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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   RA  and RB 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

RA + RB - 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:

∑ MA = 0

RB 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 :

RB  =  W (a/L)

And finally, substituting RB in to the second equation,  and found RA 

RA= W - RB  = W  -  W (a/L) 

RA  =  (1- a/L) W = (L-a) W/L

RA  =  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  HA , RA  and RB , 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 HA.

Thus the first equilibrium equation is: ∑ Fx = 0

HA 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.

HA    = 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

RA + RB - 30  - 15 = 0      

R+ R = 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:

∑ MA = 0

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

  From the third equation we can directly obtain :

RB  =  20 KN

And finally, substituting RB in the second equation, we can found RA 

RA= 45 - RB  = 45 - 20  

R=  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  HA , RA  and RC , 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 Hin longitudinal axis, this reaction will always be zero.

HA    = 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 RA , R in upwards directions.  

Thus the second equilibrium equation is: ∑ Fy = 0

R+ RC - P  - Q = 0      

R+ R = 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:

∑ MA = 0

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

  From the third equation we can directly obtain :

R =  {PxL/2 + Q x(L+a) }/L

R= P/2 + Q (1 + a/L)

And finally, substituting RC in the second equation, we can found RA 

RA  =  P + Q -  R 

R= 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. 

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