)+ 961020010400

15.0F = 23×10 -6×0.15×(80-15)- 96101.7310600

15.0F

0.0052949×10 -6F=0.00010725

F=20255 N

b=b A

F = 61040020255

b=50.637655 MPa

sl=sl A

F = 61060020255

sl=33.758436 MPa

mm150

T sl )(δ

F sl )(δ

T b )(δ

F b )(δδ

Position Initial

PositionFinal

bF slF


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Example 17: The rigid bar shown is fixed to the top of the three posts made of A-36 steeland 2014-T6 aluminum. The posts each have a length of 250 mm when no load is appliedto the bar, and the temperature is T 1=20ºC . Determine the force supported by each posts if the bar is subjected to a uniform distributed load of 150 KN/m and the temperature is

raised to T 2=80ºC . For steel =12×10-6

1/ºC , E=200 GPa , for aluminum =23×10-6

1/ºC , E=73.1 GPa .

Steel Aluminum Steel

0 yF

2F st+F al=90000 …………….(1)

=( st)T-( st)F=( al)T-( al)F

[ × T×L- AE

LF st ]st=[ × T×L- AE

LF al ]al

12×10 -6×0.25×(80-20)- =23×10 -6×0.25×(80-20)-

1.20956×10 -9Fal-0.994718×10-9Fst=0.000165 ………………..(2)


Fst=-16444.7 N

Fal=122888.8 N

mm300 mm300

mKN / 150

mm250mm40

mm60

mm40

st F st F alF

KN 906.0150

Positionnitial

Positioninal

δ

T st )(δF st )(δ

T al )(δF al )(δ

923 10200)1040(4

25.0π

st F

923 101.73)1060(4

25.0π

alF


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Example 18: The rigid bar AD is pinned at A and attached to the bars BC and ED asshown. The entire system is initially stress-free and the weights of all bars are negligible.The temperature of bar BC is lowered 25ºK and that of the bar ED is raised 25ºK .Neglecting any possibility of lateral buckling, find the normal stresses in bars BC and ED .

For BC , which is brass, assume E=90 GPa , =20×10-6

1/ºK and for ED , which is steel,take =12×10 -6 1/ºK , E=200 GPa . The cross-sectional area of BC is 500 mm 2, of ED is

250 mm 2.

0 A M

Pst×600×10-3-Pbr×250×10

-3=0

Pst=0.41666 P br ………..(1)

600250st br δδ

600250st st

st

br br

br

E A

LPT L

E A

LPT L

αα

8.333×10 -12 Pst+26.666×10 -12 Pbr=475×10 -9 …………..( 2)

From equations (1) and (2)Pbr=15760.5 N , P st=6566.77 N

br= MPa521.3110500

5.157606

Pst= MPa267.261025077.6566

6

mm300

st P

br P y A

x A

A B

C

D

E

mm250 mm350

mm250

br δ T st )(δ

T br )(δF br )(δ

F st )(δ st δ

6001020010250

1025025102501012

250109010500

1030025103001020 96

336

96

336 st br PP


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Torsion:

Torque is a moment that tends to twist a member about its longitudinal axis.When the torque is applied, the circles and longitudinal grid lines originally marked on theshaft tend to distort into the pattern shown below.

Before deformation After deformation

Twisting causes the circles to remain circles and each longitudinal grid linedeforms into a helix that intersects the circles at equal angles. Also, the cross sections at

the ends of the shaft remain flat that is, they do not warp or bulge in or out and radial lineson these ends remain straight during the deformation.The Torsion Formula:

Consider a uniform circular shaft is subjected to a torque it can be shownthat every section of the shaft is subjected to a state of pure shear.

J T ρ

τ

T

T

T

T

τmaxτ

r ρ


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: The torsional shearing stress.T: The resultant internal torque acting on the cross section.

: The distance from the centre (radial position).J: The polar moment of inertia of the cross sectional area.

J Tr

maxτ

max : The maximum shear stress in the shaft, which occurs at the outer surface.r: The outer radius of the shaft.

4

2

r J π

4

32 D J

π

for a hollow shaft

J Tr

τ

J Tr o

maxτ

)(32

)(2

4444ioio D Dr r J

ππ

Angle of Twist ( θ):

If a shaft of length L is subjected to a constant twisting moment along itslength, then the angle of twist through which one end of the shaft will twist relative to theother is:

GJ TL

θ

G: The shear modulus of elasticity ormodulus of rigidity.

: Angle of twist, measured in rad

or ir

r

T

T

r

θ

L

A


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If the shaft is subjected to several different torques or the cross sectional area orshear modulus changes from one region to the next. The angle of twist of one end of theshaft with respect to the other is then found from:

GJ TLθ

In order to apply the above equation, we must develop a sign convention for theinternal torque and the angle of twist of one end of the shaft with respect to the other end.To do this, we will use the right hand rule, whereby both the torque and angle of twist willbe positive, provided the thumb is directed outward from the shaft when the fingers curl togive the tendency for rotation.

m N .10

m N .60m N .150

AB L

BC L

CD L

A

B

C

D

m N .70

m N .80

A

B

AB L

B

BC LC

m N .80

m N .80

m N .80

m N .150m N .70

C

D

AB L

m N .60

m N .10


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GJ

L

GJ

L

GJ

L CD BC AB D A

107080 / θ

Power Transmission (P):Shaft and tubes having circular cross sections are often used to transmit

power developed by a machine.

dt d

T P θ , ωθ

dt d

: The shaft's angular velocity (rad/s).

ωT PIn SI units power is expressed in (watts) when torque is measured in (N.m) and

in (rad/s).1 W=1 N.m/s

In the foot-pound-second or FPS system the units of power are (ft.lb/s); howeverhorsepower (hp) is often used in engineering practice where:

1 hp=550 ft.lb/sFor machinery the frequency of a shaft's rotation f is often reported. This is a

measured of the number of revolutions or cycles the shaft per second and is expressed inhertz (1 Hz=1 cycle/s), 1 cycle=2 rad, then =2 f

P=2 fT

Example 19: If a twisting moment of 1 KN.m is impressed upon a 50 mm diameter shaft,what is the maximum shearing stress developed? Also what is the angle of twist in a 1 mlength of the shaft? The material is steel, for which G=85 GPa .

J Tr

maxτ

6434 106135.0)1050(3232

ππ D J m4

6

33

max 106135.01025101

τ

max =40.74979 MPa


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GJ TL

θ

69

3

106135.010851101θ

=0.01917 rad.

Example 20:The pipe shown has an inner diameter of

80 mmand an outer diameter of

100 mm . If its end is tightened against the support at A using a torque wrench at B, determinethe shear stress developed in the material at the inner and outer walls along the centralportion of the pipe when the 80 N forces are applied to the wrench.

T=80×200×10 -3+80×300×10 -3

T=40 N.m

J Tr

τ

)(32

44io D D J π

4343 )1080()10100(32 π

J

J=5.7962×10 -6 m4

Inner walls r i=40 mm

J

Tr iτ = 63

107962.5104040

=0.276042 MPa

Outer walls r o=50 mm

J

Tr oτ = 63

107962.5105040

=0.345053 MPa


29/144

Example 21: The gear motor can developed 0.1 hp when it turns at 80 rev/min . If theallowable shear stress for the shaft is τ allow =4 ksi , determine the smallest diameter of theshaft that can be used.

