Gicen strain at C=-250
since C is at the neutral axis pof the section, the strain at C is only due to axial load P
sress at C=-250*10-6*50*109 =-12.5MPa= -12.5 N/mm2
cross sectional area of beam=20*40=800 mm2
axial load = -12.5*800/1000=-10 kN
P=10 kN (compressive)
given strain at A=3500
the strain at A is due to axia load P as well as due to bending moment at that section
Strain at A due to P=-250
strain at A due to bending moment = 3500-(-250)=3750
stress at A due to bending moment = 3750*10-6*50*109=187.5 MPa=187.5 N/mm2
bending moment at the section = stress*section modulus of beam = 187.5*20*402/6*10-6= 1 kNm
the bending moment at the section in terms of w = w*12/2=0.5w
0.5w=1
w=2 kN/m
5. A beam 2 meters long is supported as shown below. An unknown distributed load with...
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is subject to a uniform distributed load with
magnitude from to and a moment
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The corresponding shear force diagram is illustrated below
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(a) Draw the bending moment diagram.
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In Appendix C, see the simply supported beam with
a uniformly distributed load. Be careful with units and the sign
convention. For this calculation, the overhung part of the beam
from C to D can be ignored, and the beam is
treated as a simply supported beam of length
2L1. Be careful with units and the sign
convention.
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The beam has the rectangular cross section shown. A beam of
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roller-supported 2 meters from the right end. The beam has a
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Part A If w = 4 kN/m , determine the maximum bending stress in
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