mechanics of materials solutions chapter08 probs65 81

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8.65 A beam with a box cross section is subjected toa resultant moment magnitude of 2,100 N-m acting at the angle shown in Fig. P8.65. Determine: (a) the maximum tension and the maximumcompression bending stresses in the beam. (b) the orientation of the neutral axis relative to the+z axis. Show its location on a sketch of the crosssection.

Fig. P8.65

Solution Section properties

3 34

3 34

(90 mm)(55 mm) (80 mm)(45 mm) 640,312.5 mm12 12

(55 mm)(90 mm) (45 mm)(80 mm) 1,421,250.0 mm12 12

y

z

I

I

= − =

= − =

Moment components

(2,100 N-m)sin 30 1,050 N-m

(2,100 N-m)cos30 1,818.65 N-my

z

MM

= ° =

= − ° = −

(a) Maximum bending stresses For a shape having at least one axis of symmetry, Eq. (8-24) can be used to determine bending stresses. Compute normal stress at y = 45 mm, z = 27.5 mm:

4 4

(1,050 N-m)(27.5 mm)(1,000 mm/m) ( 1,818.65 N-m)(45 mm)(1,000 mm/m)640,312.5 mm 1, 421, 250.0 mm

45.0952 MPa 57.5827 MPa

102.6779 MPa 102.7 MPa (T)

y zx

y z

M z M yI I

σ = −

−= −

= +

= = Ans. Compute normal stress at y = −45 mm, z = −27.5 mm:

4 4

(1,050 N-m)( 27.5 mm)(1,000 mm/m) ( 1,818.65 N-m)( 45 mm)(1,000 mm/m)640,312.5 mm 1, 421, 250.0 mm

45.0952 MPa 57.5827 MPa

102.6779 MPa 102.7 MPa (C)

y zx

y z

M z M yI I

σ = −

− − −= −

= − −

= − = Ans.


(b) Orientation of neutral axis For a shape having at least one axis of symmetry, Eq. (8-25) can be used to determine the orientation of the neutral axis:

4

4

(1,050 N-m)(1,421,250.0 mm )tan 1.2815( 1,818.65 N-m)(640,312.5 mm )

52.03 (i.e., 52.03 CCW from axis)

y z

z y

M IM I

z

β

β

= = = −−

∴ = − ° ° + Ans.


8.66 The moment acting on the cross section of theT-beam has a magnitude of 22 kip-ft and is orientedas shown in Fig. P8.66. Determine: (a) the bending stress at point H. (b) the bending stress at point K. (c) the orientation of the neutral axis relative to the+z axis. Show its location on a sketch of the crosssection.

Fig. P8.66

Solution Section properties Centroid location in y direction:

Shape Width b Height h Area Ai

yi (from bottom) yi Ai

(in.) (in.) (in.2) (in.) (in.3) top flange 7.00 1.25 8.7500 8.375 73.28125 stem 0.75 7.75 5.8125 3.875 22.52344 14.5625 95.80469

3

2

95.80469 in. 6.5789 in. (from bottom of shape to centroid)14.5625 in.

2.4211 in. (from top of shape to centroid)

i i

i

y Ay

AΣ

= = =Σ

=

Moment of inertia about the z axis:

Shape IC d = yi – y d²A IC + d²A (in.4) (in.) (in.4) (in.4)

top flange 1.1393 1.7961 28.2273 29.3666 stem 29.0928 −2.7039 42.4956 71.5884

Moment of inertia about the z axis (in.4) = 100.9550 Moment of inertia about the y axis:

3 3

4(1.25 in.)(7.00 in.) (7.75 in.)(0.75 in.) 36.0016 in.12 12yI = + =

Moment components

(22 kip-ft) cos55 12.6187 kip-ft 151.4242 kip-in.

(22 kip-ft)sin 55 18.0213 kip-ft 216.2561 kip-in.y

z

MM

= − ° = − = −

= − ° = − = −


(a) Bending stress at H For a shape having at least one axis of symmetry, Eq. (8-24) can be used to determine bending stresses. To compute the normal stress at H, use the (y, z) coordinates y = 2.4211 in. and z = −3.5 in.:

4 4

( 151.4242 kip-in.)( 3.50 in.) ( 216.2561 kip-in.)(2.4211 in.)36.0016 in. 100.9550 in.

14.7211 ksi 5.1862 ksi

19.9074 ksi 19.91 ksi (T)

y zx

y z

M z M yI I

σ = −

− − −= −

= +

= = Ans. (b) Bending stress at K To compute the normal stress at K, use the (y, z) coordinates y = −6.5789 in. and z = 0.375 in.:

4 4

( 151.4242 kip-in.)(0.375 in.) ( 216.2561 kip-in.)( 6.5789 in.)36.0016 in. 100.9550 in.

1.5773 ksi 14.0927 ksi

15.6700 ksi 15.67 ksi (C)

y zx

y z

M z M yI I

σ = −

− − −= −

= − −

= − = Ans. (c) Orientation of neutral axis For a shape having at least one axis of symmetry, Eq. (8-25) can be used to determine the orientation of the neutral axis:

4

4

( 151.4242 kip-in.)(100.9550 in. )tan 1.9635( 216.2561 kip-in.)(36.0016 in. )

63.01 (i.e., 63.01 CW from axis)

y z

z y

M IM I

z

β

β

−= = =−

∴ = ° ° + Ans.


8.67 A beam with a box cross section is subjected toa resultant moment magnitude of 75 kip-in. acting at the angle shown in Fig. P8.67. Determine: (a) the bending stress at point H. (b) the bending stress at point K. (c) the maximum tension and the maximumcompression bending stresses in the beam. (d) the orientation of the neutral axis relative to the+z axis. Show its location on a sketch of the crosssection.

