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 (80 mm)(45 mm)3 Iy = − = 640,312.5 mm 4 12 12 3 (55 mm)(90 mm) (45 mm)(80 mm)3 Iz = − = 1, 421, 250.0 mm 4 12 12 Moment components M y = (2,100 N-m) sin 30° = 1, 050 N-m M z = −(2,100 N-m) cos 30° = −1,818.65 N-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 = 45 mm, z = 27.5 mm: M z M y σx = y − z Iy Iz (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 4 1, 421, 250.0 mm 4 = 45.0952 MPa + 57.5827 MPa =
= 102.6779 MPa = 102.7 MPa (T)
Ans.
Compute normal stress at y = −45 mm, z = −27.5 mm: M z M y σx = y − z Iy Iz (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 4 1, 421, 250.0 mm 4 = −45.0952 MPa − 57.5827 MPa =
= −102.6779 MPa = 102.7 MPa (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 I (1, 050 N-m)(1, 421, 250.0 mm 4 ) = −1.2815 tan β = y z = M z I y (−1,818.65 N-m)(640,312.5 mm 4 )
∴ 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. (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 cross section.
Fig. P8.66
Solution Section properties Centroid location in y direction:
Shape top flange stem
Width b (in.) 7.00 0.75
Height h (in.) 1.25 7.75
Σyi Ai
95.80469 in.3 y= = = 6.5789 in. 14.5625 in.2 ΣAi = 2.4211 in.
Area Ai (in.2) 8.7500 5.8125 14.5625
yi (from bottom) (in.) 8.375 3.875
yi Ai (in.3) 73.28125 22.52344 95.80469
(from bottom of shape to centroid) (from top of shape to centroid)
Moment of inertia about the z axis: d = yi – y Shape d²A IC 4 (in. ) (in.) (in.4) top flange 1.1393 1.7961 28.2273 stem 29.0928 −2.7039 42.4956 Moment of inertia about the z axis (in.4) =
IC + d²A (in.4) 29.3666 71.5884 100.9550
Moment of inertia about the y axis: (1.25 in.)(7.00 in.)3 (7.75 in.)(0.75 in.)3 Iy = + = 36.0016 in.4 12 12 Moment components M y = −(22 kip-ft) cos 55° = −12.6187 kip-ft = −151.4242 kip-in. M z = −(22 kip-ft) sin 55° = −18.0213 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 Act without the permission of the copyright owner is 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 stresses. To compute the normal stress at H, use the (y, z) coordinates y = 2.4211 in. and z = −3.5 in.: M z M y σx = y − z Iy Iz
(−151.4242 kip-in.)( − 3.50 in.) (−216.2561 kip-in.)(2.4211 in.) − 36.0016 in.4 100.9550 in.4 = 14.7211 ksi + 5.1862 ksi =
= 19.9074 ksi = 19.91 ksi (T)
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.: M z M y σx = y − z Iy Iz
(−151.4242 kip-in.)(0.375 in.) (−216.2561 kip-in.)( − 6.5789 in.) − 36.0016 in.4 100.9550 in.4 = −1.5773 ksi − 14.0927 ksi =
= −15.6700 ksi = 15.67 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 I (−151.4242 kip-in.)(100.9550 in.4 ) tan β = y z = = 1.9635 (−216.2561 kip-in.)(36.0016 in.4 ) MzIy
∴ 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. (b) the bending stress at 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.)(6 in.)3 (3.25 in.)(5.25 in.)3 Iy = − = 32.8096 in.4 12 12 3 (6 in.)(4 in.) (5.25 in.)(3.25 in.)3 Iz = − = 16.9814 in.4 12 12 Moment components M y = (75 kip-in.) cos 20° = 70.4769 kip-in. M z = (75 kip-in.) sin 20° = 25.6515 kip-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 stresses. To compute the normal stress at H, use the (y, z) coordinates y = −2.0 in. and z = −3.0 in.: M z M y σx = y − z Iy Iz
(70.4769 kip-in.)( − 3.0 in.) (25.6515 kip-in.)( − 2.0 in.) − 32.8096 in.4 16.9814 in.4 = −6.4442 ksi + 3.0211 ksi =
= −3.4231 ksi = 3.42 ksi (C)
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.: M z M y σx = y − z Iy Iz
(70.4769 kip-in.)(3.0 in.) (25.6515 kip-in.)(2.0 in.) − 32.8096 in.4 16.9814 in.4 = 6.4442 ksi − 3.0211 ksi =
= 3.4231 ksi = 3.42 ksi (T)
Ans.
