8.1 During fabrication of a laminated timber arch, one of the 10 in. wide by 1 in. thick Douglas fir E = [E = 1,900 ksi] planks is bent to a radius of curvature of 12 ft. Determine the maximum bending stress developed in the plank.
Solution From Eq. (8.3): E σ x = − y = − ρ
1, 900 ksi (12 ft)(12 in./ft)
(±0.5 in.) = ±6.597 ksi = 6.60 ksi
Ans.
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8.2 A high-strength steel [ E E = = 200 GPa] tube having an outside diameter of 80 mm and a wall thickness of 3 mm is bent into a circular curve having a 24-m radius of curvature. Determine the maximum bending stress developed in the tube.
Solution From Eq. (8.3): E σ x = − y = − ρ
200,000 MPa (24 m)(1,000 mm/m)
(±80 mm / 2) 2) = ±333.333 MPa = 333 MPa
Ans.
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8.3 A high-strength steel [ E = = 200 GPa] band saw blade wraps around a pulley that has a diameter o 350 mm. Determine the maximum maximum bending stress developed in the the blade. The blade is 12-mm wide and 1-mm thick.
Solution The radius of curvature of the band saw blade is: 350 mm 1 mm ρ = + = 175.5 mm 2 2 From Eq. (8.3): 200,000 MPa E σ x = − y = − (±0.5 mm) = ±569.801 MPa = 570 MPa ρ 175.5 mm
Ans.
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8.4 The boards for a concrete form are to be bent into a circular shape having an inside radius of 8 m. What maximum thickness can be used for the boards if the normal stress is not to exceed 7 MPa? Assume that the modulus of elasticity for the wood is 12 GPa.
Solution The radius of curvature of the concrete form is dependent on the board thickness: t ρ = 8,000 mm + 2 From Eq. (8.3): E 12,000 MPa ⎛ t ⎞ σ x = − y = − ≤ ±7 MPa t ⎞ ⎝⎜ 2 ⎠⎟ ρ ⎛ ⎜ 8,000 mm + 2 ⎟ ⎝ ⎠ Solve for t: t ⎞ ⎛t⎞ ⎛ 12,000 MPa ⎜ ⎟ ≤ 7 MPa ⎜ 8,000 mm + ⎟ 2⎠ ⎝2⎠ ⎝ 6, 000t ≤ 56, 000 + 3.5t 5, 99 996.5t ≤ 56, 00 000 ∴
t ≤ 9.39 mm
Ans.
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8.5 A beam having a tee-shaped cross section is subjected to equal 18 kN-m bending moments, as shown in Fig. P8.5a P8.5a. The cross-sectional dimensions of the beam are shown in Fig. P8.5b P8.5b. Determine: (a) the centroid location, the moment of inertia about the z the z axis, and the controlling section modulus about the z the z axis. axis. (b) the bending stress at point H point H . State whether the normal stress at H at H is is tension or tension or compression. (c) the maximum bending stress produced in the cross section. State whether the stress is tension or compression.
Fig. P8.5a P8.5a
Fig. P8.5b P8.5b
Solution (a) Centroid location in y direction: (reference axis at bottom of tee shape) yi Shape Area A Area Ai (from bottom) yi Ai 2 3 (mm ) (mm) (mm ) top flange 2,500.0 162.5 406,250.0 stem 3,750.0 75.0 281,250.0 2 3 6,250.0 mm 687,500.0 mm
y
=
Σ yi Ai Σ Ai
=
687,500.0 mm3 6,250.0 mm2
=
110.0 mm (measured upward from bottom edge of stem)
Moment of inertia about the z axis: axis: d = y = yi – Shape I C C 4
top flange stem
(mm ) 130,208.33 7,031,250.00
(mm) 52.50 −35.00
d²A 4 (mm ) 6,890,625.00 4,593,750.00 4
Moment of inertia about the z the z axis axis (mm ) =
Ans.
I C C + d²A 4 (mm ) 7,020,833.33 11,625,000.00 18,645,833.33
I z = 18, 18, 656, 656, 000 mm4
Ans.
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Section moduli: I z 18,645,833.33 18,645, 833.33 mm mm4 3 S top = = = 286,858.974 mm mm − 110 110 mm mm) ctop (175 mm
S bot
=
I z c bot
∴ S =
=
18,645,833.33 18,645, 833.33 mm mm4 110 mm
= 169,507.576
mm3
169,500 mm3
Ans.
