8.52 A steel pipe assembly supports a concentrated load of 17 kN as shown in Fig. P8.52. The outside diameter of the pipe is 142 mm and the wall thickness is 6.5 mm. Determine the normal stresses produced at points H points H and K and K .
Fig. P8.52
Solution Section properties d = D − 2t = 142 mm − 2(6.5 mm) = 129 mm π
π
⎡⎣ D 2 − d 2 ⎤⎦ = ⎡⎣ (142 mm mm)2 − (129 mm mm)2 ⎤⎦ = 2, 76 766.958 mm2 4 4
A = I z =
π
π
⎡⎣ D 4 − d 4 ⎤⎦ = ⎡⎣ (142 mm)4 − (129 mm)4 ⎤⎦ = 6, 36 364, 86 867 mm mm4 64 64
Internal forces and moments F = 17 kN = 17, 00 000 N
M z = (17,00 17,000 0 N)(3 N)(370 70 mm) mm) = 6,290, 6,290, 000 000 NN-mm Stresses σ axial
=
σ bending
F A =
=
17,000 N 2,766.958 mm2
M z c I z
=
=
6.144 MPa (C)
(6, 290,000 N-mm)( N-mm)(142 142 mm/2) 6,364,867 mm4
= ±70.165
MPa
Normal stress at H By inspection, the bending stress at H at H will will be compression; therefore, the normal stress at H at H is: is: σ H
= −6.144
MPa − 70.165 MPa = − 76.309 MPa = 76.3 MPa (C)
Ans.
Normal stress at K By inspection, the bending stress at K at K will will be tension; therefore, the normal stress at K at K is: is: σ K
= −6.144
MPa + 70.165 MPa = + 64.021 MPa = 64.0 MPa (T)
Ans.
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8.53 The screw of a clamp exerts a compressive force of 300 lb on the wood blocks. Determine the normal stresses produced at points H and K . The clamp cross-sectional dimensions at the section o interest are 1.25 in. by 0.375 in. thick.
Fig. P8.53
Solution Section properties (0.375 5 in.)( in.)(1. 1.250 250 in.) in.) = 0.468 0.468750 750 in. in.2 A = (0.37
I z =
(0.375 in.)(1.250 in.)3 12
=
0.061035 in.4
Internal forces and moments F = 300 lb
M z = (300 (300 lb) lb)(3 (3.7 .75 5 in. in. + 1.25 1.25 in./ in./2) 2) = 1,312.5 ,312.50 0 lblb-in in.. Stresses σ axial
=
σ bending
F
300 lb
= 640 psi (T) A 0.468750 in.2 M z c (1, 312.50 lb-in.)(1.250 in./2) =
=
I z
=
0.061035 in.4
= ±13,440
psi
Normal stress at H By inspection, the bending stress at H at H will will be tension; therefore, the normal stress at H at H is: is: σ H
=
640 psi + 13,440 ,440 psi = 14,08 ,080 ps psi = 14,08 ,080 psi (T)
Ans.
Normal stress at K By inspection, the bending stress at K at K will will be compression; therefore, the normal stress at K at K is: is: σ K
=
640 ps psi − 13, 44 440 ps psi = − 12, 80 800 ps psi = 12, 80 800 ps psi (C)
Ans.
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8.54 Determine the normal stresses produced at points H points H and K and K of of the pier support shown in Fig. P8.54a P8.54a.
Fig. P8.54a P8.54a
Fig. P8.54b P8.54b Cross section a–a
Solution Section properties A = (250 (250 mm)( mm)(50 500 0 mm) mm) = 125 125, 000 000 mm mm2
I z =
(250 mm)(500 mm)3 12
=
2.60417 × 109 mm4
Internal forces and moments F = 250 kN + 400 kN = 650 kN
M z = (25 (250 kN kN)(3.2 3.25 m) − (40 (400 kN kN)(2.2 2.25 m) = − 87.50 kN-m Stresses σ axial
=
σ bending
F A =
=
650, 650, 000 N 125,000 mm2
M z c I z
=
= 5.20
MPa (C)
(87.5 kN-m)(500 mm/2)(1,000)2 2.60417 × 109 mm4
= ±8.40
MPa
Normal stress at H By inspection, the bending stress at H at H will will be tension; therefore, the normal stress at H at H is: is: σ H
= −5.20
MPa + 8.40 MPa = 3.20 MPa = 3.20 MPa (T)
Ans.
