Degree of static indeterminacy of a rigid-jointed plane frame having 15 members, 3 reaction components and 14 joints is
A. 2
B. 3
C. 6
D. 8
Answer: Option C
Solution(By Examveda Team)
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.
This Question Belongs to Civil Engineering >> Structural Analysis
M=15
R=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
static indeterminacy =static internal indeterminacy+ static external indeterminacy
i.e. Ds=DSi+ DSe
Dsi=Re-3 and Dsi=3m-(3j-3)
use the above prescribed formulae you will get answer as 6
3m+r-3j u will get
3(15)+3-3(14)=6
For a rigid jointed 2D plane, the formula is 3m+re-3j
Where, m= no of members
Re= reaction
J= no of joints
Re=3
M=15
J=14
Ds=(Re+3m)-3j
=(3+3×15)-3×14
=6
=3m+r-3j
=(3*15)+3-(3*14)
=6
Degree of static indeterminacy of rigid joint plane frame having 20 members, 7 reaction components and 15 joint is
3m+r-3j-no. of releases[due to inernat pin joint]
3m+r-3j-ec(ec=no.of internal hybrid joint or internal hinge)
3m+r-3j-(no of releases)
=3*15+3-3*14-0=6
No hybrid joints present hence the release is 0
3m+r-3j=15*3+3-3*14=48- 42=6
Ds=3m+r-3j {3×15+3-3×14}=6
m=member
r= reaction
J=joints
3 b +r _3j
3×15+3 _3×14 = 6
Any one could explain me