Question

strength and materials

heerhang beam with the loading shown oxcomms Identify the ums, maximum, and jumps in values on the diagrams Shear Diagram Moment Diagram

Answer 1

Let left end of the beam be 'C' and R_a and R_b be the vertical reactions at A and B in upwards direction

Taking moments about B

Clockwise moments as +ve and Anticlockwise as -ve

R_ax12 - 50 - 1.5x12x12/2 = 0

=> R_a = 13.167 kip (upwards)

Also

R_a + R_b = 1.5x12

=> R_b= 4.833 kip (upwards)

Shear force: Upwards as +ve and Downwards as -ve

S.F at C = 0

S.F at left of A = 0

S.F at right of A = 13.167 kip

S.F at left of B = 13.167 - 1.5x12 = - 4.833 kip

S.F at right of B = - 4.833 + 4.833 = 0

Bending Moment: C.W as +ve and A.C.W as -ve

M_c = - 50 kip-ft

M_a = - 50 kip-ft

M_b = - 50 + 13.167x12 - 1.5x12x6 = 0

Maximum Bending moment will occur at which Shear force is zero

Let S.F be zero at a position 'x' from left end C

V_x = 13.167 - 1.5(x-3)

0 = 13.167 - 1.5(x-3)

=> x = 11.778 ft from C

= 8.778 ft from A

M_max = - 50 + 13.167(8.778) - 1.5(8.778²)/2

= 7.79 kip-ft
13.167 7.79 +ve tve -ve 11.778 ft 11.778 -ve 4.833 50 50 S.F diagram in kip B.M diagram in kip-ft