Normal stress at K By inspection, the bending stress at K at K will will be compression; therefore, the normal stress at K at K is: is: σ K
= −5.20
MP MPa − 8.40 MP MPa = − 13.60 MP MPa = 13.60 MP MPa (C)
Ans.
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8.55 A tubular steel column CD CD supports horizontal cantilever arm ABC arm ABC , as shown in Fig. P8.55. Column CD CD has an outside diameter o 10.75 in. and a wall thickness of 0.365 in. Determine the maximum compression stress at the base of column CD. CD.
Fig. P8.55
Solution Section properties d = D − 2t = 10.750 in. − 2(0.365 in.) = 10.020 in.
A =
π
I z =
π
⎡⎣ D 2 − d 2 ⎤⎦ = ⎡⎣(10.750 in in.)2 − (10.020 in in.)2 ⎤⎦ = 11.908 in.2 4 4 π
π
⎡⎣ D 4 − d 4 ⎤⎦ = ⎡⎣(10.750 in in.)4 − (10.020 in in.)4 ⎤⎦ = 160.734 in in.4 64 64
Internal forces and moments F = 700 lb + 900 lb = 1, 60 600 lb
M = (70 (700 lb lb)(13 ft ft) + (900 lb) lb)((23 ft ft) = 29,800 ,800 lb lb-ft = 357,600 ,600 lb lb-in. Stresses σ axial
=
σ bending
F
1, 600 lb
= 134.36 psi (C) A 11.908 in.2 M c (357,600 lb-in.)(10.75 in./2) =
=
I
=
160.734 in.4
= ±11,958.27
psi
Maximum compression stress at base of column σ compression
= −134.36
ps psi − 11,95 ,958.27 ps psi = − 12,09 ,092.63 ps psi = 12.09 ks ksi (C (C)
Ans.
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8.56 Determine the normal stresses acting at points H points H and K and K for for the structure shown in Fig. P8.56a P8.56a. The cross-sectional dimensions of the vertical member are shown in Fig. P8.56b P8.56b.
Fig. P8.56b P8.56b Cross section
Fig. P8.56a P8.56a
Solution Section properties A = (4 in.)(8 in in.) = 32 in.2
I z =
(4 in.)(8 in.)3 12
= 170.6667
in.4
Internal forces and moments F = 1, 00 000 lb lb + 2, 20 200 lb lb = 3, 20 200 lb lb
M z = (1,000 lb) lb)((12 in. + 8 in./ in./2) 2) = 16,000 16,000 lblb-in in.. Stresses σ axial
=
σ bending
F
3, 200 lb
= 100 psi (C) A 32 in.2 M z c (16,000 lb-in.)(8 in./2) =
=
I z
=
170.6667 170.6667 in.4
= ±375
psi
Normal stress at H By inspection, the bending stress at H at H will will be compression; therefore, the normal stress at H at H is: is: σ H
= −100
psi − 375 psi = −475 psi = 475 psi (C)
Ans.
Normal stress at K By inspection, the bending stress at K at K will will be tension; therefore, the normal stress at K at K is: is: σ K
= −100
psi + 375 psi = 275 psi = 275 psi (T)
Ans.
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8.57 A W18 × 35 standard steel shape is subjected to a tension force P force P that that is applied 15 in. above the bottom surface of the wideflange shape as shown in Fig. P8.57. If the tension normal stress of the upper surface o the W-shape must be limited to 15 ksi, determine the allowable force P force P that that may be applied to the member.
Fig. P8.57
Solution Section properties (from Appendix B) Depth d = 17.7 in.
A = 10.3 in.2 I z = 510 in.4 Stresses σ axial
=
σ bending
F A =
=
P 10.3 in.2
M z c I z
=
P (15 in in. − 17.7 in in./2)(17.7 in in./2) 510 in.4
=
P (6.15 in in.)(8.85 in in.) 510 in.4
=
P (54.4275 in in.2 ) 510 in.4
Normal stress on the upper surface of the W-shape The tension normal stress on the upper surface is equal to the sum of the axial and bending stresses. Since these stresses are expressed in terms of the unknown force P , the tension normal stress is given by: P P (54. (54.42 4275 75 in.2 ) σ upper surface
=
10.3 in.2
+
= P (0.097087
510 in.4 in in.−2 + 0.106721 in.−2 )
in. in.−2 ) P The normal stress on the upper surface of the W-shape must be limited to 15 ksi; k si; therefore, −2 (0.203808 in. ) P ≤ 15 ksi = (0.2 (0.203 0380 808 8
∴ P ≤
15 ksi 0.203808 in.−2
=
73.6 kips
Ans.
