8.65 A beam with a box cross section is subjected to a resultant moment magnitude of 2,100 N-m acting at the angle shown in Fig. P8.65. Determine: (a) the maximum tension and the maximum compression 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 cross section.
Fig. P8.65
Solution Section properties (90 mm)(55 mm)3 I y = 12
−
3
I z =
(55 mm)(90 )(90 mm) 12
−
(80 mm)(45 mm)3 12 (45 (45 mm mm)(80 )(80 mm)3 12
=
640,312.5 mm4
421, 250.0 = 1, 421,
mm4
Moment components M y = (2,1 (2,10 00 N-m)sin )sin 30° = 1,050 ,050 N-m
M z = −(2,100 NN-m)co )cos30 s30° = − 1,81 ,818.65 NN-m (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 at y = = 45 mm, z mm, z = = 27.5 mm: M y z M z y
σ x
=
= =
I y
−
I z
(1, 050 N-m) N-m)(27 (27.5 .5 mm) mm)(1, (1,000 000 mm/m mm/m)) 640, 312.5 mm4
−
(− 1,818.65 N-m N-m)( )(45 45 mm) mm)(1, (1,000 000 mm/m mm/m)) 1, 421, 250.0 mm4
45.0 45.095 952 2 MPa MPa + 57.5 57.582 827 7 MPa MPa
102.67 6779 79 = 102.
MPa = 102. 102.7 7 MPa MPa (T)
Ans.
Compute normal stress at y at y = = −45 mm, z mm, z = = −27.5 mm: M y z M z y
σ x
=
=
I y
−
I z
(1,050 NN-m)( − 27.5 27.5 mm)(1,00 1,000 0 mm mm/m) /m)
45.095 952 2 = −45.0
640, 312.5 mm4
−
(− 1,818. ,818.65 65 NN-m)( − 45 mm)(1,00 1,000 0 mm mm/m) /m) 1, 421, 250.0 mm4
MPa MPa − 57.5 57.582 827 7 MPa MPa
102.67 6779 79 = −102.
MPa MPa = 102. 102.7 7 MPa MPa (C) (C)
Ans.
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(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: M y I z (1, 050 N-m)( N-m)(1 1, 421, 421, 250.0 mm mm4 ) tan β = = = −1.2815 M z I y (−1,818.6 ,818.65 5 N-m) N-m)(6 (640 40,312 ,312.5 .5 mm mm4 ) ∴ β = −52.03°
(i.e., 52.03° CCW from + z axis)
Ans.
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8.66 The moment acting on the cross section of the T-beam has a magnitude of 22 kip-ft and is oriented as shown in Fig. P8.66. Determine: (a) the bending stress at point H point H . (b) the bending stress at point K point K . (c) the orientation of the neutral axis relative to the + z axis. Show its location on a sketch of the cross section.
Fig. P8.66
Solution Section properties Centroid location in y direction:
Shape
Width b (in.) 7.00 0.75
top flange stem
y
=
Σ yi Ai Σ Ai
=
95.80469 in.3 14.5625 in.2
Height h (in.) 1.25 7.75
(from bo bottom of shap hape to to ce centroid oid)
2.4211 in.
(from top of shape to to ce centroid)
Moment of inertia about the z axis: axis: d = y = yi – Shape I C C
d²A 4 (in. ) (in.) (in. ) 1.1393 1.7961 28.2273 29.0928 42.4956 −2.7039 4 Moment of inertia about the z the z axis axis (in. ) = 4
top flange stem
yi Ai 3 (in. ) 73.28125 22.52344 95.80469
6.5789 in i n.
= =
Area A Area Ai 2 (in. ) 8.7500 5.8125 14.5625
yi (from bottom) (in.) 8.375 3.875
Moment of inertia about the y axis: (1.25 1.25 in.) in.)(7 (7.00 .00 in.) in.)3 (7.7 (7.75 5 in.) in.)(0 (0.7 .75 5 in.) in.)3 I y = + 12 12
= 36.0016
I C + d²A C + 4 (in. ) 29.3666 71.5884 100.9550
in.4
Moment components M y = −(22 ki kip-ft)co )cos55 s55° = −12.6187 ki kip-ft = − 151.4242 kip-in.
M z = −(22 ki kip-ft)si )sin55° = −18.0213 ki kip-ft = − 216.2561 kip-in. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Copyright Act without the permission of the copyright copyright owner is unlawful. unlawful.
(a) Bending stress at H For a shape having at least one axis of symmetry, Eq. (8-24) can be used to determine bending bend ing stresses. To compute the normal stress at H at H , use the ( y, y, z ) coordinates y coordinates y = = 2.4211 in. and z and z = = −3.5 in.: M y z M z y
σ x
=
=
I y
−
I z
(−151. 151.42 4242 42 kipkip-in in.) .)(( − 3.50 3.50 in.) in.)
36.0016 in.4 14.721 211 1 ksi ksi + 5.18 5.1862 62 ksi ksi = 14.7 = 19.9 19.907 074 4
−
(− 216. 216.25 2561 61 kipkip-in in.) .)(2 (2.4 .421 211 1 in.) in.) 100.9550 in.4
ks ksi = 19.9 19.91 1 ksi ksi (T) (T)
Ans.
(b) Bending stress at K To compute the normal stress at K at K , use the ( y, y, z ) coordinates y coordinates y = = −6.5789 in. and z and z = = 0.375 in.: M y z M z y
σ x
=
=
I y
−
I z
(−151. 151.42 4242 42 kipkip-in in.) .)(0 (0.3 .375 75 in.) in.) 4
36.0016 in. = −1.57 1.5773 73 ksi ksi − 14.0 14.092 927 7 ksi ksi 15.670 700 0 = −15.6
−
(− 216. 216.25 2561 61 kipkip-in in.) .)(( − 6.57 6.5789 89 in.) n.) 100.9550 in.4
ks ksi = 15.6 15.67 7 ksi ksi (C)
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: M y I z (−151.4 151.424 242 2 kip-in kip-in.) .)((100.9 100.955 550 0 in.4 ) tan β = = = 1.9635 M z I y (−216.2 216.2561 561 kipkip-in. in.)( )(36 36.00 .0016 16 in. in.4 ) ∴ β =
63.01°
(i.e., 63.01° CW from + z axis)
Ans.
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8.67 A beam with a box cross section is subjected to a resultant moment magnitude of 75 kip-in. acting at the angle shown in Fig. P8.67. Determine: (a) the bending stress at point H point H . (b) the bending stress at point K point K . (c) the maximum tension and the maximum compression 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 cross section.
