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Support reactions in beams due to an applied couple or moment

 

couple of Forces
A couple is defined as two parallel forces with the same magnitude    but opposite in direction separated by a perpendicular distance d.
The net external effect of a couple is that the net force equals zero and the magnitude of the net moment equals = F.d 
The resultant of a couple at any point is always a pure constant moment  and equal to 
  
  M = F d
Problem : 2.7
A 6 m-long simply supported beam carries a anti clockwise couple 400 N-m and a point load as shown in Fig. Calculate support reactions

Solution:
Let the vertical reaction at support A is Ra and at support B is Rb.
Here the vertical forces are upwards reactive forces Ra ,Rb and downward point load of 1000 N. The horizontal pair of forces are formed as a couple and its resultant is a moment of 400N-m

Apply the  equilibrium equation is:

∑ Fy = 0

R+ Rb - 1000 = 0      

R+ Rb  = 1000  -----(1)

Taking moments of the forces about end A,

∑ MA = 0

-  400  + 1000 x 4 - 6.Rb  = 0

Rb  = 3600/6 = 600 N

substitute Rvalue in equ.1

R  = 1000 -  Rb = 1000- 600 

R= 400N

Problem : 2.8
A simply supported beam carries a anti clockwise couple M as shown in Fig. Calculate support reactions
Let the vertical reaction at support A is R and at support B is Rb
The equilibrium  equation is:

∑ Fy = 0

R+ Rb  = 0      

R+ Rb  = 0  -----(1)

Taking moments of the forces about end A,

∑ MA = 0

-  M - L.Rb  = 0

Rb  = - M / L (downward reaction)

substitute Rvalue in equ.1

R  = + M/L (upward reaction)

Note: The application of a moment at any point in the beam, the support reactions doesn't change 

Problem : 2.9
A 5 m-long simply supported overhanging beam carries a clockwise moment of 6 N-m at point C and a concentrated load of 4KN at point B as shown in Fig. Calculate support reactions

Solution:
Let the vertical reaction at support A is Ra and at support D is Rd.

Apply the  equilibrium equation is:

∑ Fy = 0

R+ Rd - 4 = 0      

R+ Rd  = 4 KN  -----(1)

Taking moments of the forces about end A,

∑ MA = 0

 6  + 4x5 - 4.Rb  = 0

Rd  = 26/4 = 6.5 N (upward reaction)

substitute Rvalue in equ.1

R  = 4 -  Rd = 4 - 6.5  

R= -2.5 KN (downward reaction)

Assignment :

Q 1. Determine the support reactions of a simply supported overhang beam loaded as shown in fig.

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.

Comments

  1. 1.Ra=-14.3kN, Rb=67.3
    2.Rb=32.8kN/m, Re=77.2kN/m
    3.Ra=-436N, Rb=656N

    ReplyDelete
  2. Ra=-14.3kN,Rb=67.3
    Rb=34.8kN/M,Re=75.8KN/M
    Ra=-436N,Rb=656N

    ReplyDelete
  3. 1.Ra=-14.3KN,Rb=67.3
    2.Re=77.2KN/m,Rb=32.8KN/m
    3.Ra=-436N,Rb=656N

    ReplyDelete
  4. Ra=-14.3KN,Rb=67.3
    2.Re=77.2KN/m,Rb=32.8KN/m
    3.Ra=-436N,Rb=656N

    ReplyDelete
  5. Ra=-14.32KN,RB=67.9N
    RB=32.6Kn/M,Re=76.95KN/M
    Ra=-436N,RB=656N

    ReplyDelete
  6. Ra=-14.32KN,RB=67.9N
    RB=32.6Kn/M,Re=76.95KN/M
    Ra=-436N,RB=656N

    ReplyDelete
  7. Ra=-14.32KN,RB=67.9N
    RB=32.6Kn/M,Re=76.95KN/M
    Ra=-436N,RB=656N

    ReplyDelete
  8. Ra=-14.32KN,RB=67.9N
    RB=32.6KN/M,RE=76.95KN/M
    RA=-436N,RB=656N

    ReplyDelete
  9. Ra=-14.32KN,RB=67.9N
    RB=32.6KN/M,RE=76.95KN/M
    RA=-436N,RB=656

    ReplyDelete
  10. Ra=-14.3#KN, RB=67.9N
    RB=32.6KN/M, RE=76.95KN/M
    RA=-436N, RB=656

    ReplyDelete
  11. Ra=-14.3#KN, RB=67.9N
    RB=32.6KN/M, RE=76.95KN/M
    RA=-436N, RB=656

    ReplyDelete

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