Question

A simply supported beam of span 8m cames
UDL of 16 kN/m at a point out of
60 kN acting at it's centre. Calculate the
maximum shear force.​

Answer 1

Explanation:

solution:

I. Support Reactions:

∑MA=0

5×1.5+7×3.5−RB×5=0

5×RB=32

RB=6.4kN

∑Fy=0

RA+RB−5+7=0

RA+RB=12

RA=5.6kN

II. SF calculations

SFatA=+5.6kN

CL=+5.6kN

CR=5.6−5=0.6kN

DL=+0.6kN

DR=+0.6−7=−6.4kN

BL=−6.4kN

B=+6.4−6.4=0kN(∴ ok

B.M. calculation:-

B.M at A and B= 0 Since support A and B are simple.

B.M at C = 5.6 × 1.5=8.4 kN-m

B.M at D = 6.4× 1.5=9.6 kN-m