J Tr allow τ

4

2r J

π

4

2r

Tr allow π

τ

3

2r T

allow πτ

32

allow

T r

πτ

P=T.ωP

T

P=0.1×550=55 lb/s=80×2 /60=8.377 rad/s

377.855

T =6.5655 lb.ft

T=6.5655×12=78.786 lb.in

33104

786.782π

r =0.2323 in

d=0.4646 in


30/144

Example 22: The assembly conof a tube using a rigid disk atand in the tube. The tube has an

The rodT=50 N.mr=7.5×10 -3 m

4

2r J

π

43 )1075(2

π J

J=4.97009×10 -9 m4

J Tr

r τ

9

3

1097009.4105.750

r τ =7

The tubeT=80 N.mr=15×10 -3 m

)(2

44io r r J

π

43 1()1015(2

π J

J

Tr ot τ =

109495.46101580

sists of a solid 15 mm diameter rod c. Determine the absolute maximum souter diameter of 30 mm and a wall t

5.4512 MPa

43 )10 =46.9495×10 -9 m4

9

3

=25.559 MPa

12

mm15

nnected to the inside hear stress in the rod

ickness of 3 mm .

mm


31/144

Example 23: The tapered shaftG. Determine the angle of twist

L xd d

y L

d d x y

221212

d=d 1+2y=d 1+ L x

d d )( 12 4

1214 ])([

3232)(

L x

d d d d x J ππ

L L

d d d G

dxT x J G

dxT

0121

0 )([32

)( πφ

12112 )([

1)(3

32

L

xd d d d d G

TLπ

φ

31

32

32

31

12 .)(332

d d d d

d d GTL

πφ

31

32

2221

21

.332

d d d d d d

GTL

πφ

1d

shown below is made of a material haof its end B when subjected to the tor

L

L x

d d d

dxGT

0 4121

4 ])([

32

] π

31

3212

0

3

11)(3

32

] d d d d GTL

L

π

d

y

x

ing a shear modulus ue.

2d

22

12 d d


32/144

Example 24: The gears attachshown. If the shear modulus of

mm , determine the displacemen

Segment AC

150-T AC=0TAC=150 N.m

Segment CD

150-280+T CD=0TCD=130 N.m

Segment DE

-130-40+T DE=0TDE=170 N.m

GJ TL

θ

= TLGJ 1

=107(

21080

1

39 π

=-0.21211 rad

d to the fixed end steel shaft are subelasticity is G=80 GPa and the shaftt of the tooth P on gear A.

5.01703.01304.01504

jected to the torques has a diameter of 14


33/144

Example 25: A steel shaft A B C between gears A and B and abetween gears B and C . Both bN . m , T 2 = 5 4 0 N .m , and T 3 = 3 0modulus of elasticity for the shad if the allowable shear stress idiameter d if the angle of twist

TAB=240 N.mTBC=300 N.ma)

1 . Fo r s o l i d b a r A B

J= 432

d π

4

32

2 / 2 /

d

d T J d T AB AB

πτ

4

6

32

2 / 2401080

d

d π

63 108024016

πd

d=0.02481 md=24.81 mm

2 . Fo r h o l l o w b a r B C

J= ])25.1[(32

][32

4444 d d d d io ππ

)25.1[(32

25.12 / 25.1

4d

T

J

d T BC BC π

τ

])25.1[(32

2 / 25.13001080

44

6

d d

d π

d=0.0255 md=25.5 mm Answer

b)1 . Fo r s o l i d b a r A B

GJ LT ABθ

connecting three gears consists of a sollow bar of outside diameter 1 . 2 5 d

ars have length 0 . 6 m . The gears tra0 N . m acting in the directions shown ift is 8 0 G Pa . a ) what is the minimu

the shaft is 8 0 M P a ?. b ) what is theetween any two gears is limited to 4

]

2 /

4d

4414.1108025.130016

63

πd

olid bar of diameter d nd inside diameter d smit torques T 1 = 2 4 0

the figure. The shear permissible diameter

minimum permissible ?.


34/144

49

321080

6.0240180

4d

ππ

d=0.02263 md=22.63 mm

2 . Fo r h o l l o w b a r B C

GJ

LT BC θ

])25.1[(32

1080

6.0300180

4449 d d π

π

d=0.02184 md=21.84 mm

d=0.02263 md=22.63 mm A n s w e r

Example 26: The shaft is subjected to a distributed torque along its length of t=10x 2

N.m/m , where x is in meters. If the maximum stress in the shaft is t o remain constant at 80 MPa , determine the required variation of the radius r of the shaft for 30 x m

tdxT

dx x 210 33

10 x

J Tr

maxτ

3

36

4

3

6

320

1080

2

310

1080r x

r

r x

ππ


35/144

x x

r 002982.010803

203

6

3

πm

r=2.982 x mm

Example 26: The shaft has a radius 50 mm and is subjected to a torque per unit length of 100 N.m which is distributed uniformly over the shafts entire length 2 m . If it is fixed at itsfar end A, determine the angle of twist of end B. The shear modulus is 73.1 GPa .

T(x)=100x

2

0 439

2

0 )1050(2

101.73

100)(π

θ xdx

GJ dx xT

2

0

2

439 2)1050(101.73200 xπ

439 )1050(101.73400π

410786.2 θ rad

=0.01596 o

m2

mm N / .100


36/144

Statically Indeterminate:

TA-T+T B=0TA +T B=T

A/B =0

0GJ

LT

GJ

LT BC B AC A

If T 1>T 2

-TA-T2+T 1-TB=0

)( 221GJ

LT T

GJ

LT A B

T2

T2

03GJ

LT A

T1

L1

T1

T2

L2

L3

T2


37/144

Example 27: The solid steel shtwo torques, determine the reac

-TB+800-500-T A=0

TB+T A=300 ……

)500(2.0 GJ

T

GJ

T A B

-0.2T B+1.5T A+750+

1.8T A-0.2T B=-750

TA=-345 N.m

TB=645 N.m

ft shown has a diameter of 20 mm . Iions at the fixed supports A and B.

………………………….. ..(1)

03.05.1

GJ

T A

0.3T A=0

………………………...…(2)

it is subjected to the


38/144

Example 28: The shaft shown below is made from steel tube, which is bonded to a brasscore. If a torque of T=250 lb.ft is applied at its end, plot the shear stress distribution along aradial line of its cross sectional area. Take G st11.4×10

3 ksi, G br=5.2×103 ksi.

Tst+T br=250×12=3000 lb.in………….(1)θ=θ st=θ br

])5.0()1[(2

102.5])5.0()1[(2

104.11 446446 ππ

LT LT br st

Tst=32.88T br……………………….(2)From (1) and (2)Tst=2911 lb.in=242.6 lb.ftTbr=88.5 lb.in=7.38 lb.ft

psibr 451)5.0(

2

5.05.88)(

4max πτ

psist 1977])5.0()1[(

2

12911)(

44max πτ

psist 988

])5.0()1[(2

5.02911)(

44

min πτ

rad G

366 100867.0104.11

988102.5

451 τγ


39/144

Torsion of Solid Noncircul

Shape of cross section

Square

Equilateraltriangle

Ellipse

Example 28: The 2014-T6 aluhas a 2 in by 2 in square crossdetermine the reactions at theG=3.9×103 ksi .

