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.
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)
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.
Mohammed yasin said:
8 years ago
Simply moment at pin and roller support is zero, but if you want to how to get its value.
Ray+Rby=1*10,Ray+Rby=10.
Then, take moment at support a,Ma =0,Rby*10-10*5=0, Rby=5
Then, Ray+Rby=10,Ray=5.
Take moment at A,Ma=10*5-Rby * 10,
Ma=10*5-5*10,
Ma=0.
Ray+Rby=1*10,Ray+Rby=10.
Then, take moment at support a,Ma =0,Rby*10-10*5=0, Rby=5
Then, Ray+Rby=10,Ray=5.
Take moment at A,Ma=10*5-Rby * 10,
Ma=10*5-5*10,
Ma=0.
Tayu said:
8 years ago
Assuming Ra, Rb And Rb from right to left respectively,
The beam is hyperstatic and the normal equations of static alone does not suffice, using other methods the reactions will be 3T/16, 5T/16 and 3T/16 respectively, so the right answer is not given.
The beam is hyperstatic and the normal equations of static alone does not suffice, using other methods the reactions will be 3T/16, 5T/16 and 3T/16 respectively, so the right answer is not given.
Sunil said:
10 years ago
Right answer is 2.5 T because in this question A is starting support B is secondary or mid support and is last support. So according to calculation purpose total force are comes on the span AB is 5 and at point a = 2.5 T.
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)
Ankush kohli said:
9 years ago
Suppose two spot is A $ B
Vertical reaction,
A+B = 1 * 10 * (10 ÷ 2)
= 50T -------------> 1
Moment of support A
Rb = 1 * 10 * (10 ÷ 2)
= 50
Rb value put on reaction -1
Ra = 0t. $ Rb = 0t.
Vertical reaction,
A+B = 1 * 10 * (10 ÷ 2)
= 50T -------------> 1
Moment of support A
Rb = 1 * 10 * (10 ÷ 2)
= 50
Rb value put on reaction -1
Ra = 0t. $ Rb = 0t.
DANISH KASSAR said:
9 years ago
Reaction.
Ra + Rb = 1 * 10 = 10..........(1).
Rb * 10 = 10.
Rb = 1.
Put, Rb in above equation then,
Ra = 9.
The moment at A point.
M = 1 * 10 - 1 * 10 = 0.
The moment at A is zero this is answer.
Ra + Rb = 1 * 10 = 10..........(1).
Rb * 10 = 10.
Rb = 1.
Put, Rb in above equation then,
Ra = 9.
The moment at A point.
M = 1 * 10 - 1 * 10 = 0.
The moment at A is zero this is answer.
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)
Siva said:
9 years ago
This problem can be solved by using propped methods.
First find middle support reaction using super imposition principle (compatability eqn) at deflection where the middle support =0.
First find middle support reaction using super imposition principle (compatability eqn) at deflection where the middle support =0.
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