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8.58 A WT305 × 41 standard steel shape is subjected to a tension force P force P that that is applied 250 mm above the bottom surface of the tee shape, as shown in Fig. P8.58. If the tension normal stress of the upper surface of the WT-shape must be limited to 120 MPa, determine the allowable force P force P that that may be applied to the member.
Fig. P8.58
Solution Section properties (from Appendix B) Depth d = 300 mm
Centroid y
= 88.9
mm (from flange to centroid)
A = 5,230 mm2 I z = 48.7 × 106 mm4 Stresses σ axial
=
σ bending
F
P
=
−4
= P (1.9120 × 10
2
mm−2 )
A 5,230 mm M z c P (25 (250 mm mm − 88. 88.9 mm mm)(300 mm mm − 88.9 mm mm) =
I z
=
=
48.7 × 106 mm4 P (161.1 mm)(211.1 mm) 48.7 × 106 mm4 −4
= P (6.9832 × 10
mm−2 )
Normal stress on the upper surface of the WT-shape The tension normal stress on the upper surface is equal to the sum of the axial and bending stresses. Since these stresses are expressed in terms of the unknown force P , the tension normal stress is given by: −4 mm−2 ) + P (6.9832 × 10−4 mm−2 ) σ upper surface = P (1.9120 × 10 −4
mm−2 ) P The normal stress on the upper surface of the W-shape must be limited to 15 ksi; k si; therefore, −4 −2 (8.8953 × 10 mm ) P ≤ 120 MPa = (8.8953 × 10
∴ P ≤
120 N/mm2 8.8953 × 10−4 mm−2
,903 = 134,90
N = 134.9 kN kN
Ans.
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8.59 A pin support consists of a vertical plate 60 mm wide by 10 mm thick. The pin carries a load of 750 N. Determine the normal stresses acting at points H points H and K and K for for the structure shown in Fig. P8.59.
Fig. P8.59
Solution Section properties A = (60 (60 mm mm)(10 )(10 mm) = 600 600 mm mm2
I =
(60 mm)(10 mm)(10 mm)3 12
= 5,000
mm4
Internal forces and moments F = 750 N
M = (75 (750 N) N)(30 mm + 10 mm mm/2) = 26,250 ,250 NN-mm Stresses σ axial
=
σ bending
F
750 N
= 1.25 MPa (T) A 600 mm2 M c (26, 250 N-mm)(10 N-mm)(10 mm/2) =
=
I
=
5,000 mm4
= ±26.25
MPa
Normal stress at H By inspection, the bending stress at H at H will will be compression; therefore, the normal stress at H at H is: is: σ H
= 1.25
MPa − 26.25 MPa = − 25.00 MPa = 25.0 MPa (C)
Ans.
Normal stress at K By inspection, the bending stress at K at K will will be tension; therefore, the normal stress at K at K is: is: σ K
= 1 .2 5
MPa + 26.25 MPa = 27.50 MPa = 27.5 MPa (T)
Ans.
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8.60 The tee shape shown in Fig. P8.60b P8.60b is used as a short post to support a load of P of P = = 2,000 lb. The load P load P is is applied at a distance of 5 in. from the surface of the flange, as shown in Fig. P8.60a P8.60a. Determine the normal stresses at points H points H and K and K , which are located on section a–a.
Fig. P8.60b P8.60b Cross-sectional dimensions Fig. P8.60a P8.60a
Solution Centroid location in x direction: direction:
Shape
width b (in.) 12 2
flange stem
x
=
Σ xi Ai Σ Ai
=
164 in.3 44 in.2
height h (in.) 2 10
3.7273 73 in. = 3.72 = 8.2727
in in.