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(c) Maximum bending stresses The maximum tension normal stress occurs at the (y, z) coordinates y = −2.0 in. and z = 3.0 in.: M z M y σx = y − z Iy Iz
(70.4769 kip-in.)(3.0 in.) (25.6515 kip-in.)( − 2.0 in.) − 32.8096 in.4 16.9814 in.4 = 6.4442 ksi + 3.0211 ksi =
= 9.4653 ksi = 9.47 ksi (T)
Ans.
The maximum compression normal stress occurs at the (y, z) coordinates y = 2.0 in. and z = −3.0 in.: M z M y σx = y − z Iy Iz (70.4769 kip-in.)( − 3.0 in.) (25.6515 kip-in.)(2.0 in.) − 32.8096 in.4 16.9814 in.4 = −6.4442 ksi − 3.0211 ksi =
= −9.4653 ksi = 9.47 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 I (70.4769 kip-in.)(16.9814 in.4 ) = 1.4220 tan β = y z = M z I y (25.6515 kip-in.)(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 = 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 +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: d = yi – y Shape d²A IC 4 (mm ) (mm) (mm4) top flange 59,062.5 97.5 29,944,687.5 web 4,860,000 0 0 bottom flange 59,062.5 −97.5 29,944,687.5 Moment of inertia about the z axis (mm4) =
IC + d²A (mm4) 30,003,750 4,860,000 30,003,750 64,867,500
Moment of inertia about the y axis: (15 mm)(210 mm)3 (180 mm)(10 mm)3 Iy = 2 + = 23,167,500 mm 4 12 12 Moment components M y = (12 kN-m) sin 35° = 6.8829 kN-m = 6.8829 ×106 N-mm
M z = (12 kN-m) cos 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, use the (y, z) coordinates y = 105 mm and z = −105 mm: M z M y σx = y − z Iy Iz
(6.8829 ×106 N-mm)( − 105 mm) (9.8298 ×106 N-mm)(105 mm) − 23,167,500 mm 4 64,867,500 mm 4 = −31.1948 MPa − 15.9114 MPa =
= −47.1062 MPa = 47.1 MPa (C)
Ans.
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(b) Bending stress at K To compute the normal stress at K, use the (y, z) coordinates y = −105 mm and z = 105 mm: M z M y σx = y − z Iy Iz
(6.8829 ×106 N-mm)(105 mm) (9.8298 ×106 N-mm)( − 105 mm) − 23,167,500 mm 4 64,867,500 mm 4 = 31.1948 MPa + 15.9114 MPa =
= 47.1062 MPa = 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 I (6.8829 kN-m)(64,867,500 mm 4 ) tan β = y z = = 1.9605 M z I y (9.8298 kN-m)(23,167,500 mm 4 )
∴ 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: d = yi – y Shape d²A IC 4 (mm ) (mm) (mm4) top flange 59,062.5 97.5 29,944,687.5 web 4,860,000 0 0 bottom flange 59,062.5 −97.5 29,944,687.5 Moment of inertia about the z axis (mm4) =
IC + d²A (mm4) 30,003,750 4,860,000 30,003,750 64,867,500
Moment of inertia about the y axis: (15 mm)(210 mm)3 (180 mm)(10 mm)3 Iy = 2 + = 23,167,500 mm 4 12 12 Moment components M y = M sin 35°
M z = M cos 35°
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: M z M y M sin 35°(105 mm) M cos 35°( − 105 mm) σx = y − z = − ≤ 165 MPa Iy Iz 23,167,500 mm 4 64,867,500 mm 4
M ⎡⎣ 2.59957 ×10−6 mm −3 + 1.32595 × 10−6 mm −3 ⎤⎦ ≤ 165 N/mm 2 ∴ 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 Mz = 20 kip-in. that acts at the orientation shown 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 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=
Σyi Ai ΣAi
=
Height h (in.) 4.000 0.375
Area Ai (in.2) 1.5000 0.9844 2.4844
yi (from bottom) (in.) 2.00 0.1875
yi Ai (in.3) 3.00 0.18457 3.18457
3.18457 in.3 = 1.2818 in.(from bottom of shape to centroid) 2.4844 in.2 = 2.7182 in.
