Determination of Support Reactions in Beams due to distributed loads

If the beams are subjected to distributed loads like uniformly distributed load (U.D.L.) or distributed uniformly varying load (U.V.L.),in the analysis we can replace these loads by an equivalent load. 

The magnitude of net equivalent point load is equal to total distributed load and acts at the centroid of the loading diagram.

When solving for reactions, the following steps are recommended:

• Draw the beam free body diagram with support reactions and loads.
• Replace the uniform distributed load (if any) with the equivalent point load.
• If the loading (UDL) diagram is rectangle, the equivalent point load is W = wL, and its point of application is L/2 (mid point of the figure)
• If the loading (UVL) diagram is triangle, the equivalent point load is W = wL/2, and its point of application is L/3 as shown in the figure
  
• Solve ΣFx = 0 (sum of all horizontal forces) for finding the horizontal reaction (if any) 
• Solve ΣMA = 0 (sum of moments about support A) or Solve ΣMB = 0 (sum of moments about support B) This will give you the unknown reactions.
• Check if ΣFy = 0 (sum of all vertical forces) is satisfied.

Problem : 2.4

A 6-m-long simply supported beam AB carries uniformly distributed load of 6 kN/m over CD (2m) and a concentrated load of 12kN at E, as shown in Fig. Calculate support reactions.

Solution:

convert the udl (uniformly distributed load) on beam into an equivalent point load W

 W= wl = 6 x 2 = 12kN 

CG of the load acts a distance of 1 m from point C or 2 m from support A. 

Taking moments of the forces about end A,

∑ MA = 0

- 6 x RB  + 12 x2 + 12 x5  = 0    ------(1) 

  6 x RB  = 12 x2 + 12 x5  = 84

                                RB  = 84/6 = 14 KN

Apply the second equilibrium equation is:

∑ Fy = 0

RA + RB - 12  - 12 = 0      

RA + RB  = 24    -----(2)

Substitute the RB   in the above equation

RA  = 24 - RB  = 24 - 14 = 10 KN ans

Problem : 2.5

A 8 m-long simply supported beam carries uniformly distributed load of 3 kN/m over length of 5.5m and uniformly distributed load of 7.5 kN/m over length of 2.5 m, as shown in Fig.Calculate support reactions.

Solution:

convert the udl (uniformly distributed load) on beam into an equivalent point loads P1 and P2

The equivalent point load P1  = 3 x 5.5 =16.5 KN

CG of the load P1 acts a distance of 2.75 m from support A. 

The equivalent point load P2= 7.5 x 2.5 =18.75 KN

CG of the load P2 acts a distance of 1.25 m from  from support B or 6.75m form support A

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

∑ Fx = 0

Rcx    = 0                 ---- (1)

Taking moments of the forces about end A,

∑ MA = 0

- 8 x Rcy  + P1x2.75 + P1 x6.75 = 0    ------(1) 

- 8 x Rcy  + 16.5 x2.75 + 18.75 x6.75 = 0

Rcy  = 21.492 KN 

Apply the second equilibrium equation is: ∑ Fy = 0

RA + Rcy - 16.5  - 18.75 = 0      

RA + Rcy  = 35.25    -----(2)

Substitute the Rcy   in the above equation

RA  = 35.25- Rcy  = 35.25 - 21.492 = 13.758KN

Problem : 2.6

A simply supported beam carries uniformly varying load of w(x) over length as shown in Fig. Calculate support reactions.

Solution:

convert the uvl (uniformly varying load) on beam into an equivalent point load P

The rate of triangular load = w(x)

The ordinate of the load at B = W = w(L)

The total equivalent point load P 

P = Area of the triangular load = 1/2 . WL

This load will act at its centroid, ie. L/3 form support B ,or 2L/3 form support A

Apply the second equilibrium equation is:

 ∑ Fy = 0

RA + RB - P = 0 

RA + RB = P = WL/2 -------(1)

Taking moments of the forces about end A,

∑ MA = 0

∑ MA = {WL/2 }2L/3 - RB L   ----(2)

RB  = WL/3

and RA =  WL/6  (Form eq 1)

Assignment :

Q 1. Determine the support reactions of a simply supported beam loaded with a UDL of 7.7kN/m as shown in fig.

Q 2. Determine the support reactions of a simply supported beam, is loaded with a udl and several concentrated loads as shown in fig.

Q 3. Determine the support reactions of a simply supported beam, with a point load and udl as shown in fig.

Q 4. Determine the support reactions of a simply supported over hanging beam, with a point load and uniformly varying load (UVL) as shown in fig.