A beam of uniform rectangular section 100 mm wide and 240 mm deep is simply supported at its ends. It carries a uniformly distributed load of 9.125 KN/m run over the entire span of 4m. Find the deflection at the centre of E= 1.1×10^4 N/mm^2
Given quantities:
W=9.125∗103NmW=9.125*10^3\frac{N}{m}W=9.125∗103mN
L=4mL=4mL=4m
E=1.1∗104Nmm2=1.1∗1010NmE=1.1*10^4 \frac{N}{mm^2}=1.1*10^10\frac{N}{m}E=1.1∗104mm2N=1.1∗1010mN
b=10−1mb = 10^{-1}mb=10−1m
h=0.24mh=0.24mh=0.24m
x−?x-?x−?
y−?y-?y−?
x=WL26EI=12∗WL26Ebh3=2WL2Ebh3=2∗9.125∗103∗161.1∗1010∗0.1∗(0.24)3=19.2mmx=\frac{WL^2}{6EI}=\frac{12*WL^2}{6Ebh^3}=2\frac{WL^2}{Ebh^3}=2*\frac{9.125*10^3*16}{1.1*10^{10}*0.1*(0.24)^3}=19.2mmx=6EIWL2=6Ebh312∗WL2=2Ebh3WL2=2∗1.1∗1010∗0.1∗(0.24)39.125∗103∗16=19.2mm
y=WL28EI=12∗WL28Ehb3=1.5WL2Ehb3=14.4mmy=\frac{WL^2}{8EI}=\frac{12*WL^2}{8Ehb^3}=1.5\frac{WL^2}{Ehb^3}=14.4mmy=8EIWL2=8Ehb312∗WL2=1.5Ehb3WL2=14.4mm
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