Draw the shear force diagram and bending moment diagram for the beam loaded as shown in fig

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written 6.5 years ago by

tarashankarsharma • 550

modified 5.4 years ago by

sanketshingote • 100

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strength of materials

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written 6.4 years ago by

teamques10 ★ 69k

• modified 6.4 years ago

enter image description here

S.F.D enter image description here

B.M.D enter image description here $\sum F_y=0$

$V_B+V_E=20*4+50+25*2$

$V_B+V_E+180$

and taking moment about E,

$\sum M_E = 0$

$-20*4*(\frac{4}{2}+5+5+2)+V_b*(5+5+2)-50*(5+2)-(25*2)*\frac{2}{2} = 0$

$V_B=126.67KN$

Now, $V_G = 180-V_B = 53.33KN$

B.M analysis:

BM in the section AB, distance * from A

$BM_x = -20*\frac{x^2}{2}$

At x = 0,

$BM_A = 0$

At x = 4, …

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