Fig. P8.67
Solution Section properties (4 in.) in.)(6 (6 in.) in.)3 I y = 12
−
3
I z =
(6 in. in.)( )(4 4 in.) in.) 12
−
(3.25 3.25 in.) in.)(5 (5.2 .25 5 in.) in.)3 12 (5.2 (5.25 5 in. in.)( )(3. 3.25 25 in. in.))3 12
= 32.8096
in.4
= 16.9814
in.4
Moment components M y = (75 (75 ki kip-i p-in.)cos20 n.)cos20° = 70.4 70.476 769 9 ki kip-in p-in..
M z = (75 (75 kip kip--in.)s in.)siin 20° = 25.6 25.651 515 5 ki kip-i p-in. (a) Bending stress at H For a shape having at least one axis of symmetry, Eq. (8-24) can be used to determine bending bend ing stresses. To compute the normal stress at H at H , use the ( y, y, z ) coordinates y coordinates y = = −2.0 in. and z and z = = −3.0 in.: M y z M z y
σ x
=
=
I y
−
I z
(70. (70.47 4769 69 kipkip-in in.) .)(( − 3.0 3.0 in.) in.)
32.8096 in.4 = −6.44 6.4442 42 ks ksi + 3.02 3.0211 11 ksi ksi = −3.4 3.4231 ksi ksi =
−
(25. (25.65 6515 15 kipkip-iin.) n.)( − 2.0 2.0 in.) in.) 16.9814 in.4
3.4 3.42 ks ksi (C (C)
Ans.
(b) Bending stress at K To compute the normal stress at K at K , use the ( y, y, z ) coordinates y coordinates y = = 2.0 in. and z and z = = 3.0 in.: M y z M z y
σ x
=
=
I y
−
I z
(70.476 (70.4769 9 kip-i kip-in.) n.)(3. (3.0 0 in.) in.) 4
32.8096 in. = 6.44 6.4442 42 ks ksi − 3.02 3.0211 11 ksi ksi 4231 ksi ksi = = 3.423
3.4 3.42 ks ksi (T)
−
(25.6515 (25.6515 kip-i kip-in.) n.)(2.0 (2.0 in.) in.) 16.9814 in.4 Ans.
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(c) Maximum bending stresses The maximum tension normal stress occurs at the ( y, y, z ) coordinates y coordinates y = = −2.0 in. and z and z = = 3.0 in.: M y z M z y
σ x
=
=
I y
−
I z
(70. (70.47 4769 69 kip kip--in.) in.)((3.0 3.0 in.) in.)
32.8096 in.4 6.4442 42 ks ksi + 3.02 3.0211 11 ksi ksi = 6.44 4653 = 9.465
−
(25. (25.65 6515 15 kip kip--in.) in.)(( − 2.0 2.0 in.) in.) 16.9814 in.4
ksi ksi = 9.4 9.47 ks ksi (T)
Ans.
The maximum compression normal stress occurs at the ( y, y, z ) coordinates y coordinates y = = 2.0 in. and z and z = = −3.0 in.: M y z M z y
σ x
=
=
I y
−
I z
(70. (70.47 4769 69 kip kip--in.) in.)(( − 3.0 3.0 in.) in.) 4
32.8096 in. = −6.44 6.4442 42 ks ksi − 3.02 3.0211 11 ksi ksi 9.4653 = −9.4
−
(25. (25.65 6515 15 kip kip--in.) in.)(2 (2.0 .0 in. in.)) 16.9814 in.4
ksi = 9.47 ks ksi (C)
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: M y I z (70.4769 kip-i kip-in.)( n.)(16.9814 16.9814 in. in.4 ) tan β = = = 1.4220 M z I y (25.6515 kip-in.)( kip-in.)(32.8096 32.8096 in.4 ) ∴ β =
54.88°
(i.e., 54.88° CW from + z axis)
Ans.
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8.68 The moment acting on the cross section of the wide-flange beam has a magnitude of M of M = = 12 kN-m and is oriented as shown in Fig. P8.68. P 8.68. Determine: (a) the bending stress at point H point H . (b) the bending stress at point K point K . (c) the orientation of the neutral axis relative to the + z axis. Show its location on a sketch of the cross section.
Fig. P8.68
Solution Section properties Moment of inertia about the z axis: axis: d = y = yi – Shape I C C
d²A 4 (mm ) (mm) (mm ) 59,062.5 97.5 29,944,687.5 4,860,000 0 0 59,062.5 29,944,687.5 −97.5 4 Moment of inertia about the z the z axis axis (mm ) = 4
top flange web bottom flange
Moment of inertia about the y axis: (15 mm)( mm)(21 210 0 mm mm)3 (180 180 mm mm)(10 )(10 mm)3 I y = 2 + 12 12
=
I C C + d²A 4 (mm ) 30,003,750 4,860,000 30,003,750 64,867,500
23,167, 500 mm4
Moment components M y = (12 kN-m) si sin 35 3 5° = 6.8829 kN-m = 6.8829× 106 N-mm
M z = (12 kN-m) co cos 35 35° = 9.8298 kN-m = 9.8298× 106 N-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 at H , use the ( y, coordinates y = = 105 mm and z and z = = −105 mm: y, z ) coordinates y M y z M z y
σ x
=
=
I y
−
I z
(6.8829 × 106 N-mm)( − 105 mm) 23,167, 500 mm4
31.194 948 8 = −31.1
MPa MPa − 15.9 15.911 114 4 MPa MPa
47.1062 = −47.
MPa = 47.1 MPa (C)
−
(9.8298× 106 N-mm)(105 mm) 64, 867, 500 mm4 Ans.
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(b) Bending stress at K To compute the normal stress at K at K , use the ( y, y, z ) coordinates y coordinates y = = −105 mm and z and z = = 105 mm: M y z M z y
σ x
=
=
I y
−
I z
(6.8829 × 106 N-mm)(105 mm) 23,167, 500 mm4
31.194 948 8 = 31.1 =
−
(9.8298× 106 N-mm)( − 105 mm) 64, 867, 500 mm4
MPa MPa + 15.9 15.911 114 4 MPa MPa
47.1 7.1062 MPa MPa = 47. 47.1 MPa (T)
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: M y I z (6.8829 (6.8829 kN-m) kN-m)(64 (64,867, ,867, 500 mm mm4 ) tan β = = = 1.9605 M z I y (9.8298 (9.8298 kN-m kN-m)(23 )(23,167 ,167,, 500 mm4 ) ∴ β =
62.98°
(i.e., 62.98° CW from + z axis)
Ans.
