9.11 A 1.6-m long cantilever beam supports a concentrated load of 7.2 kN, as shown below. The beam is made of a rectangular timber having a width of 120 mm and a depth of 280 mm. Calculate the maximum horizontal shear stresses at points located 35 mm, 70 mm, 105 mm, and 140 mm below the top surface of the beam. From these results, plot a graph showing the distribution of shear stresses from top to bottom of the beam.
Fig. P9.11a P9.11a Cantilever beam
Fig. P9.11b P9.11b Cross-sectional dimensions
Solution Shear force in cantilever beam: V = = 7.2 kN = 7,000 N Shear stress formula: VQ τ =
I t
Section properties: (120 mm)(280 mm)3 I = 12
=
219.52 × 106 mm4
t = = 120 mm
Distance below top surface of beam
y
Q
35 mm
105 mm
514,500 mm
70 mm
70 mm
882,000 mm
105 mm
35 mm
1,102,500 mm
140 mm
0 mm
1,176,000 mm
τ
3
140.6 kPa
3
241 kPa 3
301 kPa
3
321 kPa
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9.12 A 14-ft long simply supported timber beam carries a 6-kip concentrated load at midspan, as shown in Fig. P9.12a P9.12a. The cross-sectional dimensions of the timber are shown in Fig. P9.12b P9.12b. (a) At section a–a, a–a, determine the magnitude of the shear stress in the beam at point H. point H. (b) At section a–a, a–a, determine the magnitude of the shear stress in the beam at point K. point K. (c) Determine the maximum horizontal shear stress that occurs in the beam at any location within within the 14-ft span length. (d) Determine the maximum tension bending stress that occurs in the beam at any location within the 14-ft span length.
Fig. P9.12b P9.12b Cross-sectional dimensions
Fig. P9.12a P9.12a Simply supported timber beam
Solution Section properties: (6 in.)(15 in.)3 I = 12
= 1, 687.5
in.4
t = 6 in. (a) Shear stress at H : Q = (6 in. in.)( )(3 3 in.) in.)(6 (6 in. in.)) = 108 108 in. in.3
τ =
=
VQ I t (3, (3, 000 lb)( lb)(108 108 in.3 ) (1, 687.500 687.500 in.4 )(6 in.) in.)
=
32.0 psi
Ans.
(b) Shear stress at K : Q = (6 in.) in.)((1 in. in.)( )(7 7 in. in.)) = 42 in. in.3
τ =
=
VQ I t (3, (3, 000 lb)(42 lb)(42 in.3 ) (1, 687.500 687.500 in. in.4 )(6 in.) in.)
=
12.44 psi
Ans.
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(c) Maximum shear stress at any location: Qmax = (6 in.) in.)(7 (7.5 .5 in. in.)( )(3. 3.75 75 in.) in.) = 168.7 168.75 5 in.3
τ =
VQ I t
=
(3, (3, 000 lb)( lb)(168.75 168.75 in.3 ) (1, 687.500 687.500 in.4 )(6 in.) in.)
=
50.0 psi
Ans.
(d) Maximum bending stress at any location: kip-fft = 21,000 21,000 lblb-fft M max = 21 kip-
σ x
=
M c I
=
(21,000 lb-ft)(7.5 in.)(12 in./ft) 1, 687.500 in.4
=
1,120 psi (T) and (C)
Ans.
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9.13 A 4-m long simply supported timber beam be am carries a uniformly distributed load of 7 kN/ m, as shown in Fig. P9.13a P9.13a. The cross-sectional dimensions of the beam are shown in Fig. P9.13b P9.13b. (a) At section a–a, a–a, determine the magnitude of the shear stress in the beam at point po int H. H. (b) At section a–a, a–a, determine the magnitude of the shear stress in the beam at point K. point K. (c) Determine the maximum horizontal shear stress that occurs in the beam at any location within the 4m span length. (d) Determine the maximum compression bending stress that occurs in the beam at any location within the 4-m span length.
Fig. P9.13b P9.13b Cross-sectional dimensions
Fig. P9.13a P9.13a Simply supported timber beam
Solution Section properties: (100 mm)(300 mm)3 I = 12
=
225 × 10 106 mm4
t = 100 mm (a) Shear stress magnitude at H : Q = (100 mm)(90 mm)(105 mm) = τ =
= =
945,000 mm3 VQ I t (7,000 N)(9 N)(945, 45,00 000 0 mm3 ) (225 × 106 mm4 )(100 mm mm) 294 kPa
Ans.
