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Mechanics of Materials CIVL 3322 / MECH 3322 Shear Stress in Beams II
7 November, 1940 – Four months after the bridge's completion, the middle section of the Tacoma Narrows Bridge across the Tacoma Narrows in state of Washington collapsed in a windstorm.
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Fx∑ = MIzA '∫ ydA '− M + ΔM
IzA '∫ ydA '+ FH = 0
− ΔMIzA '∫ ydA '+ FH = 0
FH = ΔMIzA '∫ ydA '
Q = ydA 'A '∫
FH = ΔMQIz
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τ = VQIzt
t = width of the section at some depth in the beam
Q= yiAii∑
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9.9 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. P9.9a Cantilever beam Fig. P9.9b Cross-sectional dimensions
Solution Shear force in cantilever beam: V = 7.2 kN = 7,200 N Shear stress formula:
V QI t
Section properties:
3
6 4(120 mm)(280 mm)219.52 10 mm
12I
t = 120 mm
Distance below top surface of beam y Q
35 mm 105 mm 514,500 mm3 140.6 kPa
70 mm 70 mm 882,000 mm3 241 kPa
105 mm 35 mm 1,102,500 mm3 301 kPa
140 mm 0 mm 1,176,000 mm3 321 kPa
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Homework
¢ Problem P9.10 ¢ Problem P9.11 ¢ Problem P9.13
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