#428 Draw the shear and moment diagrams

Draw the shear and moment diagrams - Mechanical Engineering

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Free Chegg Question

Draw the shear and moment diagrams for the beam for the following beam:

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Free Chegg Answer

Let R_Av and R_Bv be the vertical reactions acting at point A and B in upward direction. And R_Ah be the horizontal reaction at point A.

Sign Convention used is: -

1. “Upward Position” is taken as +ve and “Downward Position” as -ve.

2. “Towards right” is taken as +ve and “Towards left” as -ve.

Balancing the horizontal forces,

R_Ah = 0.

And balancing the vertical forces,

R_Av + R_Bv = 6 + (2x2)

R_Av + R_Bv = 10 KN ...............i)

Taking moments about pt. A,

R_Bv x 4 = 6 x 1 + (2 x 2 x 3)

R_Bv = 18/4 = 4.5 KN.

Using this value in eq. i), we get

R_Av = 10-4.5 = 5.5 KN.

Shear Forces at various points

At point A, SF_A= 5.5 KN (+ve)

At point D (point where 6KN point load is acting),

SF_D = 5.5 – 6 = -0.5 KN (downwards)

At point C, SF_C = -0.5 KN (downwards)

At point B, SF_B = -0.5 – 2x2 = -4.5 KN (downward).

Bending Moments at various points

Let section is taken between point A and D,

Moment equation is M_x = R_Avx

Where “x” is the distance between section from point A.

At x=0, M_A = 0 KN-m

At x=1m, M_D=5.5x1 = 5.5 KN-m.

Now let section is between point D and C,

Moment equation is M_x = R_Avx – 6(x-1)

At x=1m, M_D= 5.5 KN-m

At x=2m, M_C = 5.5x2-6(2-1) = 5KN-m.

Again, moving section between point C and B,

Moment equation is M_x = R_Avx -6(x-1) – 2(x-2)(x-2)/2

At x=2m, M_C = 5.5x2-6(2-1) = 5KN-m.

At x=4m, MB = 5.5x4-6(4-1)-2(4-2)(4-2)/2= 0 KN-m.

Points	Shear Forces (KN)	Bending Moment (KN-m)
A	5.5 KN	0 KN-m
D	-0.5 KN	5.5 KN-m
C	-0.5 KN	5 KN-m
B	-4.5 KN	0 KN-m

SF and Moment diagrams are shown below :-