Fig. P8.67

Solution Section properties

3 34

3 34

(4 in.)(6 in.) (3.25 in.)(5.25 in.) 32.8096 in.12 12

(6 in.)(4 in.) (5.25 in.)(3.25 in.) 16.9814 in.12 12

y

z

I

I

= − =

= − =

Moment components

(75 kip-in.) cos 20 70.4769 kip-in.

(75 kip-in.)sin 20 25.6515 kip-in.y

z

MM

= ° =

= ° =

(a) Bending stress at H For a shape having at least one axis of symmetry, Eq. (8-24) can be used to determine bending stresses. To compute the normal stress at H, use the (y, z) coordinates y = −2.0 in. and z = −3.0 in.:

4 4

(70.4769 kip-in.)( 3.0 in.) (25.6515 kip-in.)( 2.0 in.)32.8096 in. 16.9814 in.

6.4442 ksi 3.0211 ksi

3.4231 ksi 3.42 ksi (C)

y zx

y z

M z M yI I

σ = −

− −= −

= − +

= − = Ans. (b) Bending stress at K To compute the normal stress at K, use the (y, z) coordinates y = 2.0 in. and z = 3.0 in.:

4 4

(70.4769 kip-in.)(3.0 in.) (25.6515 kip-in.)(2.0 in.)32.8096 in. 16.9814 in.

6.4442 ksi 3.0211 ksi

3.4231 ksi 3.42 ksi (T)

y zx

y z

M z M yI I

σ = −

= −

= −

= = Ans.


(c) Maximum bending stresses The maximum tension normal stress occurs at the (y, z) coordinates y = −2.0 in. and z = 3.0 in.:

4 4

(70.4769 kip-in.)(3.0 in.) (25.6515 kip-in.)( 2.0 in.)32.8096 in. 16.9814 in.

6.4442 ksi 3.0211 ksi

9.4653 ksi 9.47 ksi (T)

y zx

y z

M z M yI I

σ = −

−= −

= +

= = Ans. The maximum compression normal stress occurs at the (y, z) coordinates y = 2.0 in. and z = −3.0 in.:

4 4

(70.4769 kip-in.)( 3.0 in.) (25.6515 kip-in.)(2.0 in.)32.8096 in. 16.9814 in.

6.4442 ksi 3.0211 ksi

9.4653 ksi 9.47 ksi (C)

y zx

y z

M z M yI I

σ = −

−= −

= − −

= − = Ans. (d) Orientation of neutral axis For a shape having at least one axis of symmetry, Eq. (8-25) can be used to determine the orientation of the neutral axis:

4

4

(70.4769 kip-in.)(16.9814 in. )tan 1.4220(25.6515 kip-in.)(32.8096 in. )

54.88 (i.e., 54.88 CW from axis)

y z

z y

M IM I

z

β

β

= = =

∴ = ° ° + Ans.


8.68 The moment acting on the cross section of thewide-flange beam has a magnitude of M = 12 kN-m and is oriented as shown in Fig. P8.68. Determine: (a) the bending stress at point H. (b) the bending stress at point K. (c) the orientation of the neutral axis relative to the +zaxis. Show its location on a sketch of the cross section.

Fig. P8.68

Solution Section properties Moment of inertia about the z axis:

Shape IC d = yi – y d²A IC + d²A (mm4) (mm) (mm4) (mm4)

top flange 59,062.5 97.5 29,944,687.5 30,003,750 web 4,860,000 0 0 4,860,000 bottom flange 59,062.5 −97.5 29,944,687.5 30,003,750

Moment of inertia about the z axis (mm4) = 64,867,500 Moment of inertia about the y axis:

3 3

4(15 mm)(210 mm) (180 mm)(10 mm)2 23,167,500 mm12 12yI = + =

Moment components

6

6

(12 kN-m)sin 35 6.8829 kN-m 6.8829 10 N-mm

(12 kN-m)cos35 9.8298 kN-m 9.8298 10 N-mmy

z

M

M

= ° = = ×

= ° = = ×

(a) Bending stress at H For a shape having at least one axis of symmetry, Eq. (8-24) can be used to determine bending stresses. To compute the normal stress at H, use the (y, z) coordinates y = 105 mm and z = −105 mm:

6 6

4 4

(6.8829 10 N-mm)( 105 mm) (9.8298 10 N-mm)(105 mm)23,167,500 mm 64,867,500 mm

31.1948 MPa 15.9114 MPa

47.1062 MPa 47.1 MPa (C)

y zx

y z

M z M yI I

σ = −

× − ×= −

= − −

= − = Ans.


(b) Bending stress at K To compute the normal stress at K, use the (y, z) coordinates y = −105 mm and z = 105 mm:

6 6

4 4

(6.8829 10 N-mm)(105 mm) (9.8298 10 N-mm)( 105 mm)23,167,500 mm 64,867,500 mm

31.1948 MPa 15.9114 MPa

47.1062 MPa 47.1 MPa (T)

y zx

y z

M z M yI I

σ = −

× × −= −

= +

= = Ans. (b) Orientation of neutral axis For a shape having at least one axis of symmetry, Eq. (8-25) can be used to determine the orientation of the neutral axis:

4

4

(6.8829 kN-m)(64,867,500 mm )tan 1.9605(9.8298 kN-m)(23,167,500 mm )

62.98 (i.e., 62.98 CW from axis)

y z

z y

M IM I

z

β

β

= = =

∴ = ° ° + Ans.


8.69 For the cross section shown in Fig. P8.69,determine the maximum magnitude of the bendingmoment M so that the bending stress in the wide-flange shape does not exceed 165 MPa.