(b) Bending stress at point H : ( y = y = 175 mm − 25 mm − 110 mm = 40 mm) M y σ x = −
=−
I z (18 kN-m)(40 mm)(1,000 N/kN)(1,000 mm/m) 18,654,833.33 mm4
= −38.615
MP MPa = 38.6 MP MPa (C (C)
Ans.
(c) Maximum bending stress: The maximum bending stress occurs at either the top or the bottom surface of the beam. The top of the cross section is at y y = +65 mm, and the bottom of the cross section is at y = −110 mm. The larger bending stress magnitude occurs at the larger magnitude of these two values; in this case, at the bottom of the cross section. M y σ x = −
=−
I z (18 kN-m kN-m)( )( − 110 mm)(1,000 mm)(1,000 N/kN)(1,000 N/kN)(1,000 mm/m) mm/m)
106.190 = 106
18,654,833.33 mm4 MP MPa = 106.2 6.2 MP MPa (T) (T)
Ans.
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8.6 A beam is subjected to equal 8.5 kip-ft bending moments, as shown in Fig. P8.6a P8.6a. The crosssectional dimensions of the beam are shown in Fig. P8.6b P8.6b. Determine: (a) the centroid location, the moment of inertia about the z the z axis, and the controlling section modulus about the z the z axis. axis. (b) the bending stress at point H point H , which is located 2 in. below the z the z centroidal axis. State whether the normal stress at H at H is is tension or tension or compression. (c) the maximum bending stress produced in the cross section. State whether the stress is tension or compression.
Fig. P8.6a P8.6a
Fig. P8.6b P8.6b
Solution (a) Centroid location in y direction: (reference axis at bottom of shape) yi Shape Area A Area Ai (from bottom) yi Ai 2 3 (in. ) (in.) (in. ) left side 8.0 4.0 32.0 top flange 4.0 7.5 30.0 right side 8.0 4.0 32.0 2 3 20.0 in. 94.0 in.
y
=
Σ yi Ai Σ Ai
=
94.0 in.3 20.0 in.2
=
4.70 in. (measured upward from bottom edge of section) section)
Moment of inertia about the z axis: axis: d = y = yi – Shape I C C 4
left side top flange right side
(in. ) 42.667 0.333 42.667
(in.) -0.700 2.800 -0.700
d²A 4 (in. ) 3.920 31.360 3.920 4
Moment of inertia about the z the z axis axis (in. ) =
Ans.
I C C + d²A 4 (in. ) 46.587 31.693 46.587 124.867 I z = 124.9 in.4
Ans.
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Section moduli: I z 124.867 in.4 S top = = ctop (8 in. − 4.7 in in.)
S bot
=
I z c bot
∴ S =
=
124.867 in.4 4.7 in.
=
26.5674 in.3
= 37.8384
in.3
26.6 in.3
Ans.
y = −2 in.) (b) Bending stress at point H : ( y = M y σ x = −
=−
I z (−8.5 8.5 ki kip-ft p-ft))( − 2 in.) in.)(1 (12 2 in. in./f /ft) t) 124.867 in.4
1.634 = −1.6
ks ksi = 1,634 ,634 ps psi (C) (C)
Ans.
(c) Maximum bending stress: The maximum bending stress occurs at either the top or the bottom surface of the beam. The top of the cross section is at y at y = = +3.30 in., and the bottom of the cross section is at y at y = = −4.7 in. in. The larger larger bending stress magnitude occurs at the larger magnitude of these two values; in this case, at the bottom of the cross section. M y σ x = −
=−
I z (−8.5 8.5 kipkip-fft)( t)( − 4.7 4.7 in.) in.)((12 in. in./f /ft) t)
3.839 = −3.8
124.867 in.4 ks ksi = 3,840 ,840 ps psi (C)
Ans.
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8.7 A beam is subjected to equal 375 N-m bending moments, as shown in Fig. P8.7a P8.7a. The crosssectional dimensions of the beam are shown in Fig. P8.7b P8.7b. Determine: (a) the centroid location, the moment of inertia about the z the z axis, and the controlling section modulus about the z the z axis. axis. (b) the bending stress at point H point H . State whether the normal stress at H at H is is tension or tension or compression. (c) the maximum bending stress produced in the cross section. State whether the stress is tension or compression.