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8.69 For the cross section shown in Fig. P8.69, determine the maximum magnitude of the bending moment M so that the bending stress in the wideflange shape does not exceed 165 MPa.
Fig. P8.69
Solution Section properties Moment of inertia about the z axis: axis: d = y = yi – Shape I C C
d²A 4 (mm ) (mm) (mm ) 59,062.5 97.5 29,944,687.5 4,860,000 0 0 59,062.5 29,944,687.5 −97.5 4 Moment of inertia about the z the z axis axis (mm ) = 4
top flange web bottom flange
Moment of inertia about the y axis: (15 mm)( mm)(21 210 0 mm mm)3 (180 180 mm mm)(10 )(10 mm)3 I y = 2 + 12 12 Moment components M y = M sin 35°
=
I C C + d²A 4 (mm ) 30,003,750 4,860,000 30,003,750 64,867,500
23,167, 500 mm4
M z = M cos 35°
Maximum bending moment magnitude The maximum tension bending stress should occur at point K point K , which has the ( y, y, z ) coordinates y coordinates y = = −105 mm and z and z = = 105 mm: M y z M z y M sin 35 35°(105 mm) M cos 35°( − 105 mm) σ x = − = − ≤ 165 MPa I y I z 23,167, 50 500 mm4 64, 86 867, 50 500 mm4
M ⎡⎣ 2.59957 × 10−6 mm−3 + 1.32595× 10−6 mm−3 ⎤⎦ ≤ 165 N/mm2 ∴ M ≤
42.0327 × 106 N-mm = 42.0 kN-m
Ans.
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8.70 The unequal-leg angle is subjected to a bending moment of M z = 20 kip-in. that acts at the orientation shown in Fig. P8.70. Determine: (a) the bending stress at point H point H . (b) the bending stress at point K point K . (c) the maximum tension and the maximum compression 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 cross section.
Fig. P8.70
Solution Section properties Centroid location in y direction:
Shape
Width b (in.) 0.375 2.625
upright leg bottom leg
y
=
Σ yi Ai Σ Ai
=
3.18457 in.3 2.4844 in.2
Height h (in.) 4.000 0.375
= 1.2818 =
Area A Area Ai 2 (in. ) 1.5000 0.9844 2.4844
yi (from bottom) (in.) 2.00 0.1875
yi Ai 3 (in. ) 3.00 0.18457 3.18457
in.(from bottom of shape to centroid)
2.7182 in in.
(from top of shape to to centroid)
Centroid location in z direction: direction:
Shape upright leg bottom leg
z =
Σ zi Ai Σ Ai
=
Area A Area Ai 2 (in. ) 1.5000 0.9844 2.4844 1.94243 in.3 2.4844 in.2
z i (from right edge) (in.) 0.1875 1.6875
= 0.7818 =
in in.
2.2182 in.
Moment of inertia about the z axis: axis: d = y = yi – Shape I C C
(from right edge of shape to centroid) (from left edge of shape to centroid)
d²A 4 (in. ) (in.) (in. ) 2.000 0.7182 0.77372 0.011536 1.17881 −1.0943 4 Moment of inertia about the z the z axis axis (in. ) = 4
upright leg bottom leg
z i Ai 3 (in. ) 0.2813 1.6612 1.94243
I C C + d²A 4 (in. ) 2.7737 1.1903 3.9640
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Moment of inertia about the y axis: Shape I C d = z = z i – z z d²A C 4 4 (in. ) (in.) (in. ) upright leg 0.017578 0.52979 −0.5943 bottom leg 0.565247 0.9057 0.80750 4 Moment of inertia about the y the y axis axis (in. ) = Product of inertia about the centroidal axes: Shape I y’z’ yc 4 (in. ) (in.) upright leg 0 0.7182 bottom leg 0 −1.0943
z c (in.) −0.5943 0.9057
I C + d²A C + 4 (in. ) 0.5474 1.3727 1.9201
Area A Area Ai yc z c Ai 2 4 (in. ) (in. ) 1.5000 −0.6402 0.9844 −0.9757 4 Product of inertia (in. ) =
I yz 4 (in. ) −0.6402 −0.9757 −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 b e used here. Note that the bending moment component about the y y axis is zero (i.e., M y = 0); therefore, the first first term in Eq. (8-22) is eliminated. To compute the normal stress at H at H , use ( y, y, z ) coordinates of y of y = = 2.7182 in. and z and z = = −0.4068 in.:
⎡ − I y y + I yz z ⎤ σ x = ⎢ M z 2 ⎥ I I I − ⎢⎣ y z yz ⎥⎦ ⎡ −(1.9201 201 in.4 )(2. (2.7182 in. in.) + (− 1.6159 159 in. in.4 )(− 0.4068 in.)⎤ =⎢ ⎥ (20 kip-in.) (1.9201 in.4 )(3.9640 in in.4 ) − (− 1.6159 in in.4 )2 ⎣ ⎦ ⎡ −4.5619 in.5 ⎤ (20 kip-in.) =⎢ 8 ⎥ 5.0001 in. ⎣ ⎦ = −18.2469
ksi = 18.25 ksi (C)
Ans.
(b) Bending stress at K To compute the normal stress at K at K , use ( y, y, z ) coordinates of y of y = = −0.9068 in. and z and z = = 2.2182 in.:
⎡ − I y y + I yz z ⎤ σ x = ⎢ M z 2 ⎥ I I I − ⎢⎣ y z yz ⎥⎦ ⎡ −(1.9201 201 in.4 )( − 0.906 9068 in.) + (− 1.6 1.6159 in. in.4 )(2. (2.218 2182 in.)⎤ =⎢ ⎥ (20 kip-in.) (1.9201 in.4 )(3.9640 in in.4 ) − (− 1.6159 in in.4 )2 ⎣ ⎦ ⎡ −1.8432 in.5 ⎤ (20 kip-in.) =⎢ 8 ⎥ 5.0001 in. ⎣ ⎦ = −7.3728
ksi = 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:
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tan β =
M y I z
+ M z I yz
M z I y
+ M y I yz
=
(20 (20 ki kip-in p-in.) .)(( − 1.61 1.6159 59 in. in.4 ) (20 kip-in.)( kip-in.)(1.9201 1.9201 in.4 )
∴ β = −40.08°
= −0.8416
(i.e., 40.08° CCW from + z axis)
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 the corner of the angle. The bending stress stress at H at H has already been computed. To compute the normal stress stress at the corner of the angle, use y, y ( , z ) coordinates of y of y = = −1.2818 in. and z and z = = −0.7818 in.