Fig. P8.69

Solution Section properties Moment of inertia about the z axis:


top flange 59,062.5 97.5 29,944,687.5 30,003,750 web 4,860,000 0 0 4,860,000 bottom flange 59,062.5 −97.5 29,944,687.5 30,003,750

Moment of inertia about the z axis (mm4) = 64,867,500 Moment of inertia about the y axis:

3 3

4(15 mm)(210 mm) (180 mm)(10 mm)2 23,167,500 mm12 12yI = + =

Moment components sin 35 cos35y zM M M M= ° = ° Maximum bending moment magnitude The maximum tension bending stress should occur at point K, which has the (y, z) coordinates y = −105 mm and z = 105 mm:

4 4

sin 35 (105 mm) cos35 ( 105 mm) 165 MPa23,167,500 mm 64,867,500 mm

y zx

y z

M z M y M MI I

σ ° ° −= − = − ≤

6 3 6 3 2

6

2.59957 10 mm 1.32595 10 mm 165 N/mm

42.0327 10 N-mm 42.0 kN-m

M

M

− − − −⎡ ⎤× + × ≤⎣ ⎦

∴ ≤ × = Ans.


8.70 The unequal-leg angle is subjected to a bendingmoment of Mz = 20 kip-in. that acts at the orientationshown in Fig. P8.70. Determine: (a) the bending stress at point H. (b) the bending stress at point K. (c) the maximum tension and the maximumcompression bending stresses in the cross section. (d) the orientation of the neutral axis relative to the +zaxis. Show its location on a sketch of the cross section.

Fig. P8.70




(in.) (in.) (in.2) (in.) (in.3) upright leg 0.375 4.000 1.5000 2.00 3.00 bottom leg 2.625 0.375 0.9844 0.1875 0.18457 2.4844 3.18457

3

2

3.18457 in. 1.2818 in.(from bottom of shape to centroid)2.4844 in.


i i

i

y Ay

AΣ

= = =Σ

=

Centroid location in z direction:

Shape Area Ai

zi (from right edge) zi Ai

(in.2) (in.) (in.3) upright leg 1.5000 0.1875 0.2813 bottom leg 0.9844 1.6875 1.6612 2.4844 1.94243

3

2

1.94243 in. 0.7818 in. (from right edge of shape to centroid)2.4844 in.

2.2182 in. (from left edge of shape to centroid)

i i

i

z Az

AΣ

= = =Σ

=



upright leg 2.000 0.7182 0.77372 2.7737 bottom leg 0.011536 −1.0943 1.17881 1.1903

Moment of inertia about the z axis (in.4) = 3.9640


Moment of inertia about the y axis:

Shape IC d = zi – z d²A IC + d²A (in.4) (in.) (in.4) (in.4)

upright leg 0.017578 −0.5943 0.52979 0.5474 bottom leg 0.565247 0.9057 0.80750 1.3727

Moment of inertia about the y axis (in.4) = 1.9201 Product of inertia about the centroidal axes:

Shape Iy’z’ yc zc Area Ai yc zc Ai Iyz (in.4) (in.) (in.) (in.2) (in.4) (in.4)

upright leg 0 0.7182 −0.5943 1.5000 −0.6402 −0.6402 bottom leg 0 −1.0943 0.9057 0.9844 −0.9757 −0.9757

Product of inertia (in.4) = −1.6159 (a) Bending stress at H Since the angle shape has no axis of symmetry, Eq. (8-21) or Eq. (8-22) must be used to determine the bending stresses. Equation (8-22) will be used here. Note that the bending moment component about the y axis is zero (i.e., My = 0); therefore, the first term in Eq. (8-22) is eliminated. To compute the normal stress at H, use (y, z) coordinates of y = 2.7182 in. and z = −0.4068 in.:

2

4 4

4 4 4 2

5

8

(1.9201 in. )(2.7182 in.) ( 1.6159 in. )( 0.4068 in.) (20 kip-in.)(1.9201 in. )(3.9640 in. ) ( 1.6159 in. )

4.5619 in. (20 kip-in.)5.0001 in.

18.2469 ks

y yzx z

y z yz

I y I zM

I I Iσ

⎡ ⎤− += ⎢ ⎥−⎢ ⎥⎣ ⎦⎡ ⎤− + − −= ⎢ ⎥− −⎣ ⎦⎡ ⎤−= ⎢ ⎥⎣ ⎦

= − i 18.25 ksi (C)= Ans. (b) Bending stress at K To compute the normal stress at K, use (y, z) coordinates of y = −0.9068 in. and z = 2.2182 in.:

2

4 4

4 4 4 2

5

8

(1.9201 in. )( 0.9068 in.) ( 1.6159 in. )(2.2182 in.) (20 kip-in.)(1.9201 in. )(3.9640 in. ) ( 1.6159 in. )

1.8432 in. (20 kip-in.)5.0001 in.

7.3728 ksi

y yzx z

y z yz

I y I zM

I I Iσ

⎡ ⎤− += ⎢ ⎥−⎢ ⎥⎣ ⎦⎡ ⎤− − + −= ⎢ ⎥− −⎣ ⎦⎡ ⎤−= ⎢ ⎥⎣ ⎦

= − 7.37 ksi (C)= Ans. (d) Orientation of neutral axis Since the angle shape has no axis of symmetry, Eq. (8-23) must be used to determine the orientation of the neutral axis:


4

4

(20 kip-in.)( 1.6159 in. )tan 0.8416(20 kip-in.)(1.9201 in. )

40.08 (i.e., 40.08 CCW from axis)

y z z yz

z y y yz

M I M IM I M I

z

β

β

+ −= = = −+

∴ = − ° ° + Ans.