Fig. P8.7a P8.7a
Fig. P8.7b P8.7b
Solution (a) Centroid location in y direction: (reference axis at bottom of U shape) yi Shape Area A Area Ai (from bottom) yi Ai 2 3 (mm ) (mm) (mm ) left side 400.0 25.0 10,000.0 bottom flange 272.0 4.0 1,088.0 right side 400.0 25.0 10,000.0 2 3 1,072.0 mm 21,088.0 mm
y
=
Σ yi Ai Σ Ai
=
21,088.0 mm3 1,072.0 mm2
=
19.67 mm (measured upward from bottom edge of section)
Moment of inertia about the z axis: axis: d = y = yi – Shape I C C 4
left side bottom flange right side
(mm ) 83,333.33 1,450.67 83,333.33
(mm) 5.33 -15.67 5.33
I C C + d²A 4 (mm ) 94,689.89 68,253.96 94,689.89
d²A 4 (mm ) 11,356.56 66,803.30 11,356.56 4
Ans.
Moment of inertia about the z the z axis axis (mm ) =
257,633.75
I z = 257,600 mm4
Ans.
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Section moduli: 257,633.75 mm4 I z 3 S top = = = 8,494.814 mm ctop (50 mm mm − 19. 19.672 mm mm)
S bot
=
I z c bot
∴ S =
=
257,633.75 mm4 19.672 mm
= 13,096.708
mm3
8,495 mm3
Ans.
(b) Bending stress at point H : ( y = y = 8 mm − 19.672 mm = −11.672 mm) M y σ x = −
=−
I z (375 (375 N-m) N-m)(( − 11.672 11.672 mm)(1 mm)(1,00 ,000 0 mm/ mm/m) m) 257,633.75 mm4
16.989 = 16.
MP MPa = 16. 16.99 MP MPa (T) (T)
Ans.
(c) Maximum bending stress: The maximum bending stress occurs at either the top or the bottom surface of the beam. The top of the cross section is at y at y = = +30.328 mm, and the bottom of the cross section is at y at y = = −19.672 mm. The larger bending stress magnitude occurs at the larger magnitude of these two values; in this case, at the top of the cross section. M y σ x = −
=−
I z (375 N-m)(30.328 mm)(1,000 mm/m)
44.145 = −44.
257,633.75 mm4 MPa = 44.1 MPa (C)
Ans.
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8.8 A beam is subjected to equal 10.5 kip-ft bending moments, as shown in Fig. P8.8a P8.8a. The crosssectional dimensions of the beam are shown in Fig. P8.8b P8.8b. Determine: (a) the centroid location, the moment of inertia about the z the z axis, and the controlling section modulus about the z the z axis. axis. (b) the bending stress at point H point H . State whether the normal stress at H at H is is tension or tension or compression. (c) the bending stress at point K point K . State whether the normal stress at K at K is is tension or tension or compression. (d) the maximum bending stress produced in the cross section. State whether the stress is tension or compression.
Fig. P8.8a P8.8a
Fig. P8.8b P8.8b
Solution (a) Centroid location in y direction: (reference axis at bottom of shape) yi Shape Area A Area Ai (from bottom) yi Ai 2 3 (in. ) (in.) (in. ) top flange 12.0000 11.0000 1 1.0000 132.0000 web 16.0000 6.0000 96.0000 bottom flange 20.0000 1.0000 20.0000 2 3 48.0000 in. 248.0000 in.
y
=
Σ yi Ai Σ Ai
=
248.0 in.3 48.0 in.2
=
5.1667 in. (measured upward from from bottom edge of bottom flange)
Moment of inertia about the z axis: axis: d = y = yi – Shape I C C
d²A 4 (in. ) (in.) (in. ) 4.0000 5.8333 408.3333 85.3333 0.8333 11.1111 6.6667 –4.1667 347.2222 4 Moment of inertia about the z the z axis axis (in. ) = 4
top flange web bottom flange
Ans.
I C C + d²A 4 (in. ) 412.3333 96.4444 353.8889 862.6667 Ans.
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Section Moduli cbot = 5.1667 in.
ctop
= 12
S bot = S top
=
in. − 5.1667 in in. = 6.8333 in.
I z
=
cbot I z ctop
=
862.6667 862.6667 in.4 5.1667 in. 862.6667 in.4 6.8333 in.