TA-40-20+T B=0

TA+T B=60 …

A/B =0

19.3212)20(1.7

109.32122121.7

464 B A T T

a

a

a

a

aabb

GaTL

B A 4 / 1.7

θ

ar Shafts:

max

inum strut is fixed between the twosection and it is subjected to the torsixed supports. Also what is the angl

….(1)

0109.32

122121.7122646

BT

381.4a

T

3

20a

T

2

2abT

π

alls at A and B. If it ional loading shown, of twist at C . Take

GaTL

4

1.7

GaTL

4

46

GbaTLba

33

22 )(π


40/144

TA-2T B=-20 ………………(2)


TB=26.666 lb.ftTA=33.333 lb.ft

Ga

LT AC 4

1.7θ

64 109.3212212333.331.7

C θ

C=0.001092 rad

C=0.06258º

Thin walled tubes having closed cross sections:Shear flow(q): is the product of the tube's thickness and the average shear stress. This value

is constant at all points along the tube's cross section. As a result, the largest average shearstress on the cross section occurs where the tube's thickness is smallest.

The forces acting on the two faces are dF A= A(tAdx) , dF B= B(tBdx), theseforces are equal for equilibrium, so that:

AtA= BtBq= avgt

AT

BT

ft lb.40

ft lb.20C

D


41/144


42/144

Example 29: The tube is madeshown below. If its subjected ttube at points A and B. Also,Take G=38 GPa .

60-25-T=0T=35 N.m

Am=(40×10-3-5×10 -

=0.001995 m 2

A= mtAT

2 = 105233

A=1.7543859 MPa

B=mtA

T 2

=1032

33

B=2.9239766 MPa

t ds

G ATL

m24θ

=

=0.0062912 rad

3)001995.0(45.060

2

of C86100 bronze and has a rectangthe two torques, determine the avera

hat is the angle of twist of end C ? T

)(60×10 -3-3×10 -3)

001995.05

001995.05

335

2[108.3)001995.0(4

5.135

]1051057

21031035

2[108.

92

3

3

3

3

9

ular cross section as ge shear stress in the

e tube is fixed at E .

]1051057

21010

3

3

3

3


43/144

Example 30: A thin tube is macross section that is triangular ait can be subjected, if the allow τ to twist no more than θ =2×10 -3

A=21 (200×10 -3)× (2

t=0.005 m

allow =mtA

T 2

=152

T=15.588 KN.m

t ds

G ATL

m24

θ

2×10 -3=)01732.0(4

T

T=500 N.m

e from three 5 mm thick A-36 steel pls shown below. Determine the maximable shear stress is τ allow =90 MPa and rad . Take G=75 GPa .

00×10 -3 sin60)=0.01732 m 2

01732.03T =90×10 6

]105

102003[

10753

3

3

92

ates such that it has a m torque T to which

τ the tube is restricted


44/144

Thin Walled Cylinder, Th

Cylindrical or sphericor tanks. When under pressurloading from all directions. Into wall thickness ratio of 10 or

1. Cylindrical Vessels:Consider the cylindric

shown below. A pressure pwhich is assumed to have n

The stresses set up in the wa. Circumferential o

2[ 1(tdy)]-p(2rdy)=0

t pr

1σ

b. Longitudinal or a

2(2 rt)-p( r2)=0

t pr 22

σ

c. Circumferential o

)(1

211 νσσε E d. Longitudinal stra

)(1

122 νσσε E

n Walled Pressure Vessels:

l vessels are commonly used in indus, the material of which they are maeneral "thin wall" refers to a vessel hore (r/t 10)

al vessel having a wall thickness tis developed within the vessel by a coegligible weight.

alls are: r hoop stress

ial stress

r hoop strain

in

ry to serve as boilers de is subjected to a

ving an inner radius

nd inner radius r as ntaining gas or fluid,


45/144

e. Change in lengthThe change in length of the cylinder may be determined from the

longitudinal strain.Change in length=longitudinal strain×original length

L=ε 2L= )(1 12 νσσ E L

L= )21(2

νtE

pr L

f. Change in diameterThe change in diameter may be found from the circumferential change.

Change in diameter=diametral strain×original diameterDiametral strain=circumferential strain

d= ε1d= )(1

21 νσσ E dd= )2(

2 ν

tE pr d

g. Change in internal volumeVolumetric strain=longitudinal strain+2diametral strain

εv= ε2+2 ε1= )(1

12 νσσ E +2 )(1 21 νσσ E

diametral strain

εv= )22(1 2112 νσσ νσσ E longitudinal strain

= )22

(1

t pr

t pr

t pr

t pr

E ν ν diametral strain

εv= )45(2

νtE

pr

change in internal volume=volumetric strain×original volumev= εvv

v= )45(2 νtE pr

v


46/144

2. Spherical Vessels:Because of the symm

pressure will be two mutuavalue and a radial stress.

1(2 rt)-p( r2)=0

t pr 21

σ

2= t pr 21 σ

Change in internal volume

change in internal volume=volumetric strain=3hoop str

εv= ε1=3 )(1

21 νσσ E

v= εvv

v = )1(23

νtE pr v

etry of the sphere the stresses setlly perpendicular hoop or circumfer

olumetric strain×original volume in

= )1(3 1 νσ E

= )1(23

νtE pr

p owing to internal ntial stress of equal


47/144

Cylindrical Vessels with H

r=d/2

a) For the cylindrical portio

ct Pr

1 σ hoop s

ct 2Pr

2 σ longitu

)(1

211 νσσε E =

ε1= )2(2

ν E t

pr c

b) For the spherical ends

st 2Pr

1 σ

)(1

211 νσσε

E =

ε1= )1(2

ν E t

pr

s

Thus equating the tw junction.

)1(2

ν E t

pr

s

= 2(2 E t

pr

c

υυ

21

c

s

t

t

emispherical Ends:

ress

dinal stress

)2PrPr

(1

cc t t E ν

hoop strain

hoop stress

)1(1 νσ

E

hoop strain

strains in order that there shall be

) ν

no distortion of the


48/144

Example 31: A thin cylinder 75 mm internal diameter, 250 mm long with walls 2.5 mmthick is subjected to an internal pressure of 7 MN/m 2. Determine the change in internaldiameter and the change in length. If in addition to the internal pressure, the cylinder issubjected to a torque of 200 N.m find the magnitude and nature of the stresses set up in the

cylinder. E=200 GN/m 2

, υ=0.3 .

d= )2(2

νtE

pr d

d= 393

36

1075]3.02[10200105.22

102

75107

d=33.468×10 -6 m=33.468 μm

L= )21(2

νtE

pr L

L= 393

36

10250]3.021[10200105.22

102

75107

L=26.25×10 -6 m=26.25 μm

t pr 1σ = 3

36

105.2

102

75107

1=105×106 N/m 2=105 MN/m 2

t pr 22

σ = 3

36

105.22

102

75107

2=52.5×106 N/m 2=52.5 MN/m 2

J Tr

τ =][

244

io r r

Tr π

=])105.37()1040[(

2

10402004343

3

π

=8.743862 MN/m 2


49/144

Example 32: A cylinder has an internal diameter of 230 mm , has walls 5 mm thick and is 1 m long. It is found to change in internal volume by 12×10 -6 m 3 when filled with a liquid ata pressure p. If E=200 GN/m 2 and υ=0.25 , and assuming rigid end plates, determine a) thevalues of hoop and longitudinal stresses b) the necessary change in pressure p to produce a

further increase in internal volume of 15% .

a) v= )45(2

νtE

pr v

12×10 -6= 1)102

230(]25.045[

102001052

102

23023

93

3

π p

p=1.255763 MN/m 2

t pr

1σ = 3

36

105102

230

10255763.1

1=28.882549 MN/m2

t pr 22

σ = 3

36

1052

102

23010255763.1

2=14.4412745 MN/m2

b) v=1.15×12×10 -6=13.8×10 -6 m3

v= )45(2

νtE

pr v

13.8×10 -6= 1)102

230(]25.045[

102001052

102

23023

93

3

π p

p=1.444128 MN/m 2

Necessary increase=1.444128-1.255763=0.188365 MN/m 2


50/144

Vessels Subjected to Fluid Pressure:

If a fluid is used as the pressurization medium the fluid itself will change involume as pressure is increased and this must be taken into account when calculating theamount of fluid which must be pumped into the cylinder in order to raise the pressure by aspecific amount.

