Civil Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 2 (Q.No. 15)
15.
The reaction at support A of the beam shown in below figure, is
Discussion:
61 comments Page 1 of 7.
S K Ahmad said:
2 years ago
Point A is not shown in the figure. Then how can anyone calculate reaction at A?
Engr.Sharif Ullah khan said:
3 years ago
T.l = 10 * 1 = 10t,
Rb * 10 = 10*1(10/2),
Rb * 10 = 50,
Rb = 50/10,
Rb = 5,
Ra + Rb = 10.
Ra = 10-5 = 5.
Rb * 10 = 10*1(10/2),
Rb * 10 = 50,
Rb = 50/10,
Rb = 5,
Ra + Rb = 10.
Ra = 10-5 = 5.
Numan khan said:
5 years ago
The given answer is right.
Let discuss it now,
RA + RB=o
where RA is Reaction at A and RB is Reaction at B.
First, let convert UDL into concentrated load which is equal to 1T x 10.
10 is the span length of the beam.
RA x 0 + 10RB = 10 T,
10RB = 10T,
RB = 10T/10,
RB = 0,
So the reaction at A will be equal to zero.
Let discuss it now,
RA + RB=o
where RA is Reaction at A and RB is Reaction at B.
First, let convert UDL into concentrated load which is equal to 1T x 10.
10 is the span length of the beam.
RA x 0 + 10RB = 10 T,
10RB = 10T,
RB = 10T/10,
RB = 0,
So the reaction at A will be equal to zero.
Anubala said:
5 years ago
Ra+Rb= wl/2 = (1*10)/2= 5T---> (1).
Moment about A,
Rb*10-(1*10*10)/2=0,
Rb=5T.
Substitute in (1)
Ra=0
The answer is correct.
Moment about A,
Rb*10-(1*10*10)/2=0,
Rb=5T.
Substitute in (1)
Ra=0
The answer is correct.
(2)
Rohit Kumar Behera said:
6 years ago
It's 1.875 as calculated above , 3 pt theorom, calculate mb as in str and then proceed for ra
U.S.lakshmi said:
6 years ago
I think to find out the reaction at A is step-1 is to separate the continuous beam as two spans as AB and AC and then find out the reactions ; step-2 is from span AC to find out Rc and from span BC find out the reaction Rb; step-3 is from the equilibrium condition is the summation of all vertical reactions should be zero, from this findout the the Ra. Ans., is 4t.
(1)
ABDUL KALAM V A said:
6 years ago
It is the reaction, not moment.
Total load is 1x10 = 10t. Taking moments about any support and equating the same, we get a reaction at end supports as 2.5t each and middle support as 5t.
Total load is 1x10 = 10t. Taking moments about any support and equating the same, we get a reaction at end supports as 2.5t each and middle support as 5t.
(1)
Bhoomi said:
6 years ago
Rb * 5 = 1 * 10 * 5.
So Rb = 10,
Ra + Rb = 10,
Therefore Ra = zero.
So Rb = 10,
Ra + Rb = 10,
Therefore Ra = zero.
Yahaya taofeek said:
6 years ago
The answer is zero. There are three support Ra, Rb and Rc. If you convert the distributed load to point load, it gives 5T and act directly opposite Rc. For equillibrium of the body Rc has to be 5T.Ra and Rb are both zero
(2)
Vaishali said:
7 years ago
2.5T is correct by equilibrium equation.
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