= 166.9677
in.3
= 126.2439
in.3
The controlling section modulus is the smaller of the two values; therefore, S = 126.2439 126.2439 in.3
Ans.
Bending stress at point H : From the flexure formula: M y (−10.5 10.5 kipkip-ft ft)( )(6. 6.833 8333 3 in. in. − 2 in.) in.)(1 (12 2 in./ in./ft ft)) σ x = −
I z
=−
862.6667 in.4
=
Bending stress at point K : From the flexure formula: M y (−10.5 0.5 ki kip-ft)( − 5.16 .1667 in. in. + 2 in in.)(12 in./ft) σ x = −
I z
=−
862.6667 in.4
0.7059 ks ksi = 706 ps psi (T)
Ans.
= −0.4625
Ans.
ks ksi = 463 ps psi (C (C)
Maximum bending stress Since ctop > cbot , the maximum bending stress occurs at the top of the the flanged shape. From the flexure formula: M y (−10.5 kip-ft)( kip-ft)(6.8333 6.8333 in.)(12 in.)(12 in./ft) in./ft) σ x = − 981 ksi = 998 ps psi (T) =− = 0.9981 Ans. I z 862.6667 in.4
Also, note that the same maximum bending stress magnitude can be calculated with the section modulus: M (10.5 kip-ft)(12 in./ft) σ x = Ans. psi = = 0.9981 ksi = 998 ps S 126.2439 in.3 The sense of the stress (either tension or compression) would be determined by inspection.
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8.9 The cross-sectional dimensions of a beam are shown in Fig. P8.9. (a) If the bending stress at point K point K is is 43 MPa (C), determine the internal bending moment M moment M z acting about the z the z centroidal centroidal axis of the beam. (b) Determine the bending stress at point H point H . State whether the normal stress at H is tension or compression.
Fig. P8.9
Solution Centroid location in y direction: (reference axis at bottom of double-tee shape) yi Shape Area A Area Ai (from bottom) yi Ai 2 3 (mm ) (mm) (mm ) top flange 375.0 47.5 17,812.5 left stem 225.0 22.5 5,062.5 right stem 225.0 22.5 5,062.5 2 3 825.0 mm 27,937.5 mm Σ yi Ai 27,937.5 mm3 y = mm = 33.9 mm mm (measured upward from bottom of section) = = 33.864 mm 825.0 mm2 Σ Ai Moment of inertia about the z axis: axis: d = y = yi – Shape I C C 4
(mm ) 781.250 37,968.750 37,968.750
top flange left stem right stem
(mm) 13.636 −11.364 −11.364
d²A 4 (mm ) 69,731.405 29,054.752 29,054.752
I C C + d²A 4 (mm ) 70,512.655 67,023.502 67,023.502
4
Moment of inertia about the z the z axis axis (mm ) = 204,559.659 (a) Determine bending moment: At point K point K , y = y = 50 mm − 5 mm − 33.864 mm = 11.136 mm. The bending stress at K at K is is σ x = −43 MPa; therefore, the bending moment magnitude can be determined from the flexure formula: M y σ x = −
I z
∴ M = −
σ x I z
y
=−
789,850 ,850.765 765 = 789
(−43 N/ N/mm 2 )(204,55 ,559.659 mm4 ) 11.136 mm
NN-mm = 790 N-m
(b) Bending stress at point H : At point H point H , y = y = −33.864 mm. The bending stress is computed with the flexure formula: M y (789 (789,850. ,850.765 765 N-mm N-mm)( )( − 33.86 33.864 4 mm) mm) σ x = − 130.75 755 5 MPa = 130. 130.8 8 MPa MPa (T) =− = 130. I z 204,559.659 mm4
Ans.
Ans.
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8.10 The cross-sectional dimensions of a beam are shown in Fig. P8.10. (a) If the bending stress at point K is is 2,600 psi (T), determine the internal bending moment M z acting about the z the z centroidal centroidal axis of the beam. (b) Determine the bending stress at point H . State whether the normal stress at H is tension or compression.
