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 3 of 7.
Roshan said:
7 years ago
If you consider a, b, c as supports from left to right, a will carry 2.5 t, b will carry 5.0t and c will carry 2.5t. Total load is 10 t.
Sarath said:
8 years ago
Yes, 15/8 is the correct answer.
AJAY said:
7 years ago
The correct answer is 5t, since SSB is symmetrical and total load is 1 * 10 = 10t, Ra=Rb=10/2= 5t.
Shankar said:
7 years ago
The answer is not zero because under loading conditions never support reaction zero. If it's zero structure will be failure.
Harsha said:
7 years ago
It is 3/16 wl only.
Vaishali said:
7 years ago
2.5T is correct by equilibrium equation.
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.
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
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.
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.
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