Area A Area Ai 2 (in. ) 24 20 2 44 in.
xi (from left) (in.) 1 7
xi Ai 3 (in. ) 24 140 3 164 in.
(fro (from m lef leftt sid sidee to to cen centr troi oid) d) (from right side to centroid)
Moment of inertia about the z axis: axis: d = x = xi – x x Shape I C d²A C 4 4 (in. ) (in.) (in. ) flange 8 178.5160 −2.7273 stem 166.6667 3.2727 214.2113 4 Moment of inertia about the z the z axis axis (in. ) =
I C C + d²A 4 (in. ) 186.5160 380.8790 567.3940
Internal forces and moments F = 2,000 lb
M z = (2,000 (2,000 lb) lb)((5 in. in. + 3.72 3.7273 73 in.) in.) = 17, 17, 454. 454.6 6 lblb-in. in.
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Stresses σ axial
=
F A
σ H ,bending
=
σ K ,bending
=
−2,000
lb
= −45.4545 psi 44 in. M z x (17, 17, 454. 454.6 6 lb-i lb-in. n.)( )( − 3.727 3.7273 3 in.) in.)
=
2
=
I z M z x
=
I z
567.3940 in.4 (17, 454.6 lb-in.)(8.2727 in.) 567.3940 in.4
=
= −114.6620 psi
254.4910 psi
Normal stress at H σ H
= −45.4545
ps psi − 114.6620 ps psi = − 160.1165 ps psi = 160.1 psi (C)
Ans.
Normal stress at K σ K
= −45.4545
psi + 254.4910 psi = 209.0365 psi = 209 psi (T)
Ans.
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8.61 The tee shape shown in Fig. P8.61b P8.61b is used as a short post to support a load of P . The load P load P is is applied at a distance of 5 in. from the surface of the flange, as shown in Fig. P8.61a P8.61a. The tension and compression normal stresses in the post must be limited to 1,000 psi and 800 psi, respectively. Determine the maximum magnitude of load P that satisfies both the tension and compression stress limits.
Fig. P8.61b P8.61b Cross-sectional dimensions Fig. P8.61a P8.61a
Solution Centroid location in x direction: direction:
Shape
width b (in.) 12 2
flange stem
x
=
Σ xi Ai Σ Ai
=
164 in.3 44 in.2
height h (in.) 2 10
3.7273 73 in. = 3.72 = 8.2727
in in.
Area A Area Ai 2 (in. ) 24 20 2 44 in.
xi (from left) (in.) 1 7
xi Ai 3 (in. ) 24 140 3 164 in.
(fro (from m lef leftt sid sidee to to cen centr troi oid) d) (from right side to centroid)
Moment of inertia about the z axis: axis: d = x = xi – x x Shape I C d²A C 4 4 (in. ) (in.) (in. ) −2.7273 flange 8 178.5160 stem 166.6667 3.2727 214.2113 4 Moment of inertia about the z the z axis axis (in. ) =
I C + d²A C + 4 (in. ) 186.5160 380.8790 567.3940
Internal forces and moments F = P z
= P (5
in. + 3.72 .7273 in.) = (8.7273 in.)P
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Stresses σ axial
=
F A
=−
σ H ,bending
=
σ K ,bending
=
P 2
22727 = −(0.0227
in. in.−2 ) P
44 in. M z x (8.7273 273 in.) P ( − 3.727 7273 in.) I z
M z x I z
=
=
4
567.3940 in.
(8.7 (8.7273 273 in. in.))P (8.27 (8.2727 27 in.) in.) 4
567.3940 in.
=
−2
(0.057 5733 331 1 in. in. = −(0.0
) P
(0.12 .127246 in in.−2 ) P
Compression stress limit (at H ) −2 −2 −2 σ H = −(0.022727 in. ) P − (0.057331 in. )P = − (0.080058 in. )P
(0.080058 in.−2 ) P ≤ 800 psi ∴ P ≤ 9,992.76
lb
Tension stress limit (at K ) −2 −2 −2 σ K = −(0.022727 in. ) P + (0.127246 in. )P = (0.104519 in. )P
(0.104519 in in.−2 ) P ≤ 1, 00 000 ps psi ∴ P ≤ 9,567.64
lb
Maximum magnitude of load P
P max
=
9,570 lb
Ans.