(from top of shape to centroid)
Centroid location in z direction:
Shape
zi (from right edge) (in.) 0.1875 1.6875
Area Ai (in.2) upright leg 1.5000 bottom leg 0.9844 2.4844 Σzi Ai 1.94243 in.3 = = 0.7818 in. z= ΣAi 2.4844 in.2 = 2.2182 in.
zi Ai (in.3) 0.2813 1.6612 1.94243
(from right edge of shape to centroid) (from left edge of shape to centroid)
Moment of inertia about the z axis: d = yi – y Shape d²A IC 4 (in.) (in.4) (in. ) upright leg 2.000 0.7182 0.77372 bottom leg 0.011536 −1.0943 1.17881 Moment of inertia about the z axis (in.4) =
IC + d²A (in.4) 2.7737 1.1903 3.9640
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Moment of inertia about the y axis: d = zi – z Shape IC d²A 4 (in.) (in.4) (in. ) upright leg 0.017578 −0.5943 0.52979 bottom leg 0.565247 0.9057 0.80750 Moment of inertia about the y axis (in.4) = Product of inertia about the centroidal axes: Shape yc zc Iy’z’ (in.) (in.) (in.4) upright leg 0 0.7182 −0.5943 bottom leg 0 −1.0943 0.9057
IC + d²A (in.4) 0.5474 1.3727 1.9201
Area Ai yc zc Ai (in.2) (in.4) 1.5000 −0.6402 0.9844 −0.9757 Product of inertia (in.4) =
Iyz (in.4) −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 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.: ⎡ − I y + I yz z ⎤ Mz σx = ⎢ y 2 ⎥ ⎣⎢ I y I z − I yz ⎦⎥ ⎡ −(1.9201 in.4 )(2.7182 in.) + (−1.6159 in.4 )(−0.4068 in.) ⎤ =⎢ ⎥ (20 kip-in.) (1.9201 in.4 )(3.9640 in.4 ) − (−1.6159 in.4 ) 2 ⎣ ⎦ 5 ⎡ −4.5619 in. ⎤ =⎢ (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, use (y, z) coordinates of y = −0.9068 in. and z = 2.2182 in.: ⎡ − I y + I yz z ⎤ Mz σx = ⎢ y 2 ⎥ ⎢⎣ I y I z − I yz ⎥⎦ ⎡ −(1.9201 in.4 )( − 0.9068 in.) + (−1.6159 in.4 )(2.2182 in.) ⎤ =⎢ ⎥ (20 kip-in.) (1.9201 in.4 )(3.9640 in.4 ) − (−1.6159 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 kip-in.)( − 1.6159 in.4 ) = −0.8416 (20 kip-in.)(1.9201 in.4 )
∴ β = −40.08°
(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 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. ⎡ − I y + I yz z ⎤ Mz σx = ⎢ y 2 ⎥ ⎢⎣ I y I z − I 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.4 ) − (−1.6159 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 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 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=
Σyi Ai ΣAi
=
Height h (in.) 4.000 0.375
Area Ai (in.2) 1.5000 0.9844 2.4844
yi (from bottom) (in.) 2.00 0.1875
yi Ai (in.3) 3.00 0.18457 3.18457
3.18457 in.3 = 1.2818 in.(from bottom of shape to centroid) 2.4844 in.2 = 2.7182 in.