(b) Shear stress magnitude at K : Q = (100 mm)(75 mm)(112.5 mm) = 843,750 τ =
= =
mm3
VQ I t (7, (7, 000 000 N)(8 N)(843 43,, 750 mm mm3 ) (225 × 106 mm4 )(100 mm mm) 263 kPa
Ans.
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(c) Maximum shear stress at any location: Qmax = (100 mm) mm)(15 (150 0 mm) mm)(75 (75 mm) mm) = 1,125, ,125, 000 mm mm3
τ =
VQ I t
=
(14,000 N)(1 N)(1,125, ,125, 000 mm3 ) (225 × 106 mm4 )(100 mm mm)
=
700 kPa
Ans.
(d) Maximum compression bending stress at any location: M max = 14 kN-m
σ x
M y
=−
I
=−
333 = −9.3333
(14 kN-m)(150 mm)(1,000 N/kN)(1,000 mm/m) 225 × 106 mm4
MPa = 9,330 ,330 kPa kPa (C)
Ans.
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9.14 A 5-m long simply supported timber beam carries two concentrated loads, as shown in Fig. P9.14a P9.14a. The cross-sectional dimensions of the beam are shown in Fig. P9.14b P9.14b. (a) At section a–a, a–a, determine the magnitude of the shear stress in the beam at point po int H. H. (b) At section a–a, a–a, determine the magnitude of the shear stress in the beam at point K. point K. (c) Determine the maximum horizontal shear stress that occurs in the beam at any location within the 5m span length. (d) Determine the maximum compression bending stress that occurs in the beam at any location within the 5-m span length.
Fig. P9.14b P9.14b Cross-sectional dimensions
Fig. P9.14a P9.14a Simply supported timber beam
Solution Section properties: (150 mm)(450 mm)3 I = 12
6
= 1,139.1× 10
mm4
t = 150 mm (a) Shear stress magnitude at H : Q = (150 mm)(150 mm)(150 mm) = τ =
= =
3,375,000 mm3 VQ I t (39,200 (39,200 N)(3 N)(3,375 ,375,, 000 000 mm3 ) (1,139 ,139..1× 106 mm4 )(150 mm mm) 774 kPa
Ans.
(b) Shear stress magnitude at K : Q = (150 mm)(100 mm)(175 mm) = τ =
= =
2,625,000 mm3 VQ I t (39,200 (39,200 N)(2 N)(2,, 625, 625, 000 000 mm mm3 ) (1,139 ,139..1× 106 mm4 )(150 mm mm) 602 kPa
Ans.
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(c) Maximum shear stress at any location: Qmax = (150 150 mm)(2 mm)(225 25 mm) mm)(1 (112 12.5 .5 mm) mm) = 3, 796,87 796,875 5 mm3
τ =
VQ I t
=
(39,200 (39,200 N)(3 N)(3,, 796,875 796,875 mm3 ) (1,139 ,139..1× 106 mm4 )(150 mm mm)
=
871 kPa
Ans.
(d) Maximum bending stress at any location: M max = 39.2 kN-m
σ x
M y
=−
I
=−
4296 = −7.7429
(39.2 kN-m)( k N-m)(225 225 mm)(1,000 N/kN)(1,000 mm/m) 1,139.1× 106 mm4 MPa = 7,740 ,740 kPa (C)
Ans.
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9.15 A laminated wood beam consists of eight 2 in. × 6-in. planks glued together to form a section 6 in. wide by 16 in. deep, as shown in Fig. P9.15a P9.15a. If the allowable strength of the glue in shear is 160 psi, determine: (a) the maximum uniformly distributed load w that can be applied over the full length of the beam if the beam is simply supported and has a span of 20 ft. (b) the shear stress in the glue joint at H , which is located 4 in. above the bottom of the beam and 3 ft from the left support. Assume the beam is subjected to the load w determined in part (a). (c) the maximum tension bending stress in the beam when the load of part (a) is applied.
Fig. P9.15b P9.15b Cross-sectional dimensions
Fig. P9.15a P9.15a Simply supported timber beam
Solution Section properties: (6 in.)(16 in.)3 I = 12
=
2, 048 in.4
t = 6 in. (a) Maximum Q: Q = (6 in. in.)( )(8 8 in.) in.)(4 (4 in.) in.) = 192 192 in. in.3 Maximum shear force V: VQ τ = ≤ 160 psi I t ∴V ≤
(160 psi)(2,048 psi)(2,048 in.4 )(6 in.) 192 in.3
= 10,240
Maximum distributed load w: wL V max = ≤ 10,240 lb 2 2(10, 2(10, 240 lb) ∴ wmax ≤ = 1,024 lb/ft 20 ft
lb
Ans.
