Question

The bending moment (in kNm units) at the mid-span location X in the beam with overhangs shown below is equal

• A. 0
• B. -10
• C. -15
• D. -20

Answer 1

The correct option is C -15


Let reaction at B be R.

$M_A=0$

$20\times 1+R\times 2-10\times 3=0$

$R\times 2=30-20$

$R=5kN$

Now, BM at X

$B{M}_X$ = 10 x 2 - 5 x 1 (hogging)

= 15 kN-m (hogging).