The bulk modulus of a fluid is defined as:

bulk modulus k=strainVolumetricstressVolumetric

volumetric stress=pressure p

volumetric strain=volumeoriginalvolumeinchange =

vvδ

k=vv

pδ = v

pvδ

change in volume of fluid under pressure=k

pv

extra fluid required to raise cylinder pressure by p

= )45(2

νtE

pr v+k

pv

extra fluid required to raise sphere pressure by p

= )1(23

νtE pr v+

k pv


51/144


52/144

Shear and Moment Diagram:

Beams are long straight members that carry loads perpendicular to theirlongitudinal axis. They are classified according to the way they are supported, e.g. simplysupported, cantilevered, or overhanging.

Simply supported beam overhanging beam

Cantilevered beam

Types of Loading:

Loads commonly applied to a beam may consist of concentrated forces(applied ata point), uniformly distributed loads, in which case the magnitude is expressed as a certainnumber of newtons per meter of length of the beam, or uniformly varying loads. A beammay also be loaded by an applied couple.

(point load)(concentrated force)

P P

P

N 100


53/144

V V

M M V V

M

Uniformly distributed load Uniformly varying load

Shearing force and bending moment diagrams show the variation of thesequantities along the length of a beam for any fixed loading condition. At every section in abeam carrying transverse loads there will be resultant forces on either side of the sectionwhich, for equilibrium, must be equal and opposite.

Shearing force at the section is defined as the algebraic sum of the forces taken onone side of the section. The bending moment is defined as the algebraic sum of themoments of the forces about the section, taken on either side of the section.Sign Convention:

Forces upwards to the left of a section or downwards to the right of a section arepositive. Clockwise moments to the left and counter clockwise to the right are positive.

- -

Procedure of Analysis:The shear and moment diagrams for a beam can be constructed using the

following procedure:-1. Determine all the reactive forces and couple moments acting on the beam, and

resolve all the forces into components acting perpendicular and parallel to the beam'saxis.

2. Specify separate coordinates x having an origin at the beam's left end extending toregions of the beam between concentrated forces and/or couple moments, or wherethere is no discontinuity of distributed loading.

3. Section the beam perpendicular to its axis at each distance x, and draw the free bodydiagram of one of the segments. Be sure V and M are shown acting in their positivesense, in accordance with the sign convention given as above.

4. The shear is obtained by summing forces perpendicular to the beam's axis.5. The moment is obtained by summing moment about the sectioned end of the

segment.

m N / 10 m N / 10

m4 m5

m N / 5

M


54/144

6. Plot the shear diagram(V versus x) and the moment diagram(M versus x). If numerical values of the functions describing V and M are positive, the values areplotted above the x-axis, whereas negative values are plotted below the axis.

Example 33: Draw the shear and moment diagrams for the beam shown below.

0 xF Ax=0

0C M P×L/2-A y×L=0Ay=P/2

0 yF Cy+A y-P=0Cy=P/2

Segment AB 0 yF

2P -V=0

V=2

P

0 M

M-2P ×x=0

M=2P x

Segment BC

0 yF

A

P

B C

2 / L 2 / L

x A

P

B

yC y A

2P

x

V M

2P

x

V M

P


55/144

2P -P-V=0

V=-2P

0 M M-

2P ×x+P(x-

2 L )=0

M=2P (L-x)

S.F. diagram

B.M. diagram

P

B

2P

2P

2P

2P

4maxPL

M


56/144


0 xF

Fx=0 0F M -A y×12+10×10-20×8+20×6

+30×2=0Ay=10 KN

0 yF 10-10+20-20-30+F y=0Fy=30 KN

Segment AB 20 x

0 yF 10-V=0V=10 KN

0 M M-10×x=0M=10x

Segment BC 42 x

0 yF 10-10-V=0V=0

0 M M-10x+10(x-2)=0M=20 KN.m

Segment CD 64 x

0 yF

A

KN 10

C

m4 m8KN 20

KN 20 KN 30

B D E F

m4 m2m2

A

KN 10

C

m4 m8KN 20

KN 20 KN 30

B D E F

m4 m2m2

y A yF

xF

KN 10

x

V M

KN 10


57/144

10-10+20-V=0V=20 KN

0 M M-10x+10(x-2)-20(x-4)=0

M=20(x-3)

Segment DE 106 x

0 yF 10-10+20-20-V=0V=0

0 M M-10x+10(x-2)-20(x-4)+20(x-6)=0M=60 KN.m

Segment EF 1210 x

0 yF 10-10+20-20-30-V=0V=-30 KN

0 M

M-10x+10(x-2)-20(x-4)+20(x-6)+30(x-10)=0M=30(12-x)

KN 10 x

V M

KN 10

KN 20

KN 10 x

V M

KN 10

KN 20

KN 20

KN 10 x

V M

KN 10

KN 20

KN 20 KN 30


58/144

S.F Diagram

B.M. Diagram

A

KN 10

C

m4 m8KN 20

KN 20 KN 30

B D E F

m4 m2m2

xF

KN 10 KN 30

KN 10

KN 20

KN 30

mKN .20

mKN .60


59/144


0 x

F

Ax=0 0 B M

wL2

L -A yL=0

Ay=2

wL

0 yF

2

wL +B y-wL=0

By=2

wL

0 yF

2wL -wx-V=0

V=-w(x-2

L )

0 M

M- 2wL x+wx( 2

x )=0

M=2w (xL-x 2)

A B

L

w

A B

L

wL

y A y B

x A

2wL

x

V M

wx2 / x

w

Maximum moment occur when 0dx

dM

2

02

0)2(2

L x

x L

x Lw

dxdM

Location of maximum moment


60/144

S.F Diagram

B.M Diagram

A B

L

w

2wL

2wL

2wL

2 / L

2 / L

8

2

max

wL M


61/144


0 x

F

Ax=0 0 yF

Ay=2 Lwo

0 A M

MA-2 Lwo L

32 =0

MA= 3

2 Lwo

0 yF

2 Lwo -

L xwo

2

2

-V=0

V= 2ow

(L- L x 2

)

Vmax=2 Lwo

0 M

M+3

2 Lwo -2 Lwo x+

L

xwo2

2

x31 =0

A B

L

ow

L32 L

31

2

Lwo

x A

y A

A M

L xwo

2

2

3

2 Lwo

2

Lwo x

L x

ww o

3 / x

V M

Maximum shear force occur at 0dxdV

0 L

xw

dxdV o

x=0


62/144

M= L

wo6

(3L 2x-x 3-2L 3)

Mmax=

S.F Diagram

L

ow

3

2 Lwo

2

Lwo

2 Lwo

3

2 Lwo

3

2 Lwo


63/144

Example 37: The horizontal beam AD is loaded by a uniform distributed load of 5 KN per meter of length and is also subjected to the concentrated force of 10 KN applied as shownbelow. Determine the shearing force and bending moment diagrams.

