Module 2: Equilibrium


Equilibrant: The force which is equal in magnitude, opposite in direction and collinear to the resultant force is called as equilibrant. 

This equilibrant cancel the effect of resultant force (or the force system) acting on the body.




Question: Find support reactions. Portion CD is an extended part of the beam AB. 



Solution:



Applying the conditions of equilibrium to beam AB, 

Σ MA = 0

(HA x 0) + (VA x 0) + 40 – (100sin300 x 3) – (100cos300 x 0.5) – (25 x 0.25) – (70 x 4) + (RBsin700 x 5) + (RBcos700 x 0) = 0

0 + 0 + 40 – 150 – 43.30 – 6.25 – 280 + (RB x 4.698) + 0 = 0

(RB x 4.698) = – 40 + 150 + 43.30 + 6.25 + 280

RB = 439.55 / 4.698

RB = 93.561 kN

 

Σ Fx = 0

HA + 100cos30 - RBcos70 + 25 = 0

HA = -100cos30 + RBcos70 - 25

       = - 86.60 + 31.99 - 25

HA  = - 79.60 kN (towards left) 

                   Or     

 HA  = 79.60 kN (towards left) 


Σ Fy = 0

VA – 100sin30 + RBsin70 – 70 = 0

VA = 100sin30 – RBsin70 + 70

       = 50 – 93.561sin70 + 70

       = 50 – 87.918 + 70

VA  = 32.082 kN (upward) 

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Question: For the beam shown in figure find the support reactions.

Solution:

Applying the conditions of equilibrium to beam AB, 

Σ MA = 0

(HA x 0) + (VA x 0) – (2 x 1.33) – (5 x 3) – 10 – (1 x 5.5) + (RB x 6) = 0

0 + 0 – 2.66 – 15 – 10 – 5.5 + (RB x 6) = 0

(RB x 6) = 33.166

RB = 33.166 / 6

RB = 5.527 kN 


Σ Fx = 0

  HA = 0

 

Σ Fy = 0

VA – 2 – 5 – 1 + RB = 0

VA = 8 – 5.527

       = 2.473 kN

VA  = 2.472 kN (upward) 

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Question: Figure shows beam AB hinged at A and roller supported at B. The L shaped portion DEF is an extended part of beam AB. For the loading shown find support reactions. 

Solution:




Applying the conditions of equilibrium to beam AB, 

Σ MA = 0

(HA x 0) + (VA x 0) - (20 x 2) – (25 x 4) + (30cos40 x 1.5) – (30sin40 x 8) + (RB x 10) = 0

0 + 0 – 40 – 100 + 34.471 – 154.269 + (RB x 10) = 0

- 259.798 +  (RB x 10) = 0

RB = 259.798/10

RB = 25.979 kN

 

Σ Fx = 0

HA - 30cos40 = 0

HA = 30cos40

HA =  22.981 kN  (towards right)  

 

Σ Fy = 0

VA – 20 – 25 – 30sin40 + RB = 0

VA – 20 – 25 – 19.283 + 25.979 = 0

VA = 38.304 kN (upward) 

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