30. PROCEDURE
• In structural analysis to analyze the
indeterminate structures
• Compatibility method or Flexibility method
or Force method
30
31. 1. In this method redundant forces are
unknowns
2. Additional equations are obtained by
considering the geometrical condition imposed
on the formations of the structure
3. Flexibility is an amount of displacement
caused due to unit force
31
32. 4. The number of equations in flexibility method
equal to the degree of static indeterminacy.
Since the redundant are support reactions
5. These method is used for analysis of static
indeterminate structure with lesser degree of
static indeterminacy
Ds ˂ Dk
32
33. 6. Various methods grouped under this category
are
a. Consistent deformation method
b. Clayperon’s theorem of three moments
c. Column analogy method
d. Elastic centre method
e. Maxwell – Mohr’s equation
f. Castigliano’s theorem of minimum strain
energy
33
35. 1. In this method displacement of the joints are
taken as unknowns
2. Equilibrium equations are expressed in terms
of moments, rotations to get the actual joint
displacements
3. Stiffness is an amount of force required
caused due to unit displacement
35
36. 4. The product of stiffness and flexibility is unity
5. The number of equations in stiffness method
equal to degree of freedom (Dk) as
displacements are taken as unknowns
6. These method is used to analysis of statically
indeterminate structures if Dk ˂ Ds
36
37. 7. Various methods grouped under this category
are
a. Moment distribution method
b. Slope deflection method
c. Kani’s method
37
38.
39. Ds = External Indeterminacy(E.I.) + Internal Indeterminacy (I.I.)
– Releases or Internal Hinge (R’)
68. To Find:
(i) End Moment (MB)
(ii) To draw SFD and BMD
Solution:
Step 1: Degree of static indeterminacy (n)
n = E.I + I.I - R’
n = 1+0-0
n = 1 E.I. = No. of unknown reaction – No. of equilibrium equation
= 4 – 3
E.I. = 1
Support A = Hinged = ∑H = ∑V = 2
Support B & C = Roller = ∑V = 2
69. Step 2: Coordinates Assigned
Support Moment at ∑MA = ∑MC = 0
Since Moment at ∑MB ≠0
70.
71.
72.
73.
74.
75.
76.
77.
78.
79.
80.
81.
82.
83.
84.
85.
86.
87.
88.
89.
90.
91.
92.
93.
94.
95. To Find Shear Force
Member AB BA BC CB CD DC
Free shear
(1)
WL/2 WL/2 W/2 W/2 Wb/L Wa/L
(20*3)/2 (20*3)/2 40/2 40/2 (40*2)/5 (40*3)/5
Shear
Correction
(2)
(MAB+MBA)/L (MBC+MCB)/L (MCD+MDC)/L
(0+12.15)/3 (-12.15+30.35)/4 (-30.35+0)/5
Final Shear
(1)+(2)
25.95 34.05 15.45 24.55 22.07 17.93
