Determination of Support Reactions in Beams due to point loads

Problem : 2.1

Determine the support reactions of a simply supported beam, with a point load W, at a point as shown in fig. 

Solution : 

Assigning the unknown support reactions   R_A  and R_B at supports as shown in the figure.

Applying the three equilibrium equations to find the three unknowns in this way:

Along direction x, there is no imposed force applied to the structure. 

Thus the first equilibrium equation is: ∑ Fx = 0

Unless there is an imposed load along the beam longitudinal x - axis, this reaction will always be zero.

Along direction y, the imposed force P is applied to the center of the beam, as well as support reactions.   Thus the second equilibrium equation is: ∑ Fy = 0

R_A + R_B - W = 0       -----(1)

For the third equation ∑ Mxy = 0

we have to choose one point around which the moments are calculated. It is often more convenient to select a point through which some of the forces are directed (such as point A in our example), because, the resulting moments for these forces would be zero. 

So, around point A, support reactions  and  have no lever arm, imposed force W  have a lever arm equal to a and support reaction  have a lever arm equal to the beam length. Assuming counter-clockwise positive rotation, the moment equation becomes:

∑ M_A = 0

R_B L - W.a = 0    ------(2)

There are three unknowns, and we have three equations, therefore it is possible to solve the system of equations and obtain the unknown support reactions.   From the second equation we can directly obtain :

R_B  =  W (a/L)

And finally, substituting R_B in to the second equation,  and found R_A 

R_A= W - R_B  = W  -  W (a/L) 

R_A  =  (1- a/L) W = (L-a) W/L

R_A  =  W (b/L)

Problem : 2.2

Determine the support reactions of a simply supported beam, with point loads as shown in fig. 

Solution : 

             Assigning the unknown support reactions  H_A , R_A  and R_B , at the supports A and B

Along direction x, there is no imposed force applied to the structure. There is only the one unknown support reaction H_A.

Thus the first equilibrium equation is: ∑ Fx = 0

H_A is directly found from the first equation equal to zero. Unless there is an imposed load along the beam longitudinal axis, this reaction will always be zero.

H_A    = 0                 ---- (1)

Along direction y, the imposed force P is applied to the center of the beam, as well as support reactions.   Thus the second equilibrium equation is: ∑ Fy = 0

R_A + R_B - 30  - 15 = 0      

R_A + R_B  = 45    -----(2)

For the third equation ∑ Mxy = 0

we have to choose one point around which the moments are calculated. It is often more convenient to select a point through which some of the forces are directed (such as point A in our example), because, the resulting moments for these forces would be zero. So, around point A, support reactions  and  have no lever arm,  Assuming clockwise moments are positive and anti clockwise moments are negative, the third equation becomes:

∑ M_A = 0

- 6 x R_B  + 30x2 + 15 x4  = 0    ------(3)

  From the third equation we can directly obtain :

R_B  =  20 KN

And finally, substituting R_B in the second equation, we can found R_A 

R_A= 45 - R_B  = 45 - 20  

R_A =  25 KN

Problem : 2.3

Determine the support reactions of a overhanging beam, with point loads as shown in figure. 

Solution : 

             Assigning the unknown support reactions  H_A , R_A  and R_C , at the supports A and C

Along direction x, there is no imposed force applied to the structure. There is only the one unknown support reaction H_A  in longitudinal axis, this reaction will always be zero.

H_A    = 0                 ---- (1)

Along direction y, the imposed a point load P is applied at B, and Q applied at free end D are in downwards directions and support reactions R_{A ,} R_{C  in upwards directions.}  

Thus the second equilibrium equation is: ∑ Fy = 0

R_A + R_C - P  - Q = 0      

R_A + R_C  = P + Q    -----(2)

For the third equation ∑ Mxy = 0

Taking moments about point A, and Assuming clockwise moments are positive, anti clockwise moments are negative, the third equation becomes:

∑ M_A = 0

 - L x R_C  + PxL/2 + Q x(L+a)  = 0    ------(3)

  From the third equation we can directly obtain :

R_C  =  {PxL/2 + Q x(L+a) }/L

R_C = P/2 + Q (1 + a/L)

And finally, substituting R_C in the second equation, we can found R_A 

R_A  =  P + Q -  R_C  

R_A = P/2  - Q{a/L}

Assignment :

Q 1. Determine the support reactions of a simply supported beam, with point loads as shown in fig.

 Q 2. Determine the support reactions of a simply supported beam, with point loads as shown in fig. 

Q 3. Determine the support reactions of a overhanging supported beam, with point loads as shown in fig.