(from top of shape to centroid)
Centroid location in z direction:
Shape
zi (from right edge) (in.) 0.1875 1.6875
Area Ai (in.2) upright leg 1.5000 bottom leg 0.9844 2.4844 Σzi Ai 1.94243 in.3 = = 0.7818 in. z= ΣAi 2.4844 in.2 = 2.2182 in.
zi Ai (in.3) 0.2813 1.6612 1.94243
(from right edge of shape to centroid) (from left edge of shape to centroid)
Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (in. ) (in.) (in.4) upright leg 2.000 0.7182 0.77372 bottom leg 0.011536 −1.0943 1.17881 Moment of inertia about the z axis (in.4) =
IC + d²A (in.4) 2.7737 1.1903 3.9640
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Moment of inertia about the y axis: d = zi – z Shape IC d²A 4 (in. ) (in.) (in.4) upright leg 0.017578 −0.5943 0.52979 bottom leg 0.565247 0.9057 0.80750 Moment of inertia about the y axis (in.4) = Product of inertia about the centroidal axes: Shape Iy’z’ yc zc (in.4) (in.) (in.) upright leg 0 0.7182 −0.5943 bottom leg 0 −1.0943 0.9057
IC + d²A (in.4) 0.5474 1.3727 1.9201
Area Ai yc zc Ai (in.2) (in.4) 1.5000 −0.6402 0.9844 −0.9757 Product of inertia (in.4) =
Iyz (in.4) −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 I + M z I yz (20 kip-in.)( − 1.6159 in.4 ) = = −0.8416 tan β = y z (20 kip-in.)(1.9201 in.4 ) M z I y + M y I yz
∴ 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 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.: ⎡ − I y + I yz z ⎤ ⎡ −(1.9201 in.4 )(2.7182 in.) + (−1.6159 in.4 )(−0.4068 in.) ⎤ = σx = ⎢ y M ⎥ z ⎢ ⎥ Mz 2 (1.9201 in.4 )(3.9640 in.4 ) − (−1.6159 in.4 ) 2 ⎢⎣ I y I z − I yz ⎥⎦ ⎣ ⎦ ⎡ −4.5619 in.5 ⎤ =⎢ M z = (−0.9124 in.−3 )M z 8 ⎥ 5.0001 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 Act without the permission of the copyright owner is unlawful.
Therefore, based on an allowable bending stress of 24 ksi at H, the maximum magnitude of Mz is: (0.9124 in.−3 )M z ≤ 24 ksi ∴ M z ≤ 26.3054 kip-in.
(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. ⎡ − I y + I yz z ⎤ ⎡ −(1.9201 in.4 )( − 1.2818 in.) + (−1.6159 in.4 )(−0.7818 in.) ⎤ = σx = ⎢ y M z ⎢ ⎥ Mz 2 ⎥ (1.9201 in.4 )(3.9640 in.4 ) − (−1.6159 in.4 )2 ⎢⎣ I y I z − I yz ⎥⎦ ⎣ ⎦ ⎡ 3.7245 in.5 ⎤ =⎢ M z = (0.7449 in.−3 ) M z 8⎥ ⎣ 5.0001 in. ⎦
Therefore, based on the bending stress at the corner of the angle, the maximum magnitude of Mz is: (0.7449 in.−3 )M z ≤ 24 ksi
∴ M z ≤ 32.2197 kip-in.
(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: M z ≤ 26.3 kip-in. Ans.