⎡ − I y y + I yz z ⎤ M z σ x = ⎢ 2 ⎥ − I I I ⎢⎣ y z yz ⎥⎦ ⎡ −(1.9201 in.4 )( − 1.2818 in.) + (− 1.6159 in.4 )(− 0.7818 in.)⎤ =⎢ ⎥ (20 kip-in.) (1.9201 in.4 )(3.9640 in in.4 ) − (− 1.6159 in in.4 )2 ⎣ ⎦ ⎡ 3.7245 in.5 ⎤ (20 kip-in.) =⎢ 8 ⎥ 5.0001 in. ⎣ ⎦ = 14.8977
ksi = 14.90 ksi (T)
Therefore, the maximum compression bending stress is:
σ x
=
18.25 ksi (C)
Ans.
and the maximum tension bending bend ing stress is:
σ x
=
14.90 ksi (T)
Ans.
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8.71 For the cross section shown in Fig. P8.71, determine the maximum magnitude of the bending moment M moment M so so that the bending stress in the unequalleg angle shape does not exceed 24 ksi.
Fig. P8.71
Solution Section properties Centroid location in y direction:
Shape
Width b (in.) 0.375 2.625
upright leg bottom leg
y
=
Σ yi Ai Σ Ai
=
3.18457 in.3 2.4844 in.2
Height h (in.) 4.000 0.375
= 1.2818 =
Area A Area Ai 2 (in. ) 1.5000 0.9844 2.4844
yi (from bottom) (in.) 2.00 0.1875
yi Ai 3 (in. ) 3.00 0.18457 3.18457
in.(from bottom of shape to centroid)
2.7182 in in.
(from top of shape to to centroid)
Centroid location in z direction: direction:
Shape upright leg bottom leg
z =
Σ zi Ai Σ Ai
=
Area A Area Ai 2 (in. ) 1.5000 0.9844 2.4844 1.94243 in.3 2.4844 in.2
z i (from right edge) (in.) 0.1875 1.6875
= 0.7818 =
in in.
2.2182 in.
Moment of inertia about the z axis: axis: d = y = yi – Shape I C C
(from right edge of shape to centroid) (from left edge of shape to centroid)
d²A 4 (in. ) (in.) (in. ) 2.000 0.7182 0.77372 0.011536 1.17881 −1.0943 4 Moment of inertia about the z the z axis axis (in. ) = 4
upright leg bottom leg
z i Ai 3 (in. ) 0.2813 1.6612 1.94243
I C C + d²A 4 (in. ) 2.7737 1.1903 3.9640
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Moment of inertia about the y axis: Shape I C d = z = z i – z z d²A C 4 4 (in. ) (in.) (in. ) upright leg 0.017578 0.52979 −0.5943 bottom leg 0.565247 0.9057 0.80750 4 Moment of inertia about the y the y axis axis (in. ) = Product of inertia about the centroidal axes: Shape I y’z’ yc 4 (in. ) (in.) upright leg 0 0.7182 bottom leg 0 −1.0943
z c (in.) −0.5943 0.9057
I C + d²A C + 4 (in. ) 0.5474 1.3727 1.9201
Area A Area Ai yc z c Ai 2 4 (in. ) (in. ) 1.5000 −0.6402 0.9844 −0.9757 4 Product of inertia (in. ) =
I yz 4 (in. ) −0.6402 −0.9757 −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: M y I z + M z I yz (20 (20 ki kip-in p-in.) .)(( − 1.61 1.6159 59 in. in.4 ) tan β = = = −0.8416 M z I y + M y I yz (20 kip-in.)( kip-in.)(1.9201 1.9201 in.4 ) ∴ β = −40.08°
(i.e., 40.08° CCW from + z axis)
Allowable moments based on maximum tension and compression bending stresses Sketch the orientation of the the neutral axis. By inspection, the points on the angle cross section that are farthest from the neutral axis are point H point H and and the corner of the angle. To compute the normal stress at H , use ( y, y, z ) coordinates of y of y = = 2.7182 in. and z and z = = −0.4068 in.:
⎡ − I y y + I yz z ⎤ σ x = ⎢ 2 ⎥ I I I − ⎣⎢ y z yz ⎦⎥
z
⎡ −(1.92 .9201 in.4 )(2. (2.7182 182 in. in.)) + (− 1.6159 in in.4 )(−0.4068 in.) n.)⎤ =⎢ ⎥ M z 4 4 4 2 ( 1 . 9 2 0 1 i n . ) ( 3 . 9 6 4 0 in i n . ) ( 1 . 6 1 5 9 in i n . ) − − ⎣ ⎦
⎡ −4.5619 in.5 ⎤ M z =⎢ 8 ⎥ 5.0001 in. ⎣ ⎦
= ( −0.9124
in in.−3 )M z
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Therefore, based on an allowable bending stress of 24 ksi at H at H , the maximum magnitude of M of M z is: is: 3 (0.9124 in.− ) M z ≤ 24 ksi ∴ M z ≤
26.3054 kip-in.
(a)
To compute the normal stress at the corner of the angle, use ( y, y, z ) coordinates of y of y = = −1.2818 in. and z and z = = −0.7818 in.
⎡ − I y y + I yz z ⎤ σ x = ⎢ z 2 ⎥ ⎢⎣ I y I z − I yz ⎥⎦ ⎡ 3.7245 in.5 ⎤ M z =⎢ 8 ⎥ ⎣ 5.0001 in. ⎦
⎡ −(1.9201 in.4 )( − 1.2818 in.) + (− 1.6159 in.4 )(− 0.7818 in.)⎤ =⎢ ⎥ M z 4 4 4 2 ( 1 . 9 2 0 1 i n . ) ( 3 . 9 6 4 0 in i n . ) ( 1 . 6 1 5 9 in i n . ) − − ⎣ ⎦ =
(0.7449 in. in.−3 )M z
Therefore, based on the bending stress at the corner of the angle, the maximum magnitude of M of M z is: −3 (0.7449 in. ) M z ≤ 24 ksi ∴ M z ≤ 32.2197
kip-in.
(b)
Maximum bending moment M z Compare the results in Eqs. (a) and (b) to find that the maximum bending moment that can be applied to the angle shape is:
M z ≤ 26.3 kip-in.
Ans.