Fig. P8.10
Solution Centroid location in y direction: (reference axis at bottom of inverted-tee shape) yi Shape Area A Area Ai (from bottom) yi Ai 2 3 (in. ) (in.) (in. ) bottom flange 0.56250 0.12500 0.07031 stem 0.56250 1.37500 0.77344 2 3 1.12500 in. 0.84375 in. Σ yi Ai 0.84375 in.3 (measured upward from bottom edge of section) = = 0.750 in. y = 1.1250 in.2 Σ Ai Moment of inertia about the z axis: axis: d = y = yi – Shape I C C 4
(in. ) 0.00293 0.23730
bottom flange stem
(in.) −0.62500 0.62500
d²A 4 (in. ) 0.21973 0.21973
I C + d²A C + 4 (in. ) 0.22266 0.45703
4
Moment of inertia about the z the z axis axis (in. ) = 0.67969 (a) Determine bending moment: At point K point K , y = y = 2.50 in. − 0.75 in. = 1.750 in. The bending stress at K at K is is σ x = +2,600 psi; therefore, the bending moment magnitude can be determined from the flexure formula: M y σ x = −
I z
∴ M = −
σ x I z
y
,009.820 = −1,00
=−
(2,600 psi)(0 psi)(0.67 .67967 967 in.4 )
lb-in. =
1.750 in. − 1,010
lb-in. = − 84.2 lb-ft
(b) Bending stress at point H : At point H point H , y = y = −0.75 in. The bending stress is computed with the flexure formula: M y (−1,009.8 ,009.820 20 lb-in b-in.) .)(( − 0.75 0.75 in.) in.) σ x = − ,114.2 .286 86 psi psi = 1,114 ,114 psi psi (C) (C) =− = −1,114 4 I z 0.67969 in.
Ans.
Ans.
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8.11 The cross-sectional dimensions of a boxshaped beam are shown in Fig. P8.11. If the maximum allowable bending stress is σ b = 15,000 psi, determine the maximum internal bending moment M z magnitude that can be applied to the beam.
Fig. P8.11
Solution Moment of inertia about z axis: axis: 3 (3 in. in.))(2 in.) n.) (2.5 (2.5 in.) in.)((1 in. in.))3 I z = − 12 12
= 1.791667
in.4
Maximum internal bending moment M z : M z c σ x =
I z
∴ M =
σ x I z
c
=
(15, 15, 000 psi)(1. psi)(1.7916 791667 67 in.4 ) 1 in.
=
26,875 ,875 lblb-iin. = 2,240 ,240 lblb-ft
Ans.
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8.12 The cross-sectional dimensions of a beam are shown in Fig. P8.12. The internal bending moment about the z centroidal axis is M z = +2.70 kip-ft. Determine: (a) the maximum tension bending stress in the beam. (b) the maximum compression bending stress in the beam.
Fig. P8.12
Solution Centroid location in y direction: (reference axis at bottom of shape) yi Shape Area A Area Ai (from bottom) yi Ai 2 3 (in. ) (in.) (in. ) left stem 2.000 2.000 4.000 top flange 2.500 3.750 9.375 right stem 2.000 2.000 4.000 2 3 6.500 in. 17.375 in. Σ yi Ai 17.375 in.3 = = 2.673 in. y = Σ Ai 6.500 in.2
(measured upward from bottom edge of section) sec tion) Moment of inertia about the z axis: axis: d = y = yi – Shape I C C 4
left stem top flange right stem
(in. ) 2.66667 0.05208 2.66667
(in.) −0.67308 1.07692 −0.67308
d²A 4 (in. ) 0.90607 2.89941 0.90607 4
Moment of inertia about the z the z axis axis (in. ) =
I C + d²A C + 4 (in. ) 3.57273 2.95149 3.57273 10.09696
(a) Determine maximum tension bending stress: For a positive bending moment, tension bending stresses will be created below the neutral axis. Therefore, the maximum tension bending stress will occur at point K point K (i.e., (i.e., y = y = −2.673 in.): (2.70 (2.70 kipkip-ft) ft)(( − 2.673 2.673 in.)( in.)(12 12 in./ft in./ft)) M y σ x = − 578 ksi ksi = 8.58 .58 ksi ksi (T) =− = 8.578 Ans. I z 10.09696 in.4 (b) Determine maximum compression bending stress: For a positive bending moment, compression bending stresses will be created above the neutral axis. Therefore, the maximum compression bending stress will occur at point H point H (i.e., (i.e., y = y = 4 in. − 2.673 in. = 1.327 in.): M y (2.70 kip-ft)(1.327 in.)(12 in./ft) σ x = − ksi = 4.26 ks ksi (C) =− = −4.258 ks Ans. I z 10.09696 in.4
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8.13 The cross-sectional dimensions of a beam are shown in Fig. P8.13. (a) If the bending stress at point K point K is is 35.0 MPa (T), determine the bending stress at point H . State whether the normal stress at H is tension or compression. (b) If the allowable bending stress is σ b = 165 MPa, determine the magnitude of the maximum bending moment M moment M z that can be supported by the beam.