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8.72 The moment acting on the cross section of the 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 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: d = yi – y Area Ai Shape IC d²A 4 2 (mm) (mm ) (mm4) (mm ) top flange 130,208.3 112.5 2,500 31,640,625.0 web 10,666,666.7 0 3,200 0 bottom flange 130,208.3 −112.5 2,500 31,640,625.0 Moment of inertia about the z axis (mm4) =
IC + d²A (mm4) 31,770,883.3 10,666,666.7 31,770,883.3 74,208,333.3
Moment of inertia about the y axis: d = zi – z Shape Area Ai IC + d²A IC d²A 4 2 4 (mm ) (mm) (mm ) (mm ) (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 4 Moment of inertia about the y axis (mm ) = 13,054,933.3 Product of inertia about the centroidal axes: Shape yc zc (mm) (mm) top flange 112.5 −42.0 web 0 0 bottom flange −112.5 42.0
Area Ai yc zc Ai Iyz 2 4 (mm ) (mm ) (mm4) 2,500 −11,812,500 −11,812,500 3,200 0 0 2,500 −11,812,500 −11,812,500 Product of inertia (mm4) = −23,625,000
Moment components M y = −(40 kN-m) sin15° = −10.3528 kN-m = −10.3528 ×106 N-mm
M z = −(40 kN-m) cos15° = −38.6370 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 Act without the permission of the copyright owner is 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 σx = − I y I z − I yz2 I y I z − I yz2 ⎡ (−38.6370 × 106 N-mm)(13,054,933.3 mm 4 ) + (−10.3528 ×106 N-mm)(−23,625,000 mm 4 ) ⎤ = −⎢ ⎥y (13,054,933.3 mm 4 )(74,208,333.3 mm 4 ) − (−23,625,000 mm 4 ) 2 ⎣ ⎦ 6 4 6 ⎡ (−10.3528 ×10 N-mm)(74,208,333.3 mm ) + (−38.6370 × 10 N-mm)(−23,625,000 mm 4 ) ⎤ +⎢ ⎥z (13,054,933.3 mm 4 )(74,208,333.3 mm 4 ) − (−23,625,000 mm 4 ) 2 ⎣ ⎦ = (0.63271 N/mm3 ) y + (0.35197 N/mm3 )z
To compute the normal stress at H, use (y, z) coordinates of y = 125 mm and z = −92 mm: σ x = (0.63271 N/mm3 )(125 mm) + (0.35197 N/mm3 )(−92 mm) = 46.7073 MPa = 46.7 MPa (T)
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: σ x = (0.63271 N/mm3 )(−125 mm) + (0.35197 N/mm3 )(92 mm)
= −46.7073 MPa = 46.7 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 maximum stress. M I + M z I yz tan β = y z M z I y + M y I yz
(−10.3528 kN-m)(74,208,333.3 mm 4 ) + (−38.6370 kN-m)( − 23,625,000 mm 4 ) (−38.6370 kN-m)(13,054,933.3 mm 4 ) + (−10.3528 kN-m)( − 23,625,000 mm 4 ) = −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 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: σ x = (0.63271 N/mm3 )(125 mm) + (0.35197 N/mm3 )(8 mm)
= 81.9045 MPa = 81.9 MPa (T)
Maximum tension bending stress
Ans.
To compute bending stresses at the lower point, use (y, z) coordinates of y = −125 mm and z = −8 mm: σ x = (0.63271 N/mm3 )(−125 mm) + (0.35197 N/mm3 )(−8 mm) = −81.9045 MPa = 81.9 MPa (C)
Maximum compression bending stress
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. (b) the bending stress at 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=
Σyi Ai ΣAi
=
Height h (mm) 19 181
Area Ai (mm2) 2,850 3,439 6,289
854,154.5 mm3 = 135.82 mm 6,289 mm 2 = 64.18 mm
yi (from bottom) (mm) 190.50 90.50
yi Ai (mm3) 542,925.0 311,229.5 854,154.5
(from bottom of shape to centroid) (from top of shape to centroid)
Centroid location in z direction:
Shape
Area Ai (mm2) 2,850 3,439 6,289
horizontal leg vertical leg
z=
Σ z i Ai ΣAi
=
zi (from right edge) (mm) 75.0 9.5
246, 420.5 mm3 = 39.18 mm 6,289 mm 2 = 110.82 mm
zi Ai (mm3) 213,750.0 32,670.5 246,420.5 (from right edge of shape to centroid) (from left edge of shape to centroid)
Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (mm) (mm4) (mm ) horizontal leg 85,737.50 54.68 8,522,088.15 vertical leg 9,388,756.58 −45.32 7,062,503.99 Moment of inertia about the z axis (mm4) =