0 xF Ax=0

0 A M Cy×3-30×2=0Cy=20 KN

0 yF Ay+20-30=0Ay=10 KN

Segment AB 20 x

0 yF 10-5x-V=0V=10-5x

0 M

M-10x+5x2 x =0

M=5x(2-2 x )

Segment BC 32 x

0 yF

10-5x-10-V=0V=-5x 0 M

M-10x+5x2 x +10(x-2)=0

M=20-25 x2

Segment CD 43 x

m2 m1 m1

mKN / 5

KN 10

A B

C

D

KN 10

x A

y A yC

KN 20

KN 10 x

V M

x52 / x

mKN / 5

x

V M

x5

mKN / 5

KN 10

2 / x

KN 10


64/144

0 yF 10-5x-10+20-V=0V=20-5x

0 M

M-10x+5x2 x +10(x-2)-20(x-3)=0

M=-40+20x-25 x2

S.F Diagram

B.M Diagram

`

x

V M

x5

mKN / 5

KN 10

2 / x

KN 10

KN 20

m2 m1 m1

mKN / 5

KN 10

A B

C

D

KN 10 KN 20

KN 10

KN 10

KN 15

KN 5

mKN .10

mKN .5.2


65/144

Example 38: A beam ABC is simply supported at A and B and has an overhang BC . Thebeam is loaded by two forces P and a clockwise couple of moment Pa that act through thearrangement shown. Draw the shear force and bending moment diagrams for beam ABC .

0 D M -Pa+R C(2a)-Pa=0RC=P

0 yF RD+P-P-P=0

RD=P 0 A M RB(2a)-Pa-P(3a)=0RB=2P

0 yF RA+2P-P-P=0RA=0

Segment AD a x 0 0 yF

V=0 0 M

M=0

Segment DB a xa 2

a a a a

P P

Pa

A B

C

D

P P

Pa

D R C RP

A R B R

P

V M

x

V

x

P


66/144

0 yF -V-P=0V=-P

0 M M+P(x-a)=0M=P(a-x)

Segment DB a xa 32

0 yF 2P-P-V=0V=P

0 M

M+P(x-a)-2P(x-2a)=0M=P(x-3a)

S.F. Diagram

B.M. Diagram

V

x

P

P2

P

P2

P

P

P

Pa


67/144

Graphical Method for Constructing Shear and Moment Diagram:

)( xwdxdV

Slope of shear diagram at each point=-distributed load intensity at each point.

V dx

dM

Slope of moment diagram at each point=shear at each point.When the force acts downward on the beam, V is negative so the shear will jumpdownward. Likewise, if the force acts upward, the jump will be upward.If moment M o is applied clockwise on the beam, M is positive so the momentdiagram will jump upward. Likewise, when M o acts counterclockwise, the jump willbe downward.

)( xw


68/144


0 xF Ax=0

0 yF Ay-P=0Ay=P

0 A M M-PL=0M=PL

At x=0 V=PAt x=L V=P

At x=0 M=-PLAt x=L M=0

P

L

A

P

x A

y A

M

P

PL


69/144


0 A M By×5.5-10-60×2=0By=23.63 KN

0 yF 23.63+A y-60=0Ay=36.37 Kn

0 xF Ax=0

mKN / 15mKN .10

A B

m4 m5.0 m1

mKN / 15mKN .10

x A

y B

m4 m5.0 m1

y A

mKN / 15mKN .10

KN 63.23

m4 m5.0 m1

KN 37.36

KN 37.36

KN 63.23KN 63.2363.23

437.36

x x

23.63x=36.37×4-36.37xx=2.4246 mMaximum bending moment occurwhen V=0, at x=2.4246 m

x x4

mKN .092.44

mKN .48.25mKN .665.23

mKN .665.13


70/144


0C M 80-R B×6+30×7.5+144×4=0RB=146.83 KN

0 yF RC+146.83-30-144=0RC=27.17 KN

mKN / 10

mKN / 48

A BC

mKN .80

m3 m6

KN 30

mKN .80

m2m5.1

B R C R

KN 144

mKN / 10

mKN / 48

A BC

mKN .80

m3 m6KN 83.146 KN 17.27

KN 30

KN 83.116

KN 17.27

mKN .45

mKN .125

mKN .187.47

For segment 63 xV=116.83+4(x-3) 2-48(x-3)

Maximum bending moment occur whenV=00=116.83+4(x-3) 2-48(x-3)x2-18x+74.2075=0x=6.39375 m from left end

m39375.3


71/144


0 xF Ax=0

0 A M By×6-9×7=0By=10.5 KN

0 yF Ay+10.5-9=0Ay=-1.5 KN

mKN / 3

A BC

m3 m3 m3

mKN / 3

y A y B

x A

m2

mKN / 3

KN 9

KN 5.10KN 5.1

KN 5.1

KN 75.3

KN 75.6

mKN .5.4

mKN .25.11


72/144

Stresses in Beams:

Pure bending refers to flexure of a beam under a constant bending moment.Therefore, pure bending occurs only in regions of a beam where the shear force is zero.Nonuniform bending refers to flexure in the presence of shear forces, which means that thebending moment changes as we move along the axis of the beam.

S.F. Diagram

B.M. Diagram

Nonuniform bending Pure bending

Assumptions:1. The beam is initially straight and unstressed.2. The material of the beam is perfectly homogeneous.3. The elastic limit is nowhere exceeded.

4. Young's modulus for the material is the same in tension and compression.5. Plane cross-sections remain plane before and after bending.6. Every cross-section of the beam is symmetrical about the plane of bending i.e. about

an axis perpendicular to the N.A.7. There is no resultant force perpendicular to any cross-section.

If we now considered a beam initially unstressed and subjected to a constant bendingmoment along its length, i.e. pure bending as would be obtained by applying equal couplesat each end, it will bend to a radius as shown below.

P P

a a

P

P

Pa


73/144

As a result of this bcompression and the bottom tothe two there are points at whicThe neutral axis will always pas

The length L1 of the linL1=( -y)d

d =ρdx

L1=(1-ρ y )dx

The original length of l

Strain(ε x)= lenoriginalloriginal L 1

εx=-kywhere k is the curvaturThe longitudinal norm

contraction (- εx) will occur ielongation (+ε x) will occur in fi

εx=-(1c

y ) εmax

By using Hook's law

x=-Eky=- ρ E

y

nding the top fibers of the beamension. Its reasonable to suppose, thath the stress is zero, these points is ters through the centre of area or centroi

e ef after bending takes place is:

ine ef is dx

gth

ength= dx

dxdx y

)1(ρ

= - ρ y

e. al strain will vary linearly with y fro

n fibers located above the neutralers located below the neutral axis (-y

N.A

x=Eε x

maxε

ε

ill be subjected to somewhere between

med the neutral axis. .

the neutral axis. A axis (+y), whereas

).


74/144

N.A

x=-(1c

y ) max

Normal stress will varyzero at the neutral axis to a max

dF= xdAM=

A

ydF = ydA A

x )(σ

= A

yd c y

)( max1

σ

M= A

dA yc

2

1

maxσ

A

dA y 2 =I moment of i

I Mc 1

max σ

max : The maximum normal stsectional area farthest away froM: The resultant internal momeI: The moment of inertia of thec1: The perpendicular distanceneutral axis, where max acts.

linearly with y from the neutral axis.imum value max a distance c 1 farthest

N.A.

