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Degree of static indeterminacy of a rigid-jointed plane frame having 15 members, 3 reaction components and 14 joints is
Answer & Solution
Correct Answer:
Option
C
To determine the degree of static indeterminacy (Ds) for a rigid-jointed plane frame, use the formula:
Ds = (3m + R) - (3j + A)
where:
m = number of members,
R = number of reaction components,
j = number of joints,
A = number of internal hinges (A = 0 in this case).
Given:
m = 15 (members),
R = 3 (reaction components),
j = 14 (joints),
A = 0 (no internal hinges).
Substitute these values into the formula:
Ds = (3 * 15 + 3) - (3 * 14 + 0)
Ds = (45 + 3) - (42)
Ds = 48 - 42 = 6
Thus, the degree of static indeterminacy is 6.
where:
m = number of members,
R = number of reaction components,
j = number of joints,
A = number of internal hinges (A = 0 in this case).
Given:
m = 15 (members),
R = 3 (reaction components),
j = 14 (joints),
A = 0 (no internal hinges).
Substitute these values into the formula:
Ds = (3 * 15 + 3) - (3 * 14 + 0)
Ds = (45 + 3) - (42)
Ds = 48 - 42 = 6
Thus, the degree of static indeterminacy is 6.
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LoginR=3
J=14
we know that,
Degree of static indeterminacy for plane frames (Ds)=(3m+R)-(3j+A)
but here is no internal hinge so A=0
Ds=(3x15+3)-(3x14)
Ds =6
i.e. Ds=DSi+ DSe
Dsi=Re-3 and Dsi=3m-(3j-3)
use the above prescribed formulae you will get answer as 6
Where, m= no of members
Re= reaction
J= no of joints
M=15
J=14
Ds=(Re+3m)-3j
=(3+3×15)-3×14
=6
=(3*15)+3-(3*14)
=6
=3*15+3-3*14-0=6
No hybrid joints present hence the release is 0
m=member
r= reaction
J=joints
3×15+3 _3×14 = 6