IC + d²A (mm4) 8,607,825.65 16,451,260.58 25,059,086.23
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Moment of inertia about the y axis: Shape d = zi – z IC d²A (mm) (mm4) (mm4) horizontal leg 5,343,750.00 35.82 3,656,188.87 vertical leg 103,456.58 −29.68 3,029,990.78 Moment of inertia about the y axis (mm4) = Product of inertia about the centroidal axes: Shape yc zc Iy’z’ 4 (mm) (mm) (mm ) horizontal leg 0 54.68 35.82 vertical leg 0 −45.32 −29.68
IC + d²A (mm4) 8,999,938.87 3,133,447.36 12,133,386.23
Area Ai yc zc Ai Iyz 2 4 (mm ) (mm ) (mm4) 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 σx = − I y I z − I yz2 I y I z − I yz2 ⎡ ⎤ (14 ×106 N-mm)(12,133,386.23 mm 4 ) y = −⎢ 4 4 4 2⎥ ⎣ (12,133,386.23 mm )(25,059,086.23 mm ) − (10, 207,907.81 mm ) ⎦ ⎡ ⎤ (14 × 106 N-mm)(10, 207,907.81 mm 4 ) z +⎢ 4 4 4 2⎥ ⎣ (12,133,386.23 mm )(25,059,086.23 mm ) − (10, 207,907.81 mm ) ⎦ = (−0.84997 N/mm3 ) y + (0.71509 N/mm3 )z
To compute the normal stress at H, use (y, z) coordinates of y = 45.18 mm and z = 110.82 mm: σ x = (−0.84997 N/mm3 )(45.18 mm) + (0.71509 N/mm3 )(110.82 mm) = 40.8444 MPa = 40.8 MPa (T)
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: σ x = (−0.84997 N/mm3 )(64.18 mm) + (0.71509 N/mm3 )(−39.18 mm)
= −82.5685 MPa = 82.6 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 of maximum stress. M I + M z I yz tan β = y z M z I y + M y I yz
(14 kN-m)(10, 207,907.81 mm 4 ) (14 kN-m)(12,133,386.23 mm 4 ) = 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 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: σ x = (−0.84997 N/mm3 )(−135.82 mm) + (0.71509 N/mm3 )(−20.18 mm) = 101.0129 MPa = 101.0 MPa (T)
Maximum tension bending stress
Ans.
The maximum compression bending stress is σ x = (−0.84997 N/mm3 )(64.18 mm) + (0.71509 N/mm3 )(−39.18 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 of the 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 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: d = yi – y Shape Area Ai d²A IC 4 (in.) (in.2) (in.4) (in. ) top flange 0.0260 2.75 1.25 9.4531 web 3.6458 0 1.75 0 bottom flange 0.0260 −2.75 1.25 9.4531 Moment of inertia about the z axis (in.4) =
IC + d²A (in.4) 9.4792 3.6458 9.4792 22.6042
Moment of inertia about the y axis: d = zi – z Shape Area Ai d²A IC (in.) (in.2) (in.4) (in.4) top flange 0.6510 1.075 1.25 1.4445 web 68,266.7 0 1.75 0 bottom flange 0.6510 −1.075 1.25 1.4445 Moment of inertia about the y axis (in.4) =
IC + d²A (in.4) 2.0956 0.0179 2.0956 4.2091
Product of inertia about the centroidal axes: Shape zc yc (in.) (in.) top flange 2.75 1.075 web 0 0 bottom flange −2.75 −1.075
Area Ai yc zc Ai (in.4) (in.2) 1.25 3.6953 1.75 0 1.25 3.6953 Product of inertia (in.4) =
Iyz (in.4) 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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σx = −
(M I
z y
+ M y I yz ) y
I yIz − I
2 yz
+
(M
I + M z I yz ) z
y z
I y I z − I yz2
⎡ (−4.75 kip-ft)(12 in./ft)(4.2091 in.4 ) ⎤ ⎡ (−4.75 kip-ft)(12 in./ft)(7.3906 in.4 ) ⎤ = −⎢ + z y ⎢ 4 4 4 2⎥ 4 4 4 2⎥ ⎣ (4.2091 in. )(22.6042 in. ) − (7.3906 in. ) ⎦ ⎣ (4.2091 in. )(22.6042 in. ) − (7.3906 in. ) ⎦ = (5.92065 kips/in.3 ) y − (10.39584 kips/in.3 )z
To compute the normal stress at H, use (y, z) coordinates of y = 3 in. and z = 2.325 in.: σ x = (5.92065 kips/in.3 )(3 in.) − (10.39584 kips/in.3 )(2.325 in.) = −6.4084 ksi = 6.41 ksi (C)
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.: σ x = (5.92065 kips/in.3 )(−2.50 in.) − (10.39584 kips/in.3 )( − 2.325 in.)