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8.72 The moment acting on the cross section o the unequal-leg angle has a magnitude of M of M = = 20 kip-in. and is oriented as shown in Fig. P8.72. Determine: (a) the bending stress at point H point H . (b) the bending stress at point K point K . (c) the maximum tension and the maximum compression 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 cross section.
Fig. P8.72
Solution Moment of inertia about the z axis: axis:
Shape top flange web bottom flange
I C C (mm4) 130,208.3 10,666,666.7 130,208.3
d = y = yi – Area A Area Ai I C d²A C + d²A 2 4 (mm) (mm ) (mm ) (mm4) 112.5 2,500 31,640,625.0 31,770,883.3 0 3,200 0 10,666,666.7 2,500 31,640,625.0 31,770,883.3 −112.5 4 Moment of inertia about the z the z axis axis (mm ) = 74,208,333.3
Moment of inertia about the y axis: Shape I C d = z = z i – z z Area A Area Ai d²A I C + d²A C + C 4 2 4 4 (mm ) (mm) (mm ) (mm ) (mm ) top flange 2,083,333.3 2,500 4,410,000 6,493,333.3 −42.0 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 4 Moment of inertia about the y the y axis axis (mm ) = 13,054,933.3 Product of inertia about the centroidal axes: Shape yc z c (mm) (mm) top flange 112.5 −42.0 web 0 0 bottom flange 42.0 −112.5
Area A Area Ai yc z c Ai 2 4 (mm ) (mm ) 2,500 −11,812,500 3,200 0 2,500 −11,812,500 4 Product of inertia (mm ) =
I yz 4 (mm ) −11,812,500 0 −11,812,500 −23,625,000
Moment components M y = −(40 kN kN-m) si sin 15° = −10.3528 kN-m = − 10.3528× 106 N-mm
M z = −(40 kN kN-m) co cos15° = −38.6370 kN kN-m = − 38.6370× 106 N-mm Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Copyright Act without the permission of the copyright copyright owner is unlawful. unlawful.
(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. ( M z I y + M y I yz ) y ( M y I z + M z I yz ) z
σ x
=−
I y I z
+
2
− I yz
I yIz
2
− I yz
⎡ (−38.6370 × 106 N-mm)(13,054,933.3 mm4 ) + (− 10.3528× 106 N-mm)(− 23,625,000 mm4 )⎤ ⎥ y 4 4 4 2 (13, 13,054, 054,93 933. 3.3 3 mm mm )(74 )(74,2 ,208 08,,333. 333.3 3 mm mm ) − (− 23,6 23,625 25,,000 000 mm mm ) ⎣ ⎦ ⎡ (−10.3528 ×106 N-mm)(74,208,333.3 mm4 ) + (−38.6370 × 106 N-mm)( m)(− 23,625,000 mm4 ) ⎤ +⎢ ⎥ z 4 4 4 2 ( 13,0 13 ,054 54, , 933. 93 3.3 3 mm m m ) (74, (7 4,20 208, 8,33 333. 3.3 3 mm m m ) ( 23, 23 , 625, 62 5,00 000 0 mm m m ) − − ⎣ ⎦
= −⎢
3
= (0.63271 N/mm
) y + (0.35197 N/ N/mm3 )z
To compute the normal no rmal stress at H at H , use ( y, y, z ) coordinates of y of y = = 125 mm and z and z = = −92 mm: 3 3 σ x = (0.6 (0.632 3271 71 N/m N/mm )(125 125 mm mm) + (0.3 (0.351 5197 97 N/m N/mm m )(− 92 mm) =
46.7 6.7073 MPa = 46.7 MPa MPa (T)
Ans.
(b) Bending stress at K To compute the normal no rmal stress at K at K , use ( y, y, z ) coordinates of y of y = = −125 mm and z and z = = 92 mm: σ x = (0.6 (0.632 3271 71 N/m N/mm3 )(−125 125 mm mm) + (0.3 (0.351 5197 97 N/mm /mm3 )(92 )(92 mm)
46.7073 MPa = = −46.
46.7 6.7 MP MPa (C)
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 o f maximum stress. M y I z + M z I yz tan β = M z I y + M y I yz =
(−10.3 0.3528 kN kN-m)(74,208, 08,333.3 mm4 ) + (− 38. 38.6370 kN kN-m)( − 23,625,0 5,000 mm4 ) (−38.6 8.6370 kN kN-m)(13,0 3,054,933. 33.3 mm4 ) + (− 10.3528 528 kN kN-m)( − 23,625,0 5,000 mm4 )
= −0.55629 ∴ β = −29.09°
(i.e., 29.09° CCW from + z axis)
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(c) Maximum tension and compression bending stresses Sketch the orientation of the neutral axis. By inspection, the points on the zee cross section section that are farthest from the neutral axis are on the top top and bottom surfaces at the outside corners of the web. To compute bending stresses at the upper point, use ( y, y, z ) coordinates of y of y = = 125 mm and z and z = = 8 mm: 3 3 σ x = (0.6 (0.632 3271 71 N/mm N/mm )(12 )(125 5 mm) mm) + (0.3 (0.351 5197 97 N/mm N/mm )(8 )(8 mm) mm) = 81.9045
MP MPa = 81.9 MP MPa (T (T)
Maximum te tension be bending st stress
Ans.
To compute bending stresses at the lower point, use ( y, y, z ) coordinates of y of y = = −125 mm and z and z = = −8 mm: 3 3 mm) + (0.35197 N/ N/mm )(− 8 mm mm) σ x = (0.63271 N/mm )(−125 mm 81.904 045 5 = −81.9
MPa MPa = 81.9 81.9 MPa MPa (C) (C)
Maxi Maxim mum com compre pressio ssion n ben bendi ding ng stre stresss
Ans.
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8.73 The moment acting on the cross section of the unequal-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 point H . (b) the bending stress at point K point K . (c) the maximum tension and the maximum compression 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 cross section.