(b) Shear force at x = 3 ft: V = 10,240 ,240 lb lb − (1,024 ,024 lb/ lb/fft)(3 ft ft) = 7,1 7,168 lb lb
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Q = (6 in. in.)( )(4 4 in. in.)( )(6 6 in. in.)) = 144 144 in. in.3 τ =
VQ I t
=
(7,168 (7,168 lb)( lb)(144 144 in. in.3 ) (2, (2, 048 in. in.4 )(6 )(6 in.) in.)
=
84.0 psi
Ans.
(c) Maximum tension bending stress at any location: wL2 (1, 024 lb/ft)(20 ft)2 M max = 51, 200 lb-ft = = 51, 8 8
σ x
M y
=−
I
=−
(51 (51, 200 lblb-ft ft)( )( − 8 in.)( in.)(12 12 in./f in./ft) t) 2,048 in.4
=
2, 400 psi (T)
Ans.
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9.16 A 5-ft long simply supported wood beam carries a concentrated load P load P at at midspan, as shown in Fig. P9.16a P9.16a. The cross-sectional dimensions of the beam are shown in Fig. P9.16b P9.16b. If the allowable shear strength of the wood is 80 psi, determine the maximum load P load P that that may be applied at midspan. Neglect the effects of the beam’s self weight.
Fig. P9.16a P9.16a Simply supported timber beam
Fig. P9.16b P9.16b Cross-sectional dimensions
Solution Section properties: (6 in.)(10 in.)3 I = 12
= 500
in.4
t = 6 in. Maximum Q: Q = (6 in. in.))(5 in.) in.)((2.5 2.5 in. in.)) = 75 in. in.3 Maximum shear force V: VQ τ = ≤ 80 psi I t ∴V ≤
(80 (80 psi)(50 psi)(500 0 in.4 )(6 in.) in.) 75 in.3
= 3, 200
lb
Maximum concentrated load P: P V max = ≤ 3, 200 lb 2 ∴ P max ≤
2(3,200 ,200 lb lb) = 6,400 ,400 lb lb
Ans.
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9.17 A wood beam supports the loads shown in Fig. P9.17a P9.17a. The cross-sectional dimensions of the beam are shown in Fig. P9.17b P9.17b. Determine the magnitude and location of: (a) the maximum horizontal shear stress in the beam. (b) the maximum tension bending stress in the beam. b eam.
Fig. P9.17a P9.17a Simply supported timber beam
Fig. P9.17b P9.17b Cross-sectional dimensions
Solution Section properties: (75 mm)(240 mm)3 I = 12
+2
(20 (20 mm mm)(100 mm)3 12
= 89, 733, 333
mm4
(a) Maximum shear force: V max support A max = 9.54 kN = 9,540 N @ support A Check shear stress at neutral axis: Q = (75 mm)(120 mm)(60 mm) +2(20 2(20
τ =
VQ I t
=
mm) mm)(5 (50 0 mm) mm)(2 (25 5 mm) mm) = 590,000 590,000 mm mm3 (9,540 N)(590, N)(590,000 000 mm3 ) (89,733,333 (89,733,333 mm mm4 )(115 )(115 mm) mm)
= 545
kPa
Check shear stress at top edge of cover plates: Q = (75 (75 mm)( mm)(70 70 mm) mm)(8 (85 5 mm) mm) = 446,25 446,250 0 mm3
τ =
VQ I t
=
(9, (9, 540 N)(446, N)(446,250 250 mm3 ) (89,733 (89,733,333 ,333 mm4 )(75 )(75 mm) mm)
=
633 kPa
Maximum shear stress in beam: τ H ,max
=
633 kPa
Ans.
(b) Maximum bending moment: M max support A and and point B point B)) max = 6.49 kN-m (between support A Maximum tension bending stress: M y (6.49 (6.49 kN-m) kN-m)(( − 120 mm)(1 mm)(1,000 ,000 N/kN) N/kN)(1, (1,000 000 mm/m mm/m)) σ x
=−
I
=−
905 = 8.67905
89,733,333 mm4
MPa MPa = 8,680 ,680 kPa kPa (T)
Ans.
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