Fig. P8.13
Solution Moment of inertia about the z axis: axis: Shape I C C 4 (mm ) top flange 540,000.000 web 32,518,666.667 bottom flange 540,000.000
d = y = yi – y y (mm) 160.000 0.000 −160.000
d²A 4 (mm ) 184,320,000.000 0.000 184,320,000.000 4
Moment of inertia about the z the z axis axis (mm ) =
I C C + d²A 4 (mm ) 184,860,000.000 32,518,666.667 184,860,000.000 402,238,666.667
(a) At point K point K , y = y = −90 mm, and at point H point H , y = y = −175 mm. The bending stress at K at K is is σ x = +35 MPa, and the bending stress is distributed linearly over the depth of the cross section. Therefore, the bending stress at H at H can can be found from the ratio: σ H
y H
=
σ K
yK
∴σ H = σ K
y H y K
=
(35.0 MPa)
−175 −90
mm mm
=
68.056 MPa = 68.1 MPa (T)
Ans.
(b) Maximum internal bending moment M z : M z c σ x =
I z
∴ M =
σ x I z
c
=
(165 N/mm N/mm2 )(40 )(402, 2,238 238,6 ,667 67 mm mm4 ) 175 mm
=
379,253 ,253,600 ,600 NN-mm = 379 kN kN-m
Ans.
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8.14 The cross-sectional dimensions of a beam are shown in Fig. P8.14. (a) If the bending stress at point K is is 9.0 MPa (T), determine the bending stress at point H . State whether the normal stress at H is tension or compression. (b) If the allowable bending stress is σ b = 165 MPa, determine the magnitude of the maximum bending moment M moment M z that can be supported by the beam.
Fig. P8.14
Solution Moment of inertia about the z axis: axis: Shape I C C 4 (mm ) left flange 9,720,000 web 31,680 right flange 9,720,000
d = y = yi – y y
d²A 4 (mm ) 0 0 0
(mm) 0 0 0 4
Moment of inertia about the z the z axis axis (mm ) =
I C C + d²A 4 (mm ) 9,720,000 31,680 9,720,000 19,471,680
(a) At point K point K , y = y = −60 mm, and at point H point H , y = y = +90 mm. The bending stress stress at K at K is is σ x = +9.0 MPa, and the bending stress is distributed linearly over the depth of the cross cross section. Therefore, the bending stress at H at H can can be found from the ratio: σ H
y H
=
σ K
yK
∴σ H = σ K
y H y K
=
(9.0 MPa)
90 mm −60
mm
= −13.50
MPa = 13.50 MPa (C)
Ans.
(b) Maximum bending moment M z : M z c σ x =
I z
∴ M =
σ x I z
c
=
(165 N/mm N/mm2 )(19 )(19,4 ,471, 71,68 680 0 mm mm4 ) 90 mm
35,698 8,080 ,080 = 35,69
NN-mm = 35.7 5.7 kN kN-m
Ans.
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8.15 The cross-sectional dimensions of a beam are shown in Fig. P8.15. The internal bending moment about the z centroidal axis is M z = −1.55 kip-ft. Determine: (a) the maximum tension bending stress in the beam. (b) the maximum compression bending stress in the beam.
Fig. P8.15
Solution Centroid location in y direction:
Shape
Area A Area Ai 2 (in. ) 8.0 3.0 3.0 3.0 3.0 2 20.0 in.
top flange left web left bottom flange right web right bottom flange
y
=
Σ yi Ai Σ Ai
=
54.0 in.3 20.0 in.2
=
yi (from bottom) (in.) 4.5 2.5 0.5 2.5 2 .5 0.5
yi Ai 3 (in. ) 36.0 7.5 1.5 7.5 1.5 3 54.0 in.