Fig. P8.73
Solution Section properties Centroid location in y direction:
Shape
Width b (mm) 150 19
horizontal leg vertical leg
y
=
Σ yi Ai Σ Ai
=
854,154.5 mm3 6,289 mm2
Height h (mm) 19 181
= 135. 135.82 82 = 64.18
Area A Area Ai 2 (mm ) 2,850 3,439 6,289
mm mm
mm
yi (from bottom) (mm) 190.50 90.50
yi Ai 3 (mm ) 542,925.0 311,229.5 854,154.5
(fr (from bott bottom om of shape hape to cent entroi roid) (from top of shape to centroid)
Centroid location in z direction: direction:
Shape
Area A Area Ai 2 (mm ) 2,850 3,439 6,289
horizontal leg vertical leg
z =
Σ zi Ai Σ Ai
=
246,420.5 mm3 6,289 mm2
z i (from right edge) (mm) 75.0 9.5
= 39.18
mm
= 110.82
mm mm
z i Ai 3 (mm ) 213,750.0 32,670.5 246,420.5 (from right edge of shape to centroid) (from left ed edge of sh shape to ce centroid)
Moment of inertia about the z axis: axis:
Shape horizontal leg vertical leg
d = y = yi – y y I C d²A C 4 4 (mm ) (mm) (mm ) 85,737.50 54.68 8,522,088.15 9,388,756.58 7,062,503.99 −45.32 4 Moment of inertia about the z the z axis axis (mm ) =
I C + d²A C + 4 (mm ) 8,607,825.65 16,451,260.58 25,059,086.23
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Moment of inertia about the y axis: Shape I C d = z = z i – z z d²A C 4 4 (mm ) (mm) (mm ) horizontal leg 5,343,750.00 35.82 3,656,188.87 vertical leg 103,456.58 −29.68 3,029,990.78 4 Moment of inertia about the y the y axis axis (mm ) = Product of inertia about the centroidal axes: Shape I y’z’ yc z c 4 (mm ) (mm) (mm) horizontal leg 0 54.68 35.82 vertical leg 0 −45.32 −29.68
I C C + d²A 4 (mm ) 8,999,938.87 3,133,447.36 12,133,386.23
Area A Area Ai yc z c Ai I yz 2 4 4 (mm ) (mm ) (mm ) 2,850 5,582,117.16 5,582,117.16 3,439 4,625,790.65 4,625,790.65 4 Product of inertia (mm ) = 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. ( M z I y + M y I yz ) y ( M y I z + M z I yz ) z σ x = − + 2 2 I y I z − I yz I y I z − I yz
⎡ ⎤ (14 × 106 N-mm)(12 )(12,1 ,133 33,,386. 386.2 23 mm4 ) y = −⎢ 4 4 4 2 ⎥ ( 12,1 12 ,133 33,3 ,386 86.2 .23 3 mm )(25 )( 25,0 ,059 59,0 ,086 86.2 .23 3 mm ) ( 10, 10 , 207,90 20 7,907. 7.81 81 mm ) − ⎣ ⎦ 6 4 ⎡ ⎤ (14 × 10 N-mm)( m)(10, 10, 207,9 207,907 07..81 mm ) z +⎢ 4 4 4 2 ⎥ ( 12,133 12, 133,38 ,386.2 6.23 3 mm )(25 , 059 05 9 ,086 ,0 86.2 .23 3 m m ) ( 10, 10 , 207,9 20 7,907 07. . 81 m m ) − ⎣ ⎦ = ( −0.84997
N/mm3 ) y + (0.71509 N/mm3 )z
To compute the normal stress at H at H , use ( y, y, z ) coordinates of y of y = = 45.18 mm and z and z = = 110.82 mm: 0.8499 997 7 N/m N/mm3 )(45 )(45.1 .18 8 mm) mm) + (0.7 (0.715 1509 09 N/m N/mm m3 )(11 )(110. 0.82 82 mm mm) σ x = (−0.84 =
40.8 0.8444 MP MPa = 40.8 0.8 MPa (T)
Ans.
(b) Bending stress at K To compute the normal stress at K at K , use ( y, y, z ) coordinates of y of y = = 64.18 mm and z and z = = −39.18 mm: 3 3 σ x = (−0.8499 4997 N/ N/mm )(64.18 mm mm) + (0. (0.71509 509 N/ N/mm )(− 39.1 9.18 mm mm)
82.5685 MPa = = −82.
82.6 MPa MPa (C)
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 po ints of maximum stress. M y I z + M z I yz tan β = M z I y + M y I yz = =
(14 kN-m)( kN-m)(10,207,907.81 10,207,907.81 mm mm4 ) (14 kN-m)( kN-m)(12,133, 12,133,386.23 386.23 mm4 ) 0.84131
∴ β =
40.07°
(i.e., 40.07° CW from + z axis)
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(c) Maximum tension and compression bending stresses Sketch the orientation of the 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 (at K ) and on the inside corner of the vertical leg.
To compute bending stresses at the lower point, use ( y, y, z ) coordinates of y of y = = −135.82 mm and z and z = = −20.18 mm: N/mm3 )(−135.82 mm mm) + (0.71509 N/ N/mm3 )(− 20.18 mm mm) σ x = (−0.84997 N/ = 101.0129
MP MPa = 101.0 MP MPa (T)
Maximum tension bending stress
Ans.
The maximum compression bending stress is σ x = (−0.8499 4997 N/ N/mm3 )(64.18 .18 mm mm) + (0.71509 509 N/m N/mm m3 )(− 39.18 mm mm) = −82.5685
MPa = 82.6 MPa (C)
Maximum compression bending stress
Ans.
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8.74 The moment acting on the cross section o the zee shape has a magnitude of M of M = = 4.75 kip-ft and is oriented as shown in Fig. P8.74. Determine: (a) the bending stress at point H point H . (b) the bending stress at point K point K . (c) the maximum tension and the maximum compression 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 cross section.
Fig. P8.74
Solution Moment of inertia about the z axis: axis:
Shape
d = y = yi –
Area A Area Ai d²A 2 4 (in.) (in. ) (in. ) 2.75 1.25 9.4531 0 1.75 0 1.25 9.4531 −2.75 4 Moment of inertia about the z the z axis axis (in. ) =
I C C + d²A 4 (in. ) 9.4792 3.6458 9.4792 22.6042
Moment of inertia about the y axis: Shape = z i – z Area A Area Ai I C d = z z d²A C 4 2 4 (in. ) (in.) (in. ) (in. ) top flange 0.6510 1.075 1.25 1.4445 web 68,266.7 0 1.75 0 bottom flange 0.6510 1.25 1.4445 −1.075 4 Moment of inertia about the y the y axis axis (in. ) =
I C C + d²A 4 (in. ) 2.0956 0.0179 2.0956 4.2091
top flange web bottom flange
I C C 4 (in. ) 0.0260 3.6458 0.0260
Product of inertia about the centroidal axes: Shape yc z c (in.) (in.) top flange 2.75 1.075 web 0 0 bottom flange −2.75 −1.075
Area A Area Ai yc z c Ai 2 4 (in. ) (in. ) 1.25 3.6953 1.75 0 1.25 3.6953 4 Product of inertia (in. ) =
I yz 4 (in. ) 3.6953 0 3.6953 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.
