Question

Find the support reaction at A and B in the beam as shown in fig (a).

Answer 1

First draw the FBD of the beam as shown in fig(a) In the fig (b)

6KN is the point load of UDL

W_MNQB = Weight of MNQB

= UDL × Distance(MB)

= 1 × 2

= 2KN, act at a point 1m vertically from point B

W_NPQ = Weight of Triangle NPQ

= 1/2 × MB × (BP – BQ)

= 1/2 × 2 × (3 – 1)

= 2KN and will act at MB/3 = 2/3m from point B 

Since hinged at point A and Roller at point B. let at point A R_HA and R_VA and at point B R_VB is the support reaction, Also beam is in equilibrium under action of coplanar non concurrent force system, therefore:

∑H = 0

R_AH –W_MNQB – W_NPQ = 0

R_AH – 2 – 2 = 0

R_AH = 4KN

∑V = 0

R_AV + R_BV – 5 – 6 = 0

R_AV + R_BV = 11KN .......(i)

M_A = 5 × 1 – 10 + 6 × 4.5 – R_BV × 6 – W_NPQ × (2 – 4/3) – W_MNQB × 1 = 0

5 × 1 – 10 + 6 × 4.5 – R_BV × 6 – 2 × (2 – 4/3) – 2 × 1 = 0

R_BV = 3.11KN

Putting the value of R_BV in equation (i)

R_AV = 7.99KN