Civil Engineering - Applied Mechanics - Discussion
Discussion Forum : Applied Mechanics - Section 1 (Q.No. 36)
36.
The vertical reaction at the support A of the structure shown in below figure, is
Discussion:
10 comments Page 1 of 1.
Kiran said:
4 years ago
Ra(vertical) = 2t is the right answer.
Let Ra be the reaction at A perpendicular to the support. Then, Ra(Horizontal)= Racos45
Ra(vertical)= Rasin45.
Using the equation of equilibrium,
Sum of the moment about B=0.
-Rasin45°6-Racos45° 4+3*4+2*2+2*2 = 0
Ra=2√(2).
Therefore, Ra(vertical) = Rasin45=2t.
Let Ra be the reaction at A perpendicular to the support. Then, Ra(Horizontal)= Racos45
Ra(vertical)= Rasin45.
Using the equation of equilibrium,
Sum of the moment about B=0.
-Rasin45°6-Racos45° 4+3*4+2*2+2*2 = 0
Ra=2√(2).
Therefore, Ra(vertical) = Rasin45=2t.
Fahim Ahmed Bhuiyan said:
5 years ago
We got horizontal reaction component Rbcos45°(leftward) and vertical reaction
component Rbsin45;(upward) from point B. Ok, now taking moment at A:
2t*2 + 2t*2 + 3t*4- 4*Rbsin45 -6*Rbcos45 = 0,
Or, 20 -2.828Rb -4.242Rb = 0,
Or, Rb= 2.83t.
Taking sum of load in Y direction,
Ra(up)+ Rbcos45= 2t +3t,
Ra(up) + 2.00= 5t,
Ra(up)= 3.00t -> Answer.
component Rbsin45;(upward) from point B. Ok, now taking moment at A:
2t*2 + 2t*2 + 3t*4- 4*Rbsin45 -6*Rbcos45 = 0,
Or, 20 -2.828Rb -4.242Rb = 0,
Or, Rb= 2.83t.
Taking sum of load in Y direction,
Ra(up)+ Rbcos45= 2t +3t,
Ra(up) + 2.00= 5t,
Ra(up)= 3.00t -> Answer.
Mohanraj joshi said:
5 years ago
Ra*6 = 2*2+3*4.
= 4+12.
Ra = 18÷6 = 3.
= 4+12.
Ra = 18÷6 = 3.
(1)
Manoj said:
6 years ago
But that time they are very close to each other and try to developing an internal resistance between than that we only take moment about a so we take about a equally and simultaneously than easily find the value of Ra after finding Ra putting the value of RA in equation 2nd than reaction will be fined.
Ataa alkaby said:
6 years ago
Everyone forget the horizontal reaction for joint A.
It also makes a moment about joint B. The correct answer is 5/3t=1,67t.
It also makes a moment about joint B. The correct answer is 5/3t=1,67t.
(1)
Vishu tyagi said:
7 years ago
Ra + Rb sin45= 2t+3t,
Rb cos 45 = 2t,
Ra + 2t/cos45 * sin45 = 2t + 3t.
Ra = 5t -2t,
Ra = 3t.
Rb cos 45 = 2t,
Ra + 2t/cos45 * sin45 = 2t + 3t.
Ra = 5t -2t,
Ra = 3t.
(1)
Vidhya said:
7 years ago
Hi @Jinal.
Can you explain how you multiplied 2*2?
Can you explain how you multiplied 2*2?
Navdeep singh said:
7 years ago
Ra *5 ( total downward load 3t+2t=5t).
Ra*5t= 2t*2+3t*4,
Ra*5t=4t+12t,
Ra*5t=16t,
Ra=16t/5t,
Ra = 3t(approx).
Ra*5t= 2t*2+3t*4,
Ra*5t=4t+12t,
Ra*5t=16t,
Ra=16t/5t,
Ra = 3t(approx).
Raj said:
10 years ago
RaX6 = 2tX2+3tX4.
RaX6 = 18t.
Ra = 3t.
RaX6 = 18t.
Ra = 3t.
Jinal said:
1 decade ago
Ra*6 = 3*2+2*4+2*2.
= 18/6.
= 3.
= 18/6.
= 3.
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