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σ x
=−
( M
z
Iy
M y I yz ) y
+
I y I z
−
I yz2
+
(M
I
y z
+
I yIz
M z I yz ) z 2
− I yz
⎡ (−4.75 ki ⎤ ⎡ (−4.75 ki ⎤ kip-ft)(12 in in./ft)(4.2091 in in.4 ) kip-ft)(12 in in./ft)(7.3906 in in.4 ) y = −⎢ + ⎢ ⎥ z 4 4 4 2 ⎥ 4 4 (7.3906 6 in.4 )2 ⎦ ⎣ (4.2091 in. )(22.6042 in. ) − (7.3906 in. ) ⎦ ⎣ (4.2091 in. )(22.6042 in. ) − (7.390 =
3
3
(5.92 5.9206 065 5 kips kips/i /in. n. ) y − (10.39 0.3958 584 4 kips ips/in. /in. )z
To compute the normal stress at H at H , use ( y, y, z ) coordinates of y of y = = 3 in. and z and z = = 2.325 in.: 3 3 (5.920 2065 65 kips/ kips/in in.. )(3 )(3 in.) in.) − (10.39 10.39584 584 kips kips/i /in. n. )(2.3 )(2.325 25 in.) in.) σ x = (5.9 .4084 = −6.40
ks ksi = 6.41 ksi ksi (C)
Ans.
(b) Bending stress at K To compute the normal stress at K at K , use ( y, of y = = −2.50 in. and z and z = = −2.325 in.: y, z ) coordinates of y σ x = (5.92 5.9206 065 5 kips kips//in. in.3 )(−2.50 2.50 in.) in.) − (10.3 10.395 9584 84 kips kips//in. in.3 )( − 2.32 2.325 5 in.) n.) = 9.3687
ks ksi = 9.37 ks ksi (T (T)
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 o f maximum stress. M y I z + M z I yz tan β = M z I y + M y I yz =
(−4.75 4.75 kipkip-ft ft)( )(7. 7.39 3906 06 in. in.4 ) (−4.75 4.75 kipkip-ft ft)( )(4. 4.20 2091 91 in. in.4 )
= 1.7559 ∴ β =
60.34°
(i.e., 60.34° CW from
+ z axis)
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(c) Maximum tension and compression bending stresses Sketch the orientation of the neutral axis. By inspection, the points on the zee cross section section that are farthest from the neutral axis are on the top top and bottom surfaces at the outside corners of the web. To compute bending stresses at the upper point, use ( y, y, z ) coordinates of y of y = = 3 in. and z and z = = −0.175 in.: 3 3 5.9206 065 5 kips kips/i /in. n. )(3 )(3 in.) in.) − (10.3 10.395 9584 84 kips kips/i /in. n. )( − 0.17 0.175 5 in.) in.) σ x = (5.92 = 19.5812
ksi = 19.58 ksi (T)
Maximum tension bending stress
Ans.
To compute bending stresses at the lower point, use ( y, y, z ) coordinates of y of y = = −3 in. and z and z = = 0.175 in.: σ x = (5.92 5.9206 065 5 kips kips/i /in. n.3 )(−3 in.) in.) − (10.3 10.395 9584 84 kips kips/i /in. n.3 )(0. )(0.17 175 5 in.) in.) = −19.5812
ksi = 19.58 ksi (C)
Maximum compression bending stress
Ans.
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8.75 For the cross section shown in Fig. P8.75, determine the maximum magnitude of the bending moment M so that the bending stress in the zee shape does not exceed 24 ksi.
Fig. P8.75
Solution Moment of inertia about the z axis: axis:
Shape
d = y = yi –
Area A Area Ai d²A 2 4 (in.) (in. ) (in. ) 2.75 1.25 9.4531 0 1.75 0 1.25 9.4531 −2.75 4 Moment of inertia about the z the z axis axis (in. ) =
I C C + d²A 4 (in. ) 9.4792 3.6458 9.4792 22.6042
Moment of inertia about the y axis: Shape = z i – z Area A Area Ai I C d = z z d²A C 4 2 4 (in. ) (in.) (in. ) (in. ) top flange 0.6510 1.075 1.25 1.4445 web 68,266.7 0 1.75 0 bottom flange 0.6510 1.25 1.4445 −1.075 4 Moment of inertia about the y the y axis axis (in. ) =
I C C + d²A 4 (in. ) 2.0956 0.0179 2.0956 4.2091
top flange web bottom flange
I C C 4 (in. ) 0.0260 3.6458 0.0260
Product of inertia about the centroidal axes: Shape yc z c (in.) (in.) top flange 2.75 1.075 web 0 0 bottom flange −2.75 −1.075
Area A Area Ai yc z c Ai 2 4 (in. ) (in. ) 1.25 3.6953 1.75 0 1.25 3.6953 4 Product of inertia (in. ) =
I yz 4 (in. ) 3.6953 0 3.6953 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, M problem, M y = 0 and from the sketch, M sketch, M z is is observed to be negative. The bending stress in the the zee cross section is described by:
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σ x
=−
( M
z
Iy
+
I y I z
M y I yz ) y −
2
I yz
+
(M
I
y z
+
I yIz
M z I yz ) z −
2
I yz
4 4 ⎡ ⎤ ⎡ ⎤ − M z (4.2091 in. ) − M z (7.3906 in. ) = −⎢ + y z ⎢ 4 4 4 2 ⎥ 4 4 4 2⎥ − − ( 4 . 2 0 9 1 i n . ) ( 2 2 . 6 0 4 2 i n . ) ( 7 . 3 9 0 6 i n . ) ( 4 . 2 0 9 1 i n . ) ( 2 2 . 6 0 4 2 i n . ) ( 7 . 3 9 0 6 i n . ) ⎣ ⎦ ⎣ ⎦
=
(0.103871 in.−4 ) M z y − (0.18 0.1823 2383 83 in. in.−4 ) M z z
= M z ⎡ (0.103871 in.
