Apply the equilibrium equation is:
∑ Fy = 0
Ra + Rb - 1000 = 0
Ra + Rb = 1000 -----(1)
∑ MA = 0
- 400 + 1000 x 4 - 6.Rb = 0
Rb = 3600/6 = 600 N
substitute Rb value in equ.1
Ra = 1000 - Rb = 1000- 600
Ra = 400N
∑ Fy = 0
Ra + Rb = 0
Ra + Rb = 0 -----(1)
∑ MA = 0
- M - L.Rb = 0
Rb = - M / L (downward reaction)
substitute Rb value in equ.1
Ra = + M/L (upward reaction)
Note: The application of a moment at any point in the beam, the support reactions doesn't change
Apply the equilibrium equation is:
∑ Fy = 0
Ra + Rd - 4 = 0
Ra + Rd = 4 KN -----(1)
∑ MA = 0
6 + 4x5 - 4.Rb = 0
Rd = 26/4 = 6.5 N (upward reaction)
substitute Rd value in equ.1
Ra = 4 - Rd = 4 - 6.5
Ra = -2.5 KN (downward reaction)
Assignment :
Q 2. Determine the support reactions of a simply supported overhang beam loaded as shown in fig.
Q 3. Determine the support reactions of a simply supported overhang beam loaded as shown in fig.
1.Ra=-14.3kN, Rb=67.3
ReplyDelete2.Rb=32.8kN/m, Re=77.2kN/m
3.Ra=-436N, Rb=656N
Ra=-14.3kN,Rb=67.3
ReplyDeleteRb=34.8kN/M,Re=75.8KN/M
Ra=-436N,Rb=656N
Ra=-14.3KN and Rb=67.3
ReplyDelete1.Ra=-14.3KN,Rb=67.3
ReplyDelete2.Re=77.2KN/m,Rb=32.8KN/m
3.Ra=-436N,Rb=656N
Ra=-14.3KN,Rb=67.3
ReplyDelete2.Re=77.2KN/m,Rb=32.8KN/m
3.Ra=-436N,Rb=656N
Ra=-14.32KN,RB=67.9N
ReplyDeleteRB=32.6Kn/M,Re=76.95KN/M
Ra=-436N,RB=656N
Ra=-14.32KN,RB=67.9N
ReplyDeleteRB=32.6Kn/M,Re=76.95KN/M
Ra=-436N,RB=656N
Ra=-14.32KN,RB=67.9N
ReplyDeleteRB=32.6Kn/M,Re=76.95KN/M
Ra=-436N,RB=656N
Ra=-14.32KN,RB=67.9N
ReplyDeleteRB=32.6KN/M,RE=76.95KN/M
RA=-436N,RB=656N
Ra=-14.32KN,RB=67.9N
ReplyDeleteRB=32.6KN/M,RE=76.95KN/M
RA=-436N,RB=656
Ra=-14.3#KN, RB=67.9N
ReplyDeleteRB=32.6KN/M, RE=76.95KN/M
RA=-436N, RB=656
Ra=-14.3#KN, RB=67.9N
ReplyDeleteRB=32.6KN/M, RE=76.95KN/M
RA=-436N, RB=656