(c) Maximum bending stresses Sketch the orientation of the neutral axis. By inspection, the points on the angle cross section that are farthest from the neutral axis are point H and the corner of the angle. The bending stress at H has already been computed. To compute the normal stress at the corner of the angle, use (y, z) coordinates of y = −1.2818 in. and z = −0.7818 in.

2

4 4

4 4 4 2

5

8

(1.9201 in. )( 1.2818 in.) ( 1.6159 in. )( 0.7818 in.) (20 kip-in.)(1.9201 in. )(3.9640 in. ) ( 1.6159 in. )

3.7245 in. (20 kip-in.)5.0001 in.

14.8977 ksi

y yzx z

y z yz

I y I zM

I I Iσ

⎡ ⎤− += ⎢ ⎥−⎢ ⎥⎣ ⎦⎡ ⎤− − + − −= ⎢ ⎥− −⎣ ⎦⎡ ⎤

= ⎢ ⎥⎣ ⎦

= 14.90 ksi (T)= Therefore, the maximum compression bending stress is: 18.25 ksi (C)xσ = Ans. and the maximum tension bending stress is: 14.90 ksi (T)xσ = Ans.


8.71 For the cross section shown in Fig. P8.71,determine the maximum magnitude of the bendingmoment M so that the bending stress in the unequal-leg angle shape does not exceed 24 ksi.

Fig. P8.71




(in.) (in.) (in.2) (in.) (in.3) upright leg 0.375 4.000 1.5000 2.00 3.00 bottom leg 2.625 0.375 0.9844 0.1875 0.18457 2.4844 3.18457

3

2

3.18457 in. 1.2818 in.(from bottom of shape to centroid)2.4844 in.


i i

i

y Ay

AΣ

= = =Σ

=


Shape Area Ai


(in.2) (in.) (in.3) upright leg 1.5000 0.1875 0.2813 bottom leg 0.9844 1.6875 1.6612 2.4844 1.94243

3

2

1.94243 in. 0.7818 in. (from right edge of shape to centroid)2.4844 in.

2.2182 in. (from left edge of shape to centroid)

i i

i

z Az

AΣ

= = =Σ

=



upright leg 2.000 0.7182 0.77372 2.7737 bottom leg 0.011536 −1.0943 1.17881 1.1903

Moment of inertia about the z axis (in.4) = 3.9640


Moment of inertia about the y axis:

Shape IC d = zi – z d²A IC + d²A (in.4) (in.) (in.4) (in.4)

upright leg 0.017578 −0.5943 0.52979 0.5474 bottom leg 0.565247 0.9057 0.80750 1.3727


Shape Iy’z’ yc zc Area Ai yc zc Ai Iyz (in.4) (in.) (in.) (in.2) (in.4) (in.4)

upright leg 0 0.7182 −0.5943 1.5000 −0.6402 −0.6402 bottom leg 0 −1.0943 0.9057 0.9844 −0.9757 −0.9757

Product of inertia (in.4) = −1.6159 Orientation of neutral axis Since the angle shape has no axis of symmetry, it is helpful to determine the orientation of the neutral axis from Eq. (8-23) before beginning the stress calculations:

4

4

(20 kip-in.)( 1.6159 in. )tan 0.8416(20 kip-in.)(1.9201 in. )

40.08 (i.e., 40.08 CCW from axis)

y z z yz

z y y yz

M I M IM I M I

z

β

β

+ −= = = −+

∴ = − ° ° +

Allowable moments based on maximum tension and compression bending stresses Sketch the orientation of the neutral axis. By inspection, the points on the angle cross section that are farthest from the neutral axis are point H and the corner of the angle. To compute the normal stress at H, use (y, z) coordinates of y = 2.7182 in. and z = −0.4068 in.:

4 4

2 4 4 4 2

53

8

(1.9201 in. )(2.7182 in.) ( 1.6159 in. )( 0.4068 in.)(1.9201 in. )(3.9640 in. ) ( 1.6159 in. )

4.5619 in. ( 0.9124 in. )5.0001 in.

y yzx z z

y z yz

z z

I y I zM M

I I I

M M

σ

−

⎡ ⎤− + ⎡ ⎤− + − −= =⎢ ⎥ ⎢ ⎥− − −⎢ ⎥ ⎣ ⎦⎣ ⎦⎡ ⎤−= = −⎢ ⎥⎣ ⎦


Therefore, based on an allowable bending stress of 24 ksi at H, the maximum magnitude of Mz is:

3(0.9124 in. ) 24 ksi26.3054 kip-in.

z

z

MM

− ≤∴ ≤ (a)

To compute the normal stress at the corner of the angle, use (y, z) coordinates of y = −1.2818 in. and z = −0.7818 in.

4 4

2 4 4 4 2

53

8

(1.9201 in. )( 1.2818 in.) ( 1.6159 in. )( 0.7818 in.)(1.9201 in. )(3.9640 in. ) ( 1.6159 in. )

3.7245 in. (0.7449 in. )5.0001 in.

y yzx z z

y z yz

z z

I y I zM M

I I I

M M

σ

−

⎡ ⎤− + ⎡ ⎤− − + − −= =⎢ ⎥ ⎢ ⎥− − −⎢ ⎥ ⎣ ⎦⎣ ⎦⎡ ⎤

= =⎢ ⎥⎣ ⎦

Therefore, based on the bending stress at the corner of the angle, the maximum magnitude of Mz is:

3(0.7449 in. ) 24 ksi32.2197 kip-in.

z

z

MM

− ≤∴ ≤ (b)

Maximum bending moment Mz Compare the results in Eqs. (a) and (b) to find that the maximum bending moment that can be applied to the angle shape is: 26.3 kip-in.zM ≤ Ans.