A

ertia

ess in the member, which occurs atthe neutral axis.

nt. ross sectional area computed about th

from the neutral axis to a point far

maxσ

Stress will vary from from neutral axis.

a point on the cross

e neutral axis. thest away from the


75/144

1=- I

Mc1 , 2= I

Mc 2

1=-1

S M , 2=

2S

M

S1=1c

I , S 2=2c

I

The quantities S 1 andarea.

Example 43: A simple beamintensity q=1.5 k/ft and a coallowance for the weight of theleft hand end of the beam. Thcross section of width b=8.75 i compressive stresses in the bea

ft 9

A

S2 are known as the section moduli

AB of span length L=22 ft supportcentrated load P=12 k . The uniforbeam. The concentrated load acts atbeam is constructed of glued laminand height h=27 in . Determine thedue to bending.

ft 22

k P 12

ft k q / 5.1

B

f the cross sectional

s a uniform load of m load includes an

a point 9 ft from the ted wood and has a

aximum tensile and


76/144

0 A M By×22-12×9-33×11=0By=21.409 k

0 yF

Ay+21.409-12-33=0Ay=23.591 k

S.F. Diagram

B.M. Diagram

k 12 k 33

y A y B

ft 9k P 12

ft k q / 5.1

k 591.23 k 409.21

k 591.23

k 409.21

k 091.10

k 909.1

ft k .569.151

Maximum bending momentMmax=151.569 k.ft

=151.569×12=1818.828 ksi


77/144

Example 44: The simply supported beam has the cross sectional area shown below.Determine the absolute maximum bending stress in the beam and draw the stressdistribution over the cross section at this location.

in75.8

in271c

2c

c1=c2=13.5 in

1= I

Mc1

I= 12

3bh=

43

1875.1435212)27(75.8

in

1=1875.14352

5.1310828.1818 3

=-1710.8317 psi

1=1875.14352

5.1310828.1818 3

=1710.8317 psi

A B

mKN / 5

m6

mm250

mm300

mm20

mm20

mm20

KN 30

y A y B


78/144

0 A M By×6-30×3=0By=15 KN

0 yF Ay+15-30=0Ay=15 KN

N.A

c1=c2=170 mm

I1= 23

12 Ad

bh

I1= 2333333

)10160()102010250(12

)1020(10250

I1=128.16667×10-6 m4

I3=I1=128.16667×10-6 m4

I2=12

3bh =12

)10300()1020( 333 =45×10 -6 m4

I= I 1+ I2+ I3=128.16667×10-6+128.16667×10 -6+45×10 -6

I=301.333×10 -6 m4

max= I

Mc 1 = 633

10333.301

10170105.22

=12.693598 MPa.

mKN / 5

KN 15 KN 15

KN 15

KN 15

mKN .5.22

Maximum bending moment

Mmax=22.5 KN.m

mm250

mm300

mm20

mm20

mm20

12

3


79/144

B= I

My B

= 633

10333.30110150105.22

=11.200233 MPa.

Example 45: The beam shown below has a cross section of channel shape with width b=300 mm and height h=80 mm , the web thickness is t=12 mm . Determine the maximumtensile and compressive stresses in the beam due to uniform load.

0 A M By×3-14.4×2.25=0By=10.8 KN

0 yF Ay+10.8-14.4=0Ay=3.6 KN

0 xF

Ax=0

B

m3 m5.1

mKN / 2.3

A B

mm300

mm80

mm12

x A

y A

KN 4.14

y B

mKN / 2.3


80/144

M1=2.025 KN.mM2=3.6 KN.m

A

A y yc

6

9

10523210321888

c y =61.52×10-3 m

yc=61.52 mm

I1= 23

12 Ad

bh

No. of Area A(m 2) y (m) A y (m3)1 960×10 - 40×10 - 38400×10 -

2 3312×10 - 74×10 - 245088×10 -

3 960×10 - 40×10 - 38400×10 -

A =5232×10 - A y =321888×10 -9

KN 6.3 KN 8.10

KN 6.3

KN 6

KN 8.4

mKN .025.2

mKN .6.3

mm300

mm80

mm12

12

3

mm80mm12

mm300

mmc 52.611

mmc 48.182


81/144

I1=12

)1080(1012 333

I3= I1=0.95658×10-6

I2= 23

12 Ad

bh

I2=12

)1012(10276 333

I= I 1+ I2+ I3=2.46874

4.21025.2

)( 211 I c M

t σ

46.2106.3

)(3

122 I

c M t σ

( t)max=50.462179 M

025.2)( 111 I c M

cσ

26.3

)( 222 I c M

cσ

( c)max=-89.71054 M

Composite Beams:

Composite beams areload.

Normal stress in material 1 is dNormal stress in material 2 is d

dA=dydzThe force dF acting on the area

dF= dA=( E 1ε)dydz

If the material 1 is being transfob2=nb

960×10 -6×(21.52×10 -3)2=0.95658×10 -

4

+3312×10 -6×(12.48×10 -3)2=0.55558×

10 -6 m4

462179.5010874

1052.616

33

MPa

94815.261074

1048.186

3

MPa

a

158339.151046874.

1048.18106

33

MPa

71054.891046874.

1052.61106

33

MPa

a

made from different materials in order

termined from =E 1ε termined from =E 2ε

dA of the beam is

rmed into material 2

6 m4

0-6 m4

to efficiently carry a


82/144

Ad F d σ =( E 2ε)ndyddF= F d ( E 1ε)dydz=( E 2ε)ndy

n=2

1 E

E

n: transformation factor (modul

If the material 2 is being transfo

b1= n b where n = 12

E E

For the transformed m=n σ

z

ar ratio).

rmed into material 1

aterial


83/144

Example 46: A composite beam is made of wood and reinforced with a steel strap locatedon its bottom side. It has the cross sectional area shown below. If the beam is subjected to abending moment of M=2 KN.m determine the normal stress at point B and C . Take E w=12GPa and E st=200 GPa .

st

w

E

E n

06.020012

n

wst bnb mmbst 915006.0

mm150

mm150

mm20

B

C

mm150C

mm20

mm150

mm9

B

1

2

mm150

mm150

mm9

mm20


84/144

A

A y yc

mm

m y c

379.36

10379.3610435010158250 3

3

9

I=I 1+I22

3

1 12 Ad

bh I

236333

1 )10379.26(10300012)1020(10150 I

I1=2.187554×10-6 m4

23

2 12 Ad

bh I

236333

2 )10621.58(10135012)10150(109 I

I2=7.170419×10-6 m4

I=9.35797×10 -6 m4

I My

B σ

MPa B 557689.281035797.910621.133102

6

33

σ

B B n σσ

MPa B 71346134.1)557689.28(06.0 σ

I My

B σ

MPa B 774976.71035797.910379.361026

33

σ

No. of Area A(m 2) y (m) A y (m3)1 3000×10 - 10×10 - 30000×10 -

2 1350×10 - 95×10 - 128250×10 -

A =4350×10-

A y =158250×10-9

mm150

mm150

mm9

mm20

mmc 621.1331

mmc 379.362

A N .


85/144

Shear Stresses in Beams

A

ydA It V

τ

A y ydA A

= Q

It VQ

τ

τ :- the shear stress in the member at the point located a distance y from the neutral axis.V:-the internal resultant shear force.I:-the moment of inertia of the entire cross sectional area computed about the neutral axis.t:-the width of the members cross sectional area, measured at the point where τ is to bedetermined.Q A y , where A is the top (or bottom) portion of the members cross sectional area,defined from the section where t is measured, and y is the distance to the centroid of A ,measured from the neutral axis.


