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## Saturday, January 6, 2018

For the continuous beam loaded as shown above, it is desired to find the bending moments at the critical points using force method (method of consistent deformations) and Clapeyron's theorem (3 Moment Equation). We should however note that both methods are force methods (flexibility method) since we generally solve for unknown forces.

Solution
(1) By force method
Degree of static indeterminacy neglecting horizontal forces and reactions.

D = 2m + r - 2n
D = 2(3) + 4 - 2(4) = 2
Therefore the structure is indeterminate to the 2nd order.

Basic System
A basic system is a system that is statically determinate and stable. This obtained by removing the redundant supports at A and B, and replacing them with unit loads. See the figure below.

Case 1
Bending moment on the basic system due to vertical unit virtual load at support A

Case 2
Bending moment on the basic system due to unit virtual load at support B

Case 3
Bending moment due to externally applied load on the basic system;

Influence Co-efficients

Î´11

Î´11 = [1/3 × 2L × 2L × 2L] + [1/3 × 2L × 2L × L] = 4L3

Î´22

Î´11 = 2[1/3 × L × L × L] = 2L3/3

Î´21 = Î´12

Î´11 = [1/6 × L × L(4L + L)] + [1/3 × L × L × L] = 3L3/2

Î´1P

Î´1P = [1/6 × qL2/16 × (2L + 2L) × L/2]] + [15/12 × qL2/16 × L × L/2] = 13qL4/384

Î´2P

Î´2P = [1/6 × qL2/16 × (L + L) × L/2] + [15/12 × qL2/16 × L/2× L/2] = 13qL4/768

The appropriate cannonical equation is given by;

Î´11X1 +  Î´12X2 + Î´1P = 0
Î´21X1 +  Î´22X2 + Î´2P = 0

Therefore;

(4L3)X1 +  (3L3/2)X2  = -13qL4/384
(3L3/2)X1 +  (2L3/3)X2  = -13qL4/768

On sloving the above equations simultaneously;

X1 = 13qL/1920 KN

X2 = -13qL/320 KN

The final moment values can now be computed;

Mf = M1X1 + M2X2 + MP

MB = (13qL/1920 × L) + 0 + 0 = 13qL2/1920
MC = (13qL/1920 × 2L) - (13qL/230 × L) + 0 = -13qL2/480
MD = (13qL/1920 × L) - (13qL/230 × L/2) + qL2/16  = 47qL2/960

(2) By Clapeyron's Theorem;

First of all, we draw the free bending moment diagram

By Clapeyron's three moment equation (EI = constant, no sinking of support);

MAL1 + 2MB + MCL2 + 6A1X1 + 6A2X= 0

Geometrical Properties of the free moment diagram (centroid)

SPAN A - C
MA = 0
2MB(L + L) + MCL = 0
4MB L + MCL = 0 -------------------- (1)

SPAN B-D
MD = 0
MBL + 2MC(L + L) = [-6 × 7qL3/192 × 13L/128]/L
MB L + 4MCL =  -13qL3/128 -------------------- (2)

Solving (1) and (2) simultaneously;
MB =  13qL2/1920
MC = -13qL2/480

Therefore, the bending moment due to externally applied load is given below;

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