8.72 The moment acting on the cross section ofthe unequal-leg angle has a magnitude of M = 20 kip-in. and is oriented as shown in Fig. P8.72.Determine: (a) the bending stress at point H. (b) the bending stress at point K. (c) the maximum tension and the maximumcompression bending stresses in the cross section.(d) the orientation of the neutral axis relative tothe +z axis. Show its location on a sketch of thecross section.

Fig. P8.72

Solution Moment of inertia about the z axis:

Shape IC d = yi – y Area Ai d²A IC + d²A (mm4) (mm) (mm2) (mm4) (mm4)

top flange 130,208.3 112.5 2,500 31,640,625.0 31,770,883.3 web 10,666,666.7 0 3,200 0 10,666,666.7 bottom flange 130,208.3 −112.5 2,500 31,640,625.0 31,770,883.3

Moment of inertia about the z axis (mm4) = 74,208,333.3 Moment of inertia about the y axis:

Shape IC d = zi – z Area Ai d²A IC + d²A (mm4) (mm) (mm2) (mm4) (mm4)

top flange 2,083,333.3 −42.0 2,500 4,410,000 6,493,333.3 web 68,266.7 0 3,200 0 68,266.7 bottom flange 2,083,333.3 42.0 2,500 4,410,000 6,493,333.3

Moment of inertia about the y axis (mm4) = 13,054,933.3 Product of inertia about the centroidal axes:

Shape yc zc Area Ai yc zc Ai Iyz (mm) (mm) (mm2) (mm4) (mm4)

top flange 112.5 −42.0 2,500 −11,812,500 −11,812,500 web 0 0 3,200 0 0 bottom flange −112.5 42.0 2,500 −11,812,500 −11,812,500

Product of inertia (mm4) = −23,625,000 Moment components

6

6

(40 kN-m)sin15 10.3528 kN-m 10.3528 10 N-mm

(40 kN-m)cos15 38.6370 kN-m 38.6370 10 N-mmy

z

M

M

= − ° = − = − ×

= − ° = − = − ×


(a) Bending stress at H Since the zee shape has no axis of symmetry, Eq. (8-21) or Eq. (8-22) must be used to determine the bending stresses. Equation (8-21) will be used here.

( ) ( )2 2

6 4 6 4

4 4 4 2

6

( 38.6370 10 N-mm)(13,054,933.3 mm ) ( 10.3528 10 N-mm)( 23,625,000 mm )(13,054,933.3 mm )(74,208,333.3 mm ) ( 23,625,000 mm )

( 10.3528 10 N-

z y y yz y z z yzx

y z yz y z yz

M I M I y M I M I zI I I I I I

y

σ+ +

= − +− −

⎡ ⎤− × + − × −= − ⎢ ⎥− −⎣ ⎦

− ×+4 6 4

4 4 4 2

3 3

mm)(74,208,333.3 mm ) ( 38.6370 10 N-mm)( 23,625,000 mm )(13,054,933.3 mm )(74,208,333.3 mm ) ( 23,625,000 mm )

(0.63271 N/mm ) (0.35197 N/mm )

z

y z

⎡ ⎤+ − × −⎢ ⎥− −⎣ ⎦

= + To compute the normal stress at H, use (y, z) coordinates of y = 125 mm and z = −92 mm:

3 3(0.63271 N/mm )(125 mm) (0.35197 N/mm )( 92 mm)

46.7073 MPa 46.7 MPa (T)xσ = + −

= = Ans. (b) Bending stress at K To compute the normal stress at K, use (y, z) coordinates of y = −125 mm and z = 92 mm:

3 3(0.63271 N/mm )( 125 mm) (0.35197 N/mm )(92 mm)

46.7073 MPa 46.7 MPa (C)xσ = − +

= − = Ans. (d) Orientation of neutral axis Since the zee shape has no axis of symmetry, it is helpful to determine the orientation of the neutral axis from Eq. (8-23) to help identify points of maximum stress.

4 4

4 4

tan

( 10.3528 kN-m)(74,208,333.3 mm ) ( 38.6370 kN-m)( 23,625,000 mm )( 38.6370 kN-m)(13,054,933.3 mm ) ( 10.3528 kN-m)( 23,625,000 mm )0.55629

29.09 (i.e., 29.09 CCW from

y z z yz

z y y yz

M I M IM I M I

β

β

+=

+

− + − −=− + − −

= −

∴ = − ° ° axis)z+


(c) Maximum tension and compression bending stresses Sketch the orientation of the neutral axis. By inspection, the points on the zee cross section that are farthest from the neutral axis are on the top and bottom surfaces at the outside corners of the web. To compute bending stresses at the upper point, use (y, z) coordinates of y = 125 mm and z = 8 mm:

3 3(0.63271 N/mm )(125 mm) (0.35197 N/mm )(8 mm)

81.9045 MPa 81.9 MPa (T) Maximum tension bending stressxσ = +

= = Ans. To compute bending stresses at the lower point, use (y, z) coordinates of y = −125 mm and z = −8 mm:

3 3(0.63271 N/mm )( 125 mm) (0.35197 N/mm )( 8 mm)

81.9045 MPa 81.9 MPa (C) Maximum compression bending stressxσ = − + −

= − = Ans.