86/144

Example 47: A metal beam with span L=3 ft is simply supported at points A and B. Theuniform load on the beam is q=160 lb/ in. The cross section of the beam is rectangular withwidth b=1 in and height h=4 in . Determine the normal stress and shear stress at point C ,which is located 1 in below the top of the beam and 8 in from the right hand support.

0 A M By×3×12-5760×1.5×12=0By=2880 lb

0 yF Ay+2880-5760=0Ay=2880 lb

A B

ft 3

inlbq / 160

C

lb5760

y A y B

lb2880

inlbq / 160

lb2880

in18

ink M .92.25max

lb2880

lb2880

in18

C

in8

V

182880

10V

V=1600 lb


87/144

At point C x=28 in from left endfrom shear force diagram

N.A

4

33

3333.512 )4(112 inbh

I A =1×1=1 in 2

y =1.5 inQ A y =1.5×1=1.5 in 3

ksi I

MyC 36.33333.5

117920σ

It VQ

τ

psi45013333.55.11600

τ

10160021

18288021

M

M=17.92 k.in

A y

in1

in4

in1


88/144

Example 48: Consider the cantilever beam subjected to the concentrated load shownbelow. The cross section of the beam is of T-shape. Determine the maximum shearingstress in the beam and also determine the shearing stress 25 mm from the top surface of thebeam of a section adjacent to the supporting wall.

0 A M M-50×2=0M=100 KN.m

0 yF Ay-50=0Ay=50 KN

S.F. Diagram

B.M. Diagram

From shear and bending moment diagramsV=50 KNM=100 KN.m

A B

KN 50m2

mm200

mm50

mm50

mm125

M B

KN 50m2

y A

mKN .100 B

KN 50m2

KN 50

KN 50

mKN .100

mm200

mm50

mm50

mm125

1

2


89/144

A

A y yc

mm

m y c

65.58

1065.58101625010953125 3

3

9

I=I 1+I22

3

1 12 Ad

bh I

236333

1 )1065.33(101000012)1050(10200 I

I1=13.40655833×10-6 m4

23

2 12 Ad

bh I

236333

2 )1085.53(10625012)10125(1050 I

I2=26.26191146×10-6 m4

I=39.6684×10 -6 m4

Q A y A =50×10 -3×116.35×10 -3

=0.0058175 m 2

y =58.175×10 -3 mQ=0.000338433 m 3

It VQ

maxτ

MPa5315553.81050106684.39

000338433.0105036

3

max τ

No. of Area A(m 2) y (m) A y (m3)1 10000×10 - 25×10 -3 250000×10 -9

2 6250×10 - 112.5×10 - 703125×10 -

A =16250×10-

A y =953125×10-9

mm200

mm50

mm50

mm125

1

2

A N .

mmc 65.582

mmc 35.1161

mm200

mm50

mm50

mm125

1

2

A N .

mmc 65.582

mmc 35.1161

A

y


90/144


91/144

Curved Beams

Due to the curvature of the beam, the normal strain in the beam does not vary linearly with depth as in thecase of a straight beam .As result, the neutral axis does not pass through the centroid of the cross section.

C7

If we isolate a differential segment of the beam let a strip material located at r distance has an original lengthr dθ .Due to the rotations δθ / 2, the strip’s total change in length is δθ )R -r(

Strain is a nonlinear function of r, in fact it varies in a hyperbolic fashion .Hooke's law applies,

0 RF

A A

dAr r R Ek dA 0)(0σ

A A

dAr

dA R 0

)(,)(

r r R

k d

k rd

r R ε

θδθ

θδθ

ε

)(r

r R Ek

σ


92/144

)( y R Ae My

σ

R :- The location of the neutral axis, specified from the center of curvature 0' of the member.

A :- The cross -sectional area of the member.

r :- The arbitrary position of the area element dAon the cross section, specified from the center of curvature 0' of the member.

y=R -r , e= r -R

σ :- The normal stress in the member.

M :- T he internal moment, determined from the method of sections equations of equilibrium and computed aboutthe centroidal axis .

A :- The cross -sectional area of the member.

R :- T he distance measured from the center of curvature to the neutralaxis.

r :- T he distance measured from the center of curvature to the centroidof the cross -sectional area.

r :- T he distance measured from the center of curvature to the point where the stress σ is to be determined.

A r dA

A R


93/144

)()(

Rr Ar Rr M

o

ooσ

)()(

Rr Ar r R M

i

iiσ

Normal stress at the bar's top.

Normal stress at the bar's bottom.

Example:- The curved bar has a cross -sectional area shown below .If it issubjected to bending moments of 4kN • m, determine the maximum normal stress developed in the bar.

Area A(mm 2) y-(mm) y -A(mm 3)

rectangle 2500 225 562500

triangle 750 260 195000

3250 757500

mm A

A yr 076.233

3250757500

A r dA

A

R

mm A 3250

br

r

r r

br

r

r b

RdA

1

2

12

2

1

2 ln)(

ln

= mm04389.1450250280

ln)250280(

28050200250

ln50

mm R 417.23104389.14

3250


94/144

MPa Rr Ar

r R M

i

i B 5373.116

)10417.23110076.233(10200103250)1020010417.231(104

)()(

3336

333

σ

MPa Rr Ar

Rr M

o

o A 7231.128)10417.23110076.233(10280103250

)10417.23110280(104)()(

3336

333

σ The

maximum stress at point A= MPa7231.128

Example:- The frame of a punch press is shown below. Find the stresses at the inner and outer surface atsection x-x of the frame if W=5000 N.

hbb

A oi2

248040)2

618( mm A

mmbbr

r

h

r br b

R

dAoi

i

oiooi 365.12)618(25

65ln)

40

2566518()(ln)(

mm

r dA

A R

A

8175.38365.12

480

W

W

mm100

mm50mm25

mm18mm6

mm40

x x

h

ob ib

mm65

mm25mm40

mm18mm6


95/144

Area(mm 2) A(mm 2) y-(mm) y -A(mm 3)

rectangle 720 45 32400

Triangle -120 51.666 -6200

Triangle -120 51.666 -6200480 20000

mm A

A yr 666.41

48020000

M=W×d=5000×(100×10 -3+41.666×10 -3)=708.33 N.m

MPa

AW

Rr Ar r R M

i

ii

747.296

104805000

)108175.3810666.41(102510480)1025108175.38(33.708

)()(

63336

33

σ

MPa

AW

Rr Ar Rr M

o

oo

26.198

104805000

)108175.3810666.41(106510480)108175.381065(33.708

)()(

63336

33

σ


96/144

Slop and Deflection in Beams

The elastic curve :-the deflection diagram of the longitudinal axis that passesthrough the centroid of each cross sectional area of the beam.


97/144

x-axis extends positive to the right.v-axis extends positive upward from the x-axis.

y

ε

ρ

1

E σ

ε

I My

σ

EI M

ρ1

When M is positive, ρ extends above the beam, i.e. ρ in the positive v direction.When M is negative, ρ extends below the beam, or in the negative v direction.

Integration Method

The elastic curve for a beam can be expressed mathematically as v=f(x)2 / 32

22

]) / (1[ / 1

dxdvdxvd

ρ

2 / 32

22

]) / (1[ /

dxdvdxvd

EI M

The slop of the elastic curve which is determined from dv/dx will be very small, andits square will be negligible compared with unity.