2.70 in. (measured upward from bottom edge of bottom flange)
Moment of inertia about the z axis: axis:
Shape top flange left web left bottom flange right web right bottom flange
d = y = yi – y y I C d²A C 4 4 (in. ) (in.) (in. ) 0.6667 1.8000 25.9200 2.2500 0.1200 −0.2000 0.2500 14.5200 −2.2000 2.2500 0.1200 −0.2000 0.2500 14.5200 −2.2000 4 Moment of inertia about the z the z axis axis (in. ) =
I C C + d²A 4 (in. ) 26.5867 2.3700 14.7700 2.3700 14.7700 60.8667
(a) Maximum tension bending stress: For a negative bending moment, the maximum tension bending stress will occur at the top surface of the cross section; that is, point H point H for for this shape. shape. From the flexure formula, formula, the bending stress at point H point H is: is: M y (−1.55 1.55 kip kip-f -ft) t)(5 (5.0 .0 in. in. − 2.70 2.70 in. in.)( )(12 12 in./ in./ft ft)) ksi = 703 ps psi (T (T) σ x = − Ans. =− = 0.7028 ks I z 60.8667 in.4 (b) Maximum compression bending stress: The maximum compression bending stress will occur at the bottom surface of the cross section, which is point K point K for for this shape. shape. From the flexure formula, formula, the bending stress at point K point K is: is: M y (−1.55 1.55 kipkip-ft ft)( )( − 2.70 2.70 in.) in.)((12 in./ in./ft ft)) σ x = − 251 ksi ksi = 825 psi (C) =− = −0.8251 Ans. I z 60.8667 in.4 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.
8.16 The cross-sectional dimensions of a beam are shown in Fig. P8.16. The internal bending moment about the z centroidal axis is M z = +270 lb-ft. Determine: (a) the maximum tension bending stress in the beam. (b) the maximum compression bending stress in the beam.
Solution
Fig. P8.16
Centroid location in y direction:
yi (from bottom) (in.) 0.06250 1.25000 2.43750 1.25000 2.43750
Shape
Area A Area Ai yi Ai 2 3 (in. ) (in. ) bottom flange 0.40625 0.02539 left web 0.28125 0.35156 left top flange 0.09375 0.22852 right web 0.28125 0.35156 right top flange 0.09375 0.22852 2 3 1.15625 in. 1.18555 in. Σ yi Ai 1.18555 in.3 y = = = 1.0253 in. (measured upward from bottom edge of bottom flange) 1.15625 in.2 Σ Ai Moment of inertia about the z axis: axis:
Shape bottom flange left web left top flange right web right top flange
d = y = yi – y y I C d²A C 4 4 (in. ) (in.) (in. ) 0.000529 -0.962838 0.376617 0.118652 0.224662 0.014196 0.000122 1.412162 0.186956 0.118652 0.224662 0.014196 0.000122 1.412162 0.186956 4 Moment of inertia about the z the z axis axis (in. ) =
I C + d²A C + 4 (in. ) 0.377146 0.132848 0.187079 0.132848 0.187079 1.016999
(a) Maximum tension bending stress: For a positive bending moment of M of M z = +270 lb-ft, the maximum tension bending stress will occur at the bottom surface of the cross section (i.e., y (i.e., y = = −1.0253 in.). From the flexure formula, the bending stress stress at the bottom of the cross section is: M y (270 lblb-ft) ft)(( − 1.0253 1.0253 in.)( in.)(12 12 in./f in./ft) t) σ x = − Ans. 446 ps psi = 3,270 ,270 ps psi (T) (T) =− = 3, 266.446 4 I z 1.016999 in. (b) Maximum compression bending stress: The maximum compression bending stress will occur at the top surface of the cross section (i.e., y (i.e., y = = 2.50 in. − 1.0253 in. = 1.4747 in.). From the flexure formula, the bending stress stress at the top of the cross section is: M y (270 lb-ft)(1.4747 lb-ft)(1.4747 in.)(12 in./ft) σ x = − ,698.164 psi psi = 4,700 ,700 ps psi (C) =− = −4,698 Ans. I z 1.016999 in.4 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.
8.17 Two vertical forces are applied to a simply supported beam (Fig. P8.17a P8.17a) having the cross section shown in Fig. P8.17b P8.17b. Determine the maximum tension and compression bending stresses produced in segment BC segment BC of of the beam.