8.73 The moment acting on the cross section of theunequal-leg angle has a magnitude of 14 kN-m and is oriented as shown in Fig. P8.73. Determine: (a) the bending stress at point H. (b) the bending stress at point K. (c) the maximum tension and the maximumcompression bending stresses in the cross section. (d) the orientation of the neutral axis relative to the+z axis. Show its location on a sketch of the crosssection. Fig. P8.73




(mm) (mm) (mm2) (mm) (mm3) horizontal leg 150 19 2,850 190.50 542,925.0 vertical leg 19 181 3,439 90.50 311,229.5 6,289 854,154.5

3

2

854,154.5 mm 135.82 mm (from bottom of shape to centroid)6,289 mm

64.18 mm (from top of shape to centroid)

i i

i

y Ay

AΣ

= = =Σ

=


Shape Area Ai


(mm2) (mm) (mm3) horizontal leg 2,850 75.0 213,750.0 vertical leg 3,439 9.5 32,670.5 6,289 246,420.5

3

2

246, 420.5 mm 39.18 mm (from right edge of shape to centroid)6,289 mm

110.82 mm (from left edge of shape to centroid)

i i

i

z Az

AΣ

= = =Σ

=



horizontal leg 85,737.50 54.68 8,522,088.15 8,607,825.65 vertical leg 9,388,756.58 −45.32 7,062,503.99 16,451,260.58

Moment of inertia about the z axis (mm4) = 25,059,086.23


Moment of inertia about the y axis: Shape IC d = zi – z d²A IC + d²A

(mm4) (mm) (mm4) (mm4) horizontal leg 5,343,750.00 35.82 3,656,188.87 8,999,938.87 vertical leg 103,456.58 −29.68 3,029,990.78 3,133,447.36

Moment of inertia about the y axis (mm4) = 12,133,386.23 Product of inertia about the centroidal axes:

Shape Iy’z’ yc zc Area Ai yc zc Ai Iyz (mm4) (mm) (mm) (mm2) (mm4) (mm4)

horizontal leg 0 54.68 35.82 2,850 5,582,117.16 5,582,117.16 vertical leg 0 −45.32 −29.68 3,439 4,625,790.65 4,625,790.65

Product of inertia (mm4) = 10,207,907.81 Since the angle shape has no axis of symmetry, Eq. (8-21) or Eq. (8-22) must be used to determine the bending stresses. Equation (8-21) will be used here.

( ) ( )2 2

6 4

4 4 4 2

6 4

4

(14 10 N-mm)(12,133,386.23 mm )(12,133,386.23 mm )(25,059,086.23 mm ) (10,207,907.81 mm )

(14 10 N-mm)(10, 207,907.81 mm )(12,133,386.23 mm )(25

z y y yz y z z yzx

y z yz y z yz


y

σ+ +

= − +− −

⎡ ⎤×= − ⎢ ⎥−⎣ ⎦

×+ 4 4 2

3 3

,059,086.23 mm ) (10,207,907.81 mm )

( 0.84997 N/mm ) (0.71509 N/mm )

z

y z

⎡ ⎤⎢ ⎥−⎣ ⎦

= − + To compute the normal stress at H, use (y, z) coordinates of y = 45.18 mm and z = 110.82 mm:

3 3( 0.84997 N/mm )(45.18 mm) (0.71509 N/mm )(110.82 mm)

40.8444 MPa 40.8 MPa (T)xσ = − +

= = Ans. (b) Bending stress at K To compute the normal stress at K, use (y, z) coordinates of y = 64.18 mm and z = −39.18 mm:

3 3( 0.84997 N/mm )(64.18 mm) (0.71509 N/mm )( 39.18 mm)

82.5685 MPa 82.6 MPa (C)xσ = − + −

= − = Ans. (d) Orientation of neutral axis Since the angle shape has no axis of symmetry, it is helpful to determine the orientation of the neutral axis from Eq. (8-23) to help identify points of maximum stress.

4

4

tan

(14 kN-m)(10,207,907.81 mm )(14 kN-m)(12,133,386.23 mm )0.84131

40.07 (i.e., 40.07 CW from axis)

y z z yz

z y y yz

M I M IM I M I

z

β

β

+=

+

=

=

∴ = ° ° +


(c) Maximum tension and compression bending stresses Sketch the orientation of the neutral axis. By inspection, the points on the angle cross section that are farthest from the neutral axis are on the top corner (at K) and on the inside corner of the vertical leg. To compute bending stresses at the lower point, use (y, z) coordinates of y = −135.82 mm and z = −20.18 mm:

3 3( 0.84997 N/mm )( 135.82 mm) (0.71509 N/mm )( 20.18 mm)

101.0129 MPa 101.0 MPa (T) Maximum tension bending stressxσ = − − + −

= = Ans. The maximum compression bending stress is

3 3( 0.84997 N/mm )(64.18 mm) (0.71509 N/mm )( 39.18 mm)

82.5685 MPa 82.6 MPa (C) Maximum compression bending stressxσ = − + −

= − = Ans.


8.74 The moment acting on the cross section ofthe zee shape has a magnitude of M = 4.75 kip-ft and is oriented as shown in Fig. P8.74. Determine:(a) the bending stress at point H. (b) the bending stress at point K. (c) the maximum tension and the maximumcompression bending stresses in the cross section.(d) the orientation of the neutral axis relative tothe +z axis. Show its location on a sketch of thecross section.

Fig. P8.74


Shape IC d = yi – y Area Ai d²A IC + d²A (in.4) (in.) (in.2) (in.4) (in.4)

top flange 0.0260 2.75 1.25 9.4531 9.4792 web 3.6458 0 1.75 0 3.6458 bottom flange 0.0260 −2.75 1.25 9.4531 9.4792


Shape IC d = zi – z Area Ai d²A IC + d²A (in.4) (in.) (in.2) (in.4) (in.4)

top flange 0.6510 1.075 1.25 1.4445 2.0956 web 68,266.7 0 1.75 0 0.0179 bottom flange 0.6510 −1.075 1.25 1.4445 2.0956


Shape yc zc Area Ai yc zc Ai Iyz (in.) (in.) (in.2) (in.4) (in.4)

top flange 2.75 1.075 1.25 3.6953 3.6953 web 0 0 1.75 0 0 bottom flange −2.75 −1.075 1.25 3.6953 3.6953

Product of inertia (in.4) = 7.3906 (a) Bending stress at H Since the zee shape has no axis of symmetry, Eq. (8-21) or Eq. (8-22) must be used to determine the bending stresses. Equation (8-21) will be used here.


