ΣMD=0
8R1=200(5)+400(1)
R1=175 lb
ΣMA=0
8R2=200(3)+400(7)
R2=425 lb
By ratio and proportion
yC1=21255
yC=425 lb⋅ft
From the conjugate beam
ΣMD=0
8F1+12(4)(1600)[1+23(4)]=12(3)(525)[5+13(3)]+12(5)(2125)[23(5)]
F1=1337.5 lb⋅ft2
ΣMA=0
8F2+12(4)(1600)[3+13(4)]=12(3)(525)[23(3)]+12(5)(2125)[3+13(5)]
F2=1562.5 lb⋅ft2
Consider the section to the left of B in conjugate beam
MB=12(3)(525)[13(3)]−3F1
MB=787.5−3(1337.5)
MB=−3225 lb⋅ft3
Thus, the deflection at B is
EI δB=MB
EI δB=3225 lb⋅ft3 answer
Consider the section to the right of C in conjugate beam
MC=12(1)(yC)[13(1)]−1F2
MC=12(1)(425)[13(1)]−1(1562.5)
MC=−1491.67 lb⋅ft3
Thus, the deflection at C is
EI δC=MC
EI δC=−1491.67 lb⋅ft3
EI δC=1491.67 lb⋅ft3 downward answer
dito po sa part ng solution
dito po sa part ng solution
ΣMA=0
8F2+(1/2)(4)(1600)[(1/3)(4)]=(1/2)(3)(525)[(2/3)(3)]+(1/2)(5)(2125)[3+(1/3)(5)]
F2=1562.5 lb⋅ft2
hindi po ba
ΣMA=0
8F2+(1/2)(4)(1600)[3+(1/3)(4)]=(1/2)(3)(525)[(2/3)(3)]+(1/2)(5)(2125)[3+(1/3)(5)]
F2=1566.25 lb⋅ft2
You are correct. Update has
In reply to dito po sa part ng solution by Mark123 (not verified)
You are correct, fortunately, the answer for F2 is correct. Update has been made for the moment arm. Thank you.