2

21dx

vd ρ

2

2

dxvd

EI M


98/144

dxdM

V

)()( 22

dxvd

EI dxd

xV

dxdV w

)()( 22

2

2

dxvd

EI dxd

xw

EI always positive quantity

)( 22

dxvd

EI M

)( 33

dx

vd EI V

)( 44

dxvd

EI w

Sign convention and coordinates

Positive deflection v is upward, the positive slope will be measuredcounterclockwise from the x-axis when x is positive to the right.

If positive x is directed to the left, then will be positive clockwise.

dxdv

θ


99/144

Boundary conditions

1C Mdxdxdv

EI

21][ C xC dx Mdx EIv

Example: The cantilevered beam shown is subjected to a vertical load P at its end.Determine the equation of elastic curve. EI is constant.

0 M M+Px=0M=-Px

)( 22

dxvd

EI M =-Px

12

21

C Pxdxdv

EI ……………………..(1)

213

6

1C xC Px EIv ………………………(2)

Boundary conditions


100/144

at x=L

dxdv

θ =0 and v =0

12210 C PL 21 21 PLC

233

21

61

0 C PLPL 332 21

61

PLPLC =3

3PL

22

21

21

PLPx EI θ )(2

22 x L EI P

θ

3261 323 PL x

PLPx EIv

)23(6

323 L x L x EI P

v

Example: The simply supported beam shown supports the triangular distributed loading.Determine the maximum deflection. EI is constant.

Due to symmetry we take 0 ≤ x ≤2

L

0 M

034

3

L xw

x Lw

M )34

(3

L x

x L

w M

)( 22

dxvd

EI M = )34

(3

L x

x L

w

1

42

128C

L xw

x Lw

dxdv

EI ………………….(1)

21

53

6024 C xC L xw

x Lw

EIv

……..………..(2)


101/144

Boundary conditions

at x=2

Ldxdv

θ =0

1

42

1280 C

L xw

x Lw

at x=0 v=00=0-0+0+C 2 C

)1925

6024(

353 x

L L

x x

L EI w

v

Vmax at x=2

L

)3845

1920192(

444

max L L L

EI w

v = w12

Discontinuity Method

1925 3

1

LwC

=0

EI L4

0


102/144

a x for a x

a x for a x

n

n

)(

0

C n

a xdxa x

nn

1

1

Example:- Find the moment expression using continuity equations.

M=2.75 1+1.5 0-2

3 2-6

1 3

=2.75x+1.5 0-23 2-

61 3

Example:- Determine the equation of the elastic curve for the beam shown below. EI isconstant.

0 yF Ay-40-12=0 A y=52 kN

0 A

M

MA-40×2.5-50-12×9=0 M A=258 kN.m


103/144

M=-258 0+52 1-28 2+50 0+

28 2

EI 22

dxvd =-258+52x-4x 2+50 0+4 2

EIdxdv =-258x+26x 2-

34 x3+50 1+

34 3+C 1………….….(1)

EIv=-129x 2+3

26 x3-31 x4+25 2+

31 4+C 1x+C 2………….(2)

B.C

dxdv =0 at x=0 in eq.(1)

v=0 at x=0 in eq.(2)C1=0C2=0v=

EI 1 (-129x 2+

326 x3-

31 x4+25 2+

31 4)

Moment Area Method

B

A A B dx EI

M / θ

The notation A B / θ is referred to as the angle of the tangent at B measured with respect tothe tangent at A.


104/144

Theorem 1 The angle betweenthe area under the M/EI diagra

If the area under M/EI difrom the tangent A to tangent B

If the area under M/EI diatangent A to tangent B. A B / θ wil

B

A B A dx EI

M xt /

B At / : the vertical deviation of th dA x xdA

B

A

dx EI M represents the area unde

B

A B A dx EI

M xt /

x is the distance from A to thand B.

the tangents at any two points on thebetween these two points.

gram is positive, the angle is measu.

ram is negative, the angle A B / θ is meal measured in radians.

e tangent at A with respect to the tang

r the M/EI diagram, we can also write:

e centroid of the area under the M/EI

elastic curve equals

ed counterclockwise

ured clockwise from

nt at B.

-

diagram between A


105/144

Theorem 2 The vertical deviarespect to the tangent extendedthe M/EI diagram between thwhere the vertical deviation B At /

Example: Determine the slope

Bθ = A B / θ

C θ = AC / θ

MA-PL=0

MA=PL

Ay=P

B

A

A B dx EI M

/ θ Area under the M

0 A M

0 yF

tion of the tangent at a point A on tfrom another point B equals the momse two points. This moment is comis to be determined.

f the beam shown at points B and C .

EI diagram from A to B

A M

y A

e elastic curve with ent of the area under

uted about point A

I is constant

P


106/144

Bθ =

21

22 / EI PL L

EI PL

A Bθ

= EI

PL

8

3 2 rad clockwise

C θ = AC / θ = 21

L EI

PL

=

EI PL

2

2

r

Example: Determine the displconstant.

vB= A Bt / vC= AC t /

0 yF Ay=0

0 A M M-M o=0M=M o

vB= B

A A B dx EI

M xt /

=

2221 L

EI M L o =

E M o8

vC= A

C

AC dx

EI

M xt / = L

EI

M L o

2

22 L

EI PL

d clockwise

acement of points B and C of the

L2

= EI

L M o

2

2

o M

beam shown. EI is

o M


107/144

Example: Determine the slope I=17×10 6 mm 4

C θ = AC A / θθ

Since the angle is very small Aθ

A Bt / =

)2(32

21

)6(24

)6(31

2 EI

A Bt / = EI 320

AC / θ = EI EI 8

21

)2(8

C θ = EI 8320

- EI 8

= EI 32

C θ = 69 10171020032 =0.00941

Castigliano’s Theorem Ap

L

EI dx

P M M

0

)(

Where:-=displacement of the point ca

P=external force of variable maM=internal moment in the beaP and the load on the beam.

L

EI dx

M M

M 0

)(θ

θ =the slope of the tangent at a

C θ

at point C for the steel beam shown.

=tan Aθ = Lt A B /

21

)2(24

EI

1 rad

plied to Beams

used by the real loads acting on the be gnitude applied to the beam in the dire

, expressed as a function of x and cau

oint on the elastic curve.

L

. Take E st=200 GPa ,

am. ction of . sed by both the force


108/144

M =an external couple moment

Example: Determine the dispconstant.

0 M

0)2

( Px xwx M

Px x

w M 2

2

xP

M

When P=0

2

2 xw M , x

P M

L L

B EI dx

P M

M 00

)( L

B x EI w

0

4

8=

EI wL8

4

Example: Determine the dispI=450 in 4, E st=29×10

3 ksi.

0 B M

Cy×20-60×10+15×3

10 +P

Cy =27.5-0.5P

0 yF By+27.5-0.5P-60-15-P=0

acting at the point.

lacement of point B on the beam

L

dx x EI w

dx EI

xwx

0

3

2

2

)(2

lacement of point A of the steel

10=0

P

y B

shown below. EI is

eam shown below.

yC

xC


109/144

By=47.5+1.5P 0 xF

Cx=0 0 M

M1+ )3

(203 12

1

x x +Px 1=0

M1=-20

31 x - Px 1

P M 1 =-x 1

When P=0

M1=-20

31 x

P M 1 =-x 1

0 M

M2+3x 2(2

2 x )+15×(3

10 +x 2)+P(10+x 2)-( 47.5+1.5P)x 2=0

M2= 2223

x +(32.5+0.5P)x 2-10P-50

P

M 2 =0.5x 2-10

When P=0

M2= 2223

worked exp mechanics 2.pdf

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