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5-25
:
darw the shear and bending-moment diagram for the beam and loading shown and determine the maximum normal stress due to bending
.
solution:
Fy=0
20+40-Rc-Rb=0
Mc=0
(-20*1.5)+(40*2.4)-(Rb*3.9)=0
Rb=18.33 KN Rc=41.67 KN
Fy=0
S1=-20 KN
M1=0 20X+M1=0
M1=-20X
Fy=0
20-41.67+S2=0
S2=21.67 KN
M2=0
20X-41.67(X-1.5)+M2=0
M2=21.67X-62.505
Fy=0
20-41.67+40+S3=0
S3=-18.33 KN
M3=0
20X-41.67(X-1.5)-40(X-3.9)+M3=0
M3=61.67X-218.505
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Fy=0
20+40-Rc-Rb=0
Mc=0
(-20*1.5)+(40*2.4)-(Rb*3.9)=0 Rb=18.33 KN Rc=41.67 KN
Fy=0
S1=-20 KN
M1=0 20X+M1=0 M1=-20X
Fy=0
20-41.67+S2=0 S2=21.67 KN
M2=0
20X-41.67(X-1.5)+M2=0
M2=21.67X-62.505
Fy=0
20-41.67+40+S3=0 S3=-18.33 KN
M3=0
20X-41.67(X-1.5)-40(X-3.9)+M3=0 M3=61.67X-218.505