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8.62 The tee shape shown in Fig. P8.62b P8.62b is used as a post that supports a load of P of P = = 25 kN. Note that the load P load P is is applied 400 mm from the flange of the tee shape, as shown in Fig. P8.62a P8.62a. Determine the normal stresses at points H points H and K and K .
Fig. P8.62a P8.62a
Fig. P8.62b P8.62b Cross-sectional dimensions
Solution Centroid location in x direction: direction:
Shape
width b (mm) 20 120
stem flange
x
=
Σ xi Ai Σ Ai
=
height h (mm) 130 20
505,000 mm3
= 101.0
5,000 mm2
=
mm mm
49.0 mm
Area A Area Ai 2 (mm ) 2,600 2,400 5,000
xi (from left) (mm) 65 140
xi Ai 3 (mm ) 169,000 336,000 505,000
(from left si side to centroid) (from right side to centroid)
Moment of inertia about the z axis: axis: d = x = xi – x x Shape I C I C d²A C + d²A C 4 4 4 (mm ) (mm) (mm ) (mm ) stem 3,661,666.67 3,369,600.00 7,031,266.67 −36.0 flange 80,000.00 39.0 3,650,400.00 3,730,400.00 4 Moment of inertia about the z the z axis axis (mm ) = 10,761,666.67 Internal forces and moments F = 25 kN = 25, 00 000 N
M z = −(25 (25,000 ,000 N) N)(400 mm mm + 49.0 mm mm) = − 11,225 ,225,000 ,000 NN-mm
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Stresses σ axial
=
F A
=
σ H ,bending
=
σ K ,bending
=
25, 000 −25,
N
5,000 mm2
M z x I z M z x I z
=
=
= −5
MPa
(−11, 11, 225 225,000 N-mm)( − 101. 101.0 0 mm) 10,761,666.67 mm4 (−11, 11, 225 225, 000 N-mm N-mm)( )(49. 49.0 0 mm) 10,761,666.67 mm4
= 105.35
= −51.11
MPa
MPa
Normal stress at H σ H
= −5
MPa + 105.35 MPa = 100.4 MP MPa (T)
Ans.
Normal stress at K σ K
= −5
MPa − 51.11 MPa = 56.1 MPa (C)
Ans.
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8.63 The tee shape shown in Fig. P8.63b P8.63b is used as a post that supports a load of P of P , which is applied 400 mm from the flange of the tee shape, as shown in Fig. P8.63a P8.63a. The tension and compression normal stresses in the post must be limited to 165 MPa and 80 MPa, respectively. Determine the maximum magnitude of load P load P that that satisfies both the tension and compression stress limits.
Fig. P8.63a P8.63a
Fig. P8.63b P8.63b Cross-sectional dimensions
Solution Centroid location in x direction: direction:
Shape
width b (mm) 20 120
stem flange
x
=
Σ xi Ai Σ Ai
=
height h (mm) 130 20
505,000 mm3
= 101.0
5,000 mm2
=
mm mm
49.0 mm
Area A Area Ai 2 (mm ) 2,600 2,400 5,000
xi (from left) (mm) 65 140
xi Ai 3 (mm ) 169,000 336,000 505,000
(from left si side to centroid) (from right side to centroid)
Moment of inertia about the z axis: axis: d = x = xi – x x Shape I C I C d²A C C + d²A 4 4 4 (mm ) (mm) (mm ) (mm ) stem 3,661,666.67 3,369,600.00 7,031,266.67 −36.0 flange 80,000.00 39.0 3,650,400.00 3,730,400.00 4 Moment of inertia about the z the z axis axis (mm ) = 10,761,666.67 Internal forces and moments F = P z
= − P (400
mm + 49.0 mm mm) = − (449.0 mm mm)P
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Stresses σ axial
=
F A
=−
σ H ,bending
=
σ K ,bending
=
P 5,000 mm
M z x I z M z x I z
=
=
2
= −(2 × 10
−4
mm −2 ) P
(−449 mm)P ( − 101.0 mm) 4
10,761,666.67 mm
(−449 mm mm)P (49.0 mm mm) 4
10,761,666.67 mm
=
(4.21394 × 10−3 mm−2 ) P −3
= −( 2.04439 × 10
mm−2 ) P
Tension stress limit (at H ) σ H
= −( 2 × 10 =
−4
mm−2 ) P + (4.21394 × 10−3 mm−2 )P
(4.01394 × 10−3 mm−2 ) P
(4.01394 × 10−3 mm−2 ) P ≤ 165 N/mm2 ∴ P ≤
41,106.7 41,106.7 N
Compression stress limit (at K ) −4 σ K = −( 2 × 10 mm−2 ) P − (2.04439 × 10−3 mm−2 )P = − (2.24439× 10−3 mm−2 )P
(2.24439 × 10−3 mm−2 ) P ≤ 80 N/mm2 ∴ P ≤ 35,644.43
N
Maximum magnitude of load P
P max
=
35.6 kN
Ans.
