A simply supported beam of span 5 m carries two point loads of 5 kn and 7kn as shown in fig . Draw SFD and BMD showing the important values.

 A simply supported beam of span 5 m carries two point loads of 5 kn and 7kn as shown in fig . Draw SFD and BMD showing the important values.











MA = 0

5*1.5+7*3.5-RB=0

RB = 0.4 KN

FY =0

RA+RB-5-7 = 0

RA+6.4-12 =0

RA = 5.6 KN

SF BETWEEN A AND C = 5.6 KN

OR, 5+7-6.4 = 5.6 KN

2) SF BTWN C AND D = 5.6 -5 = .6 KN

OR 7-6.4 = .6 KN

3) SF BTWN D AND B 5.6-5-7=-6.4KN

OR 

-6.4 KN

BM CALCULATION

BM AT PNT A = 0

OR, -5*1.5-7*3.5+6.4*5 = 0

2) BM AT PNT B = 0

OR 5.6*5-5*3.5-7*1.5=0

3) BM AT POINT C = 5.6 *1.5=8.4 KN/M

OR, -7*2+6.4*3.5 = 8.4 KN/M

BM AT D = 5.6 *3.5-5*2

=9.6KN/M

OR 

6.4*1.5 = 9.6 KN/M






Comments

Popular posts from this blog

what is cement in construction?

Advantage and disadvantage of RCC

1) Comparison of working stress method and limit state method