( ) ( )2 2

4 4

4 4 4 2 4 4

( 4.75 kip-ft)(12 in./ft)(4.2091 in. ) ( 4.75 kip-ft)(12 in./ft)(7.3906 in. )(4.2091 in. )(22.6042 in. ) (7.3906 in. ) (4.2091 in. )(22.6042 in. )

z y y yz y z z yzx

y z yz y z yz


y

σ+ +

= − +− −

⎡ ⎤− −= − +⎢ ⎥−⎣ ⎦4 2

3 3

(7.3906 in. )

(5.92065 kips/in. ) (10.39584 kips/in. )

z

y z

⎡ ⎤⎢ ⎥−⎣ ⎦

= − To compute the normal stress at H, use (y, z) coordinates of y = 3 in. and z = 2.325 in.:

3 3(5.92065 kips/in. )(3 in.) (10.39584 kips/in. )(2.325 in.)

6.4084 ksi 6.41 ksi (C)xσ = −

= − = Ans. (b) Bending stress at K To compute the normal stress at K, use (y, z) coordinates of y = −2.50 in. and z = −2.325 in.:

3 3(5.92065 kips/in. )( 2.50 in.) (10.39584 kips/in. )( 2.325 in.)

9.3687 ksi 9.37 ksi (T)xσ = − − −

= = Ans. (d) Orientation of neutral axis Since the zee shape has no axis of symmetry, it is helpful to determine the orientation of the neutral axis from Eq. (8-23) to help identify points of maximum stress.

4

4

tan

( 4.75 kip-ft)(7.3906 in. )( 4.75 kip-ft)(4.2091 in. )1.7559

60.34 (i.e., 60.34 CW from axis)

y z z yz

z y y yz

M I M IM I M I

z

β

β

+=

+

−=−

=

∴ = ° ° +


(c) Maximum tension and compression bending stresses Sketch the orientation of the neutral axis. By inspection, the points on the zee cross section that are farthest from the neutral axis are on the top and bottom surfaces at the outside corners of the web. To compute bending stresses at the upper point, use (y, z) coordinates of y = 3 in. and z = −0.175 in.:

3 3(5.92065 kips/in. )(3 in.) (10.39584 kips/in. )( 0.175 in.)

19.5812 ksi 19.58 ksi (T) Maximum tension bending stressxσ = − −

= = Ans. To compute bending stresses at the lower point, use (y, z) coordinates of y = −3 in. and z = 0.175 in.:

3 3(5.92065 kips/in. )( 3 in.) (10.39584 kips/in. )(0.175 in.)

19.5812 ksi 19.58 ksi (C) Maximum compression bending stressxσ = − −

= − = Ans.


8.75 For the cross section shown in Fig. P8.75,determine the maximum magnitude of the bendingmoment M so that the bending stress in the zeeshape does not exceed 24 ksi.

Fig. P8.75


Shape IC d = yi – y Area Ai d²A IC + d²A (in.4) (in.) (in.2) (in.4) (in.4)

top flange 0.0260 2.75 1.25 9.4531 9.4792 web 3.6458 0 1.75 0 3.6458 bottom flange 0.0260 −2.75 1.25 9.4531 9.4792


Shape IC d = zi – z Area Ai d²A IC + d²A (in.4) (in.) (in.2) (in.4) (in.4)

top flange 0.6510 1.075 1.25 1.4445 2.0956 web 68,266.7 0 1.75 0 0.0179 bottom flange 0.6510 −1.075 1.25 1.4445 2.0956


Shape yc zc Area Ai yc zc Ai Iyz (in.) (in.) (in.2) (in.4) (in.4)

top flange 2.75 1.075 1.25 3.6953 3.6953 web 0 0 1.75 0 0 bottom flange −2.75 −1.075 1.25 3.6953 3.6953

Product of inertia (in.4) = 7.3906 Bending stresses in the section Since the zee shape has no axis of symmetry, Eq. (8-21) or Eq. (8-22) must be used to determine the bending stresses. Equation (8-21) will be used here. For this problem, My = 0 and from the sketch, Mz is observed to be negative. The bending stress in the zee cross section is described by:


( ) ( )2 2

4 4

4 4 4 2 4 4 4 2

4

(4.2091 in. ) (7.3906 in. )(4.2091 in. )(22.6042 in. ) (7.3906 in. ) (4.2091 in. )(22.6042 in. ) (7.3906 in. )

(0.103871 in. ) (

z y y yz y z z yzx

y z yz y z yz

z z

z


M My z

M y

σ

−

+ += − +

− −

⎡ ⎤ ⎡ ⎤− −= − +⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦= − 4

4 4

0.182383 in. )

(0.103871 in. ) (0.182383 in. )z

z

M z

M y z

−

− −⎡ ⎤= −⎣ ⎦ Orientation of neutral axis Since the angle shape has no axis of symmetry, it is helpful to determine the orientation of the neutral axis from Eq. (8-23) before beginning the stress calculations:

4

4

(7.3906 in. )tan 1.7559(4.2091 in. )

60.34 (i.e., 60.34 CW from axis)

y z z yz z

z y y yz z

M I M I MM I M I M

z

β

β

+= = =

+

∴ = ° ° +

Allowable moments based on maximum tension and compression bending stresses Sketch the orientation of the neutral axis. By inspection, the points on the zee cross section that are farthest from the neutral axis are on the top and bottom surfaces at the outside corners of the web. To compute bending stresses at the upper point, coordinates of y = 3 in. and z = −0.175 in. are used. Set the bending stress at this point to the 24-ksi allowable bending stress and solve for the moment magnitude:

4 4

3

(0.103871 in. )(3 in.) (0.182383 in. )( 0.175 in.) 24 ksi

24 ksi 69.86287 kip-in. 5.82 kip-ft0.343530 in.

x z

z

M

M

σ − −

−

⎡ ⎤= − − ≤⎣ ⎦

∴ ≤ = = Ans.


