SOM-2 UNIT-1 CHAPTER-1
ILLUSTRATIVE PROBLEMS
Using Castigliano's Theorem :
1. The cantilever beam CD supports a uniformly distributed Load w. and a concentrated load P as shown in figure below. Suppose
L = 3m; w = 6KN/m ; P = 6KN and E. I = 5 MN m2 determine the deflection at D
The deflection 'Y0 ‘at the point D Where load ‘P' is applied is obtained from the relation
Since P is acting vertical and directed downward d ; represents a vertical deflection and is positions downward.
The bending moment M at a distance x from D
And its derivative with respect to ‘P' is
Substituting for M and ¶ M/ ¶ P into equation (1)
2.
Areas
a1 = 500 mm2
a2 = 1000 mm2
For the truss as shown in the figure above, Determine the vertical deflection at the joint C.
Solution:
Since no vertical load