= 9.3687 ksi = 9.37 ksi (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 maximum stress. M I + M z I yz tan β = y z M z I y + M y I yz
(−4.75 kip-ft)(7.3906 in.4 ) (−4.75 kip-ft)(4.2091 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 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.: σ x = (5.92065 kips/in.3 )(3 in.) − (10.39584 kips/in.3 )( − 0.175 in.)
= 19.5812 ksi = 19.58 ksi (T)
Maximum tension bending stress
Ans.
To compute bending stresses at the lower point, use (y, z) coordinates of y = −3 in. and z = 0.175 in.: σ x = (5.92065 kips/in.3 )(−3 in.) − (10.39584 kips/in.3 )(0.175 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: d = yi – y Shape Area Ai d²A IC 4 (in.) (in.2) (in.4) (in. ) top flange 0.0260 2.75 1.25 9.4531 web 3.6458 0 1.75 0 bottom flange 0.0260 −2.75 1.25 9.4531 Moment of inertia about the z axis (in.4) =
IC + d²A (in.4) 9.4792 3.6458 9.4792 22.6042
Moment of inertia about the y axis: d = zi – z Shape Area Ai d²A IC (in.) (in.2) (in.4) (in.4) top flange 0.6510 1.075 1.25 1.4445 web 68,266.7 0 1.75 0 bottom flange 0.6510 −1.075 1.25 1.4445 Moment of inertia about the y axis (in.4) =
IC + d²A (in.4) 2.0956 0.0179 2.0956 4.2091
Product of inertia about the centroidal axes: Shape zc yc (in.) (in.) top flange 2.75 1.075 web 0 0 bottom flange −2.75 −1.075
Area Ai yc zc Ai (in.4) (in.2) 1.25 3.6953 1.75 0 1.25 3.6953 Product of inertia (in.4) =
Iyz (in.4) 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, My = 0 and from the sketch, Mz is observed to be negative. The bending stress in the zee cross section is described by:
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σx = −
(M I
z y
+ M y I yz ) y
I yIz − I
2 yz
+
(M
I + M z I yz ) z
y z
I y I z − I yz2
⎡ ⎤ ⎡ ⎤ − M z (4.2091 in.4 ) − M z (7.3906 in.4 ) = −⎢ + y z ⎢ 4 4 4 2⎥ 4 4 4 2⎥ ⎣ (4.2091 in. )(22.6042 in. ) − (7.3906 in. ) ⎦ ⎣ (4.2091 in. )(22.6042 in. ) − (7.3906 in. ) ⎦ = (0.103871 in.−4 ) M z y − (0.182383 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 I + M z I yz M z (7.3906 in.4 ) tan β = y z = = 1.7559 M z I y + M y I yz M z (4.2091 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 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: σ x = M z ⎡⎣(0.103871 in.−4 )(3 in.) − (0.182383 in.−4 )( − 0.175 in.) ⎤⎦ ≤ 24 ksi
∴Mz ≤
24 ksi = 69.86287 kip-in. = 5.82 kip-ft 0.343530 in.−3
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 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 w 0.375 in. = = 1.50 h 0.25 in. r For K t = 1.40, ≅ 0.25 h ⎛r⎞ ∴ r = ⎜ ⎟ h = 0.25(0.25 in.) = 0.0625 in. ⎝h⎠
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 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 w 75 mm = = 1.25 h 60 mm r 6 mm = = 0.10 h 60 mm ∴ K t ≅ 1.70
Section properties 1 I = (25 mm)(60 mm)3 = 450,000 mm 4 12 Maximum bending moment Mc σ x = Kt ≤ 100 MPa I (100 N/mm 2 )(450,000 mm 4 ) ∴M ≤ = 882,353 N-mm = 882 N-m (1.70)(60 mm/2)
Ans.