8.76 A stainless-steel spring (shown in Fig.P8.76) has a width of ¾ in. and a change inheight at section B from h1 = 3/8 in. to h2 = ¼ in. Determine the minimum acceptableradius r for the fillet if the stress-concentration factor must not exceed 1.40.

Fig. P8.76

Solution From Fig. 8.17b

0.375 in. 1.500.25 in.

For 1.40, 0.25

0.25(0.25 in.) 0.0625 in.

t

wh

rKh

rr hh

= =

= ≅

⎛ ⎞∴ = = =⎜ ⎟⎝ ⎠

Ans.


8.77 An alloy-steel spring (shown in Fig.P8.77) has a width of 25 mm and a change in height at section B from h1 = 75 mm to h2 = 60 mm. If the radius of the fillet between the twosections is r = 6 mm, determine the maximummoment that the spring can resist if themaximum bending stress in the spring must notexceed 100 MPa.

Fig. P8.77

Solution From Fig. 8.17b

75 mm 1.2560 mm6 mm 0.1060 mm

1.70t

whrh

K

= =

= =

∴ ≅ Section properties

3 41 (25 mm)(60 mm) 450,000 mm12

I = =

Maximum bending moment

2 4

100 MPa

(100 N/mm )(450,000 mm ) 882,353 N-mm 882 N-m(1.70)(60 mm/2)

x tM cK

I

M

σ = ≤

∴ ≤ = = Ans.


8.78 A stainless-steel bar ¾ in. wide by ⅜ in. deep has a pair of semicircular grooves cut in the edges of the bar from top to bottom. If the grooves have a radius of 1/16 in., determine the percent reduction instrength for flexural-type loadings.

Solution From Fig. 8.17d

1/16 in. 1.001/16 in.1/16 in. 0.105/8 in.

2.30t

drrb

K

= =

= =

∴ ≅

3/ 4 in.2.30 2.763/8 in.

1 1percentage reduction (100%) 1 (100%) 1 (100%) 63.8%2.76

g t

g gR

g

R

g

BK Kb

I IM M

c K c

M MM K

σ σ

⎛ ⎞= = =⎜ ⎟⎝ ⎠

= =

⎛ ⎞− ⎛ ⎞= = − = − =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ Ans.


8.79 A timber beam 150 mm wide by 200 mm deep has a 25-mm-diameter hole drilled from top to bottom of the beam on the centerline of the cross section. Determine the percent reduction in strength for flexure about the horizontal centroidal axis.

Solution From Fig. 8.17c

200 mm 8.0025 mm25 mm 0.17

150 mm2.55t

hddb

K

= =

= =

∴ ≅

150 mm2.55 3.06150 mm 25 mm

1 1percentage reduction (100%) 1 (100%) 1 (100%) 67.3%3.06

g t

g gR

g

R

g

bK Kb d

I IM M

c K c

M MM K

σ σ

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠

= =

⎛ ⎞− ⎛ ⎞= = − = − =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ Ans.


8.80 A 3-in.-diameter hot-rolled steel [σY = 53 ksi] shaft has a reduced diameter of 2.73in. for 12 in. of its length, as shown in Fig.P8.80. If the tool used to turn down thesection had a radius of 0.25 in., determinethe maximum allowable bending load P that can be applied to the end of the shaft if afactor of safety of 3 with respect to failureby yielding is specified.

Fig. P8.80

Solution For hot-rolled steel:

allow53 ksi 17.6667 ksi

FS 3Yσσ = = =

At support A

4 4

4

(3 in.) 3.97608 in.64

(22 in.) (3 in./2) 17.6667 ksi3.97608 in.

2.1286 kips

I

Mc PI

P

π

σ

= =

= = ≤

∴ ≤ (a) At the reduced section:

3 in. 0.25 in.1.09 0.092.75 in. 2.75 in.

w rh h

= = = =

From Fig. 8.17b, 1.50tK ≅

4 4

4

(2.75 in.) 2.80738 in.64

(1.50)(12 in.) (2.75 in./2) 17.6667 ksi2.80738 in.

2.0039 kips

t

I

Mc PKI

P

π

σ

= =

= = ≤

∴ ≤ (b) Compare the values of P in Eqs. (a) and (b) to find the maximum allowable load: max 2.00 kipsP = Ans.


8.81 A hot-rolled steel [σY = 360 MPa] bar with a rectangular cross section will be loaded as a cantileverbeam. The bar has a depth of h = 200 mm and has a 25-mm-diameter hole drilled from top to bottom of the beam on the centerline of a cross section where a bending moment of 60 kN-m must be supported. If a factor of safety of 4 with respect to failure by yielding is specified, determine the minimum acceptablewidth b for the bar.

Solution For hot-rolled steel:

allow360 MPa 90 MPa

FS 4Yσσ = = =

3200 mm 8

25 mm 12g t g gg

h b bh McK K I Kd b d I

σ= = = = =−

Refer to Fig. 8.17c. Since Kg and Ig depend on b, use a trial-and-error solution:

b ratio d/b Kt Kg Ig σ (mm) (106 mm4) (MPa) 200 0.125 2.65 3.03 133.3 136.4

250 0.10 2.70 3.00 166.7 108.0

300 0.08 2.75 3.00 200.0 90.0

minTherefore: 300 mmb ≅ Ans.

mechanics of materials solutions chapter08 probs65 81

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