−4
⎣
) y − (0.182383 in.−4 )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: M y I z + M z I yz M z (7.3 (7.390 906 6 in. in.4 ) tan β = = = 1.7559 M z I y + M y I yz M z (4.2 (4.209 091 1 in. in.4 ) ∴ β =
60.34°
(i.e., 60.34° CW from
+ z axis)
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 section that are farthest from the neutral axis are on the top top and bottom surfaces at the outside corners of the web. To compute bending stresses at the upper point, coordinates of y of y = = 3 in. and z 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:
σ x = M z ⎡⎣ (0.103871 in.−4 )(3 in.) − (0.182383 in.−4 )( − 0.175 in.)⎤⎦ ≤ 24 ks ksi ∴ M z ≤
24 ksi 0.343530 in.−3
=
69.8 69.862 6287 87 kipkip-in in.. = 5.82 5.82 kipkip-ft ft
Ans.
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8.76 A stainless-steel spring (shown in Fig. P8.76) has a width of ¾ in. and a change in height at section B section B from from h1 = 3/8 in. to h2 = ¼ in. Determine the minimum acceptable radius r for the fillet if the stressconcentration factor must not exceed 1.40.
Fig. P8.76
Solution From Fig. 8.17b 8.17b w h
=
0.375 in. 0.25 in.
For K t = 1.40,
= 1.50
r h
≅
0.25
⎛ r ⎞ 0.25(0 (0.2 .25 5 in.) in.) = 0.06 0.0625 25 in. in. ⎟ h = 0.25 h ⎝ ⎠
∴r = ⎜
Ans.
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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 section B from from h1 = 75 mm to h2 = 60 mm. If the radius of the fillet between the two sections is r = = 6 mm, determine the maximum moment that the spring can resist if the maximum bending stress in the spring must not exceed 100 MPa.
Fig. P8.77
Solution From Fig. 8.17b 8.17b w h r h
=
=
75 mm 60 mm 6 mm
60 mm ∴ K t ≅ 1.70
= 1.25
=
0.10
Section properties 1 I = (25 (25 mm)(60 mm)3 12
=
450 450,000 ,000 mm4
Maximum bending moment M c σ x = K t ≤ 100 MPa I ∴ M ≤
(100 100 N/m N/mm m2 )(45 )(450,00 0,000 0 mm mm4 ) (1.70)(60 mm/2)
,353 = 882,35
N-mm = 882 NN-m
Ans.
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8.78 A stainless-steel bar ¾ in. wide by ⅜ in. deep has a pair of semicirc semicircular ular grooves grooves cut in the edges edges o the bar from top to bottom. If the grooves have a radius of 1/16 in., determine the percent reduction in strength for flexural-type loadings.
Solution From Fig. 8.17d 8.17d d
=
1/16 in.
r 1/16 in. r 1/16 in. b
=
5 /8 in. in. ∴ K t ≅ 2.30
K g
=
M
=
B K t b
σ I g c
=
= 1.00
=
0.10
in. ⎞ ⎛ 3/ 4 in. ⎟ = 2.76 3/8 in. ⎝ ⎠ σ I g
2.30 ⎜
M R
=
M
−
percentage reduction =
K g c
M R
M
⎛ 1 ⎞ 1 ⎞ ⎛ (100%) = ⎜ 1 − (100%) = ⎜ 1− ⎟ ⎟ (100%) = 63.8% ⎜ K g ⎟ 2.76 ⎝ ⎠ ⎝ ⎠
Ans.
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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 8.17c h d d
=
200 mm
=
25 mm 25 mm
b 150 mm ∴ K t ≅ 2.55
K g
=
M
=
= 8.00
=
0.17
150 mm ⎛ b ⎞ ⎛ ⎞ = 2.55 ⎜ ⎟ ⎟ = 3. 0 6 b d 1 5 0 m m 2 5 m m − − ⎝ ⎠ ⎝ ⎠ σ I g σ I g K t ⎜
c
M R
=
M
−
percentage reduction =
K g c
M R
M
⎛ 1 ⎞ 1 ⎞ ⎛ (100%) = ⎜ 1 − (100%) = ⎜ 1− ⎟ ⎟ (100%) = 67.3% ⎜ K g ⎟ 3.06 ⎝ ⎠ ⎝ ⎠
Ans.
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8.80 A 3-in.-diameter hot-rolled steel [σ Y Y = 53 ksi] shaft has a reduced diameter of 2.73 in. for 12 in. of its length, as shown in Fig. P8.80. If the tool used to turn down the section had a radius of 0.25 in., determine the maximum allowable bending load P load P that that can be applied to the end of the shaft if a factor of safety of 3 with respect to failure by yielding is specified.
Fig. P8.80
Solution For hot-rolled steel: σ Y 53 ksi
σ allow
=
FS
=
3
= 17.6667
ksi
At support A support A I =
σ =
π
(3 in.)4
64 Mc
=
= 3.97608
in.4
(22 (22 in. in.))P (3 in./ in./2) 2)
I 3.97608 in.4 ∴ P ≤ 2.1286 kips
At the reduced section: w 3 in. = = 1. 0 9 h 2.75 in.
≤ 17.6667
ksi
r h
=
0.25 in. 2.75 in.
=
(a)
0 .0 9
From Fig. 8.17b 8.17b, K t ≅ 1.50 I =
π
(2.75 in in.)4
=
2.80738 in in.4
64 Mc (1.50)( 1.50)(12 12 in.) in.)P (2.75 (2.75 in./2) in./2) = ≤ 17.6667 ksi σ = K t I 2.80738 in.4 ∴ P ≤ 2.0039 kips
(b)
Compare the values of P of P in in Eqs. (a) and (b) to find the maximum allowable load: P max
=
2.00 kips
Ans.
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8.81 A hot-rolled steel [σ Y = 360 MPa] bar with a rectangular cross section will be loaded as a cantilever Y = beam. The bar has a depth of h = 200 mm and has a 25-mm-diameter hole drilled from top top to bottom o the beam on the centerline of a cross section where a bending moment of 60 kN-m must be supported. I a factor of safety of 4 with respect to failure by yielding is specified, determine the minimum acceptable width b for the bar.
Solution For hot-rolled steel: σ Y 360 MPa
σ allow h d
=
=
FS
=
200 mm 25 mm
4
= 90
=8
MPa
K g
=
b b−d
Kt
Ig
=
b h3
σ = K g
12
Mc I g
Refer to Fig. 8.17c 8.17c. Since K Since K g and I and I g depend on b, use a trial-and-error solution: b (mm)
ratio d /b
200
0.125
2.65
250
0.10
300
0.08
Therefore:
K t
K g
I g 4 (10 mm )
(MPa)
3.03
133.3
136.4
2.70
3.00
166.7
108.0
2.75
3.00
200.0
90.0
6
bmin
≅
300 mm
σ
Ans.
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