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8.64 The tee shape shown in Fig. P8.64b P8.64b is used as a post that supports a load of P of P = = 25 kN, which is applied 400 mm from the flange of the tee shape as, shown in Fig. P8.64a P8.64a. Determine the magnitudes and locations of the maximum tension and compression normal stresses within the vertical portion BC portion BC o o the post.
Fig. P8.64a P8.64a
Fig. P8.64b P8.64b Cross-sectional dimensions
Solution Centroid location in x direction: direction:
Shape
width b (mm) 20 120
stem flange
x
=
Σ xi Ai Σ Ai
=
505,000 mm3 5,000 mm2
height h (mm) 130 20
= 101.0 =
mm mm
49.0 mm
Area A Area Ai 2 (mm ) 2,600 2,400 5,000
xi (from left) (mm) 65 140
xi Ai 3 (mm ) 169,000 336,000 505,000
(from left si side to centroid) (from right side to centroid)
Moment of inertia about the z axis: axis: d = x = xi – x x Shape I C I C d²A C C + d²A 4 4 4 (mm ) (mm) (mm ) (mm ) stem 3,661,666.67 3,369,600.00 7,031,266.67 −36.0 flange 80,000.00 39.0 3,650,400.00 3,730,400.00 4 Moment of inertia about the z the z axis axis (mm ) = 10,761,666.67 Internal forces and moments F = (25 kN kN) co cos 35 35° = 20.4788 kN kN = 20, 47 4 78.8 N
V = (25 kN kN) si sin 3 5° = 14.3394 kN kN = 14, 33 339.4 N
(vertical component) (horizontal component)
at B M z = −(20 (20,478. ,478.8 8 N)(400 mm + 49.0 mm) = − 9,194 ,194,98 ,981 1.2 N-mm at C M z = −(20, 478.8 N)(400 N)(400 mm + 49. 49.0 mm mm) + (14,339 ,339.4 N) N)(1,2 1,200 mm) = 8,012 ,012, 298.8 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.
Normal stress at H at at location B F −20,478.8 N σ axial = = = −4.0958 MPa A 5,000 mm2 σ H ,bending
σ H
=
M z x I z
= −4.0958
=
(−9,194 ,194,98 ,981. 1.2 2 NN-mm)( − 101. 101.0 0 mm mm) 10,761,666.67 mm4
σ H
=
I z
= −4.0958
=
10,761,666.67 mm4
σ K
=
I z
= −4.0958
=
10,761,666.67 mm4
σ K
=
I z
= −4.0958
MPa
= −41.8666
MPa
MPa − 41.8666 MPa = − 46.0 MPa
Normal stress at K at at location C M z x (8, (8, 012, 298.8 N-mm)(49.0 N-mm)(49.0 mm) σ K ,bending
= −75.1967
MPa − 75.1967 MPa = − 79.3 MPa
Normal stress at K at at location B 9,194,981.2 N-mm N-mm)(49 )(49.0 .0 mm) mm) M z x (−9,194,981.2 σ K ,bending
MPa
MPa + 86.2964 MP MPa = 82.2 MPa
Normal stress at H at at location C M z x (8, (8, 012, 012, 298.8 298.8 N-mm N-mm)( )( − 101.0 101.0 mm) mm) σ H ,bending
= 86.2964
=
10,761,666.67 mm4
= 36.4816
MPa
MPa + 36.4816 MPa = 32.4 MPa
Maximum tension stress σ max tension
=
82.2 MPa (T)
at location B
Ans.
at location C
Ans.
Maximum compression stress σ max compression
=
79.3 MPa (C)
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.