Fig. P8.17a P8.17a
Fig. P8.17b P8.17b
Solution Centroid location in y direction:
Shape top flange stem
y
=
Σ yi Ai Σ Ai
=
Area A Area Ai 2 (mm ) 3,000.0 1,440.0 2 4,440 mm 617,700 mm3 4,440 mm2
=
yi (from bottom) (mm) 167.5 80.0
yi Ai 3 (mm ) 502,500.0 115,200.0 3 617,700 mm
139.1216 mm (measured upward from bottom edge of stem) stem)
Moment of inertia about the z axis: axis:
Shape top top flan flange ge stem
d = y = yi – I C d²A C 4 4 (mm ) (mm) (mm ) 56,2 56,250 50.0 .00 0 28.3 28.38 8 2,41 2,415, 5,99 997. 7.08 08 3,072,000.00 5,033,327.25 −59.12 4 Moment of inertia about the z the z axis axis (mm )
I C + d²A C + 4 (mm ) 2,47 2,472, 2,24 247. 7.08 08 8,105,327.25 10,577,574.32
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Shear-force and bending-moment diagrams:
The maximum moment occurs between B between B and and C . The moment magnitude magnitude is 12 kN-m. Maximum tension bending stress: For a positive bending moment, the maximum tension bending stress will occur at the bottom surface of this cross section. From the flexure formula, the bending stress at the bottom of the tee stem is: is: M y (12 kN-m)( kN-m)( − 139.1216 mm)(1,000 mm)(1,000 N/kN)(1,000 N/kN)(1,000 mm/m) mm/m) σ x = − Ans. =− = 157.8 MPa (T) I z 10.5776 × 106 mm4 Maximum compression bending stress: The maximum compression bending stress will occur at the top of the flange: M y σ x = −
=−
I z (12 kN-m)( kN-m)(175 175 mm − 139.1216 139.1216 mm)(1,000 mm)(1,000 N/kN) N/kN)(1,000 (1,000 mm/m) mm/m)
= −40.7
10.5776 × 106 mm4 MP MPa = 40.7 MP MPa (C)
Ans.
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8.18 Two vertical forces are applied to a simply supported beam (Fig. P8.18a P8.18a) having the cross section shown in Fig. P8.18b P8.18b. Determine the maximum tension and compression bending stresses produced in segment BC segment BC of of the beam.
Fig. P8.18a P8.18a
Fig. P8.18b P8.18b
Solution Centroid location in y direction:
Shape
Area A Area Ai 2 (in. ) 0.7500 0.5000 0.7500 2 2.000 in.
left stem bottom flange right stem
y
=
Σ yi Ai Σ Ai
=
2.3125 in.3 2.000 in.2
=
yi (from bottom) (in.) 1.5000 0.1250 1.5000
yi Ai 3 (in. ) 1.1250 0.0625 1.1250 3 2.3125 in.
1.1563 in. (measured upward from bottom edge of stem) stem)
Moment of inertia about the z axis: axis:
Shape left stem bottom flange right stem
d = y = yi – I C d²A C 4 4 (in. ) (in.) (in. ) 0.56250 0.34375 0.08862 0.00260 -1.03125 0.53174 0.56250 0.34375 0.08862 4 Moment of inertia about the z the z axis axis (in. )
I C + d²A C + 4 (in. ) 0.65112 0.53434 0.65112 1.83659
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Shear-force and bending-moment diagrams:
The maximum moment occurs between B between B and and C . The moment magnitude magnitude is 600 lb-ft. Maximum tension bending stress: For a positive bending moment, the maximum tension bending stress will occur at the bottom surface of this cross section at y at y = = −1.1563 in. From the flexure formula, the the bending stress at the bottom of the U shape is: M y (600 lb-ft) lb-ft)(( − 1.1563 1.1563 in.) in.)(12 (12 in./ft in./ft)) σ x = − Ans. ,533.053 053 ps psi = 4,530 ,530 ps psi (T) (T) =− = 4,533 4 I z 1.83659 in. Maximum compression bending stress: The maximum compression bending stress will occur at the top of the U shape, where whe re y y = = 1.8438 in.: M y (600 lb-ft)(1.8438 lb-ft)(1.8438 in.)(12 in./ft) σ x = − Ans. 265 ps psi = 7,230 ,230 ps psi (C) =− = −7, 228.265 I z 1.83659 in.4
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