Question

Draw the shear and moment diagrams for the beam.

Answer 1

Calculation of support reactions

In the case of the beam shown , these 3 are the reactions :

B_x ​​​​​​, B_y and A_y

Applying equations of static equilibrium, we get :

F_{x =} 0 , B_x = 0

F_y = 0 , A_y + B_y = (-0.2*9) - 2 = 3.8 kip

Considering the z axis passing through B , and taking the moment of all forces about z axis (taking clockwise +ve and anti-clockwise -ve)

M_z = 0

A_y*18 - (2*24) - (0.2*9*4.5) = 0

A_y = 3.1167 kip

B_y = 0.683 kip

Shear force calculations

In case of point load acting at a point, we should calculate shear force on both sides. (left and right of the point)

F_Bright =0

F_B left = 0.683 kip

F_B = 0.683 - (0.2*9) = -1.12 kip

F_Aright = -1.12 kip

F_{A left} = 0.683 - 1.8 + 3.1167 = 2 kips

F_cright = 2 kips

F_{c left} = 2-2 = 0

The value of shear force is plotted in the sfd below

Bending moment calculations

M_B = 0

Let D be the point at a distance 9 ft. From B

M_D = (0.683*9) - (0.2*9*4.5) = -1.953 kip ft.

M_A = (0.683*18) - (0.2*9*13.5) = -12.006kip ft.

The values of bending moments are plotted in the BMD

Maximum bending moment will occur at the point of 0 shear force which can be calculated using the property of similar triangles of shear force diagram between B and D

0.683/x = 1.12/(9-x)

x= 3.42 ft.

Maximum +ve bending moment (at 3.42 ft.) =(0.683*3.42) - (0.2*3.42*3.42/2) = 1.17 kip

Maximum -ve bending moment = -12.006 kip

Point of contraflexure can be identified by using the equation of BM for part BD and equate it to 0.

if 'a' is the distance from B

0.683a - 0.2*9*(a-4.5) = 0

a = 7.28 ft.