A beam in Figure 2 is supported by two square planes A\' and B\'. If the allowab
ID: 1842020 • Letter: A
Question
A beam in Figure 2 is supported by two square planes A' and B'. If the allowable stress for the material under the supports at A and B 14 MPa, determine the maximum load P that can be applied to the beam. The bearing plates A' and B' have square cross sections of 150 mm x 150 mm and 250 mm x 250 mm, respectively. Propose the suitable cross-sectional dimensions of the two square plates in Figure 2 if the allowable bearing stress under the support at A and Bis 14 MPa. Given that the value of load P = 8 kN.Explanation / Answer
solution:
for problem 1
1)intensity of UVL is =(0+40/2)=20 KN
where total load act at b/3=1.5/3=.5 from point A andd it is total load=Fd=1.5*20=30 KN
hence to calculate allowable value of load P we have to take reaction at A and B are
Ra=p*Aa=1.4*150^2=31500 N=31.5 KN
Rb=p*Ab=1.4*250^2=87500 N=87.5 KN
hence or moment at A and B is
Ma=0
Fd*.5+Rb*3-P*4.5=0
on putting value we get that
P=61.666 KN
where for Mb=0
Fd*3.5-Ra*3-P*1.5=0
P=7 KN
as we have to choose minimum load for safety we choose
Pmin=7 KN
2)problem 2
as we have choose Pmin=8 KN we can have reaction at A and B as
Ma=0
Fd*.5+Rb*3-P*4.5=0
on putting value we get that
Rb=7 KN
hence surface area for same p=1.4 MPa
Aa=5000 mm2=70.71*70.71 mm2
where for Mb=0
Fd*3.5-Ra*3-P*1.5=0
Ra=31 KN
area for same bearing pressure
Ab=22142.85 mm2=148.80*148.80 mm2
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