Evaluate Bending moment for overhanging beam:

Draw the shear force & bending moment diagrams for 15 m span overhanging beam, that overhangs on both sides. This is subjected to a u.d.l. of 5 kN/m on left side overhanging portion of length 5 m and a u.d.l. of 4 kN/m on right side overhanging portion of length 4 m. Mention the numerical values at all of salient points.

Solution

To determine Reaction at B

letting right side, the moments around A,

M_A       =+ R _B × 6 - 4 × 4 × ( 6 + (4/2))

M_A = 6 R_B - 128

Letting left side, take moments around A,

M _A = ( 5 × 5 × (5/2))  = - 62.5

Equating these two values, we obtain

6 R_B - 128 = - 62.5

6 R_B = 65.5

R_B = 10.92 kN

R_A =  (5 × 5) + (4 × 4) - RB

R_A = 41 - 10.92 = 30.08 kN

Shear Force (Starting from the left end C)

 SF at C, F_C = 0

SF just left of A, F_A = 0 - 5 × 5 = - 25 kN

SF just right of A, F_A = - 25 + 30.08 = + 5.08 kN

SF just left of B, F_B = + 5.08 kN

SF just right of B, F_B = + 5.08 + 10.92 = + 16 kN

SF at D, F_D = + 16 -  (4 × 4) = 0

Figure

Bending Moment

BM at C and D, M_C  = M_D = 0

BM at A, MA = - 5 × 5 ×(5/2) = - 62.5 kN m

BM at B, MB = - 4 × 4 ×(4/2) = - 32 kN m