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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 in strength for flexural-type loadings.
Solution From Fig. 8.17d d 1/16 in. = = 1.00 r 1/16 in. r 1/16 in. = = 0.10 b 5 / 8 in. ∴ K t ≅ 2.30
B ⎛ 3/ 4 in. ⎞ = 2.30 ⎜ ⎟ = 2.76 b ⎝ 3/8 in. ⎠ σI σI M= g MR = g c Kgc K g = Kt
percentage reduction =
⎛ M − MR 1 ⎞ 1 ⎞ ⎛ (100%) = ⎜ 1 − ⎟ (100%) = ⎜ 1 − ⎟ (100%) = 63.8% ⎜ ⎟ M K 2.76 ⎝ ⎠ g ⎠ ⎝
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 h 200 mm = = 8.00 d 25 mm d 25 mm = = 0.17 b 150 mm ∴ K t ≅ 2.55 150 mm ⎛ b ⎞ ⎛ ⎞ K g = Kt ⎜ ⎟ = 2.55 ⎜ ⎟ = 3.06 ⎝b−d ⎠ ⎝ 150 mm − 25 mm ⎠ σI σI M= g MR = g c Kgc
percentage reduction =
⎛ M − MR 1 ⎞ 1 ⎞ ⎛ (100%) = ⎜ 1 − ⎟ (100%) = ⎜ 1 − ⎟ (100%) = 67.3% ⎜ ⎟ M K 3.06 ⎝ ⎠ g ⎠ ⎝
Ans.
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8.80 A 3-in.-diameter hot-rolled steel [σY = 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 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 53 ksi = 17.6667 ksi σ allow = Y = FS 3 At support A
I=
π
(3 in.) 4 = 3.97608 in.4
64 Mc (22 in.)P(3 in./2) = ≤ 17.6667 ksi σ= I 3.97608 in.4 ∴ P ≤ 2.1286 kips At the reduced section: w 3 in. = = 1.09 h 2.75 in.
(a)
r 0.25 in. = = 0.09 h 2.75 in.
From Fig. 8.17b, K t ≅ 1.50 I=
π 64
(2.75 in.) 4 = 2.80738 in.4
Mc (1.50)(12 in.)P (2.75 in./2) = ≤ 17.6667 ksi I 2.80738 in.4 ∴ P ≤ 2.0039 kips
σ = Kt
Compare the values of P in Eqs. (a) and (b) to find the maximum allowable load: Pmax = 2.00 kips
(b)
Ans.
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8.81 A hot-rolled steel [σY = 360 MPa] bar with a rectangular cross section will be loaded as a cantilever beam. 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 acceptable width b for the bar.
Solution For hot-rolled steel: σ 360 MPa = 90 MPa σ allow = Y = FS 4 h 200 mm = =8 d 25 mm
Kg =
b Kt b−d
Ig =
bh3 12
σ = Kg
Mc Ig
Refer to Fig. 8.17c. Since Kg and Ig 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:
bmin ≅ 300 mm
Kt
Kg
σ
Ig (106 mm4)
(MPa)
3.03
133.3
136.4
2.70
3.00
166.7
108.0
2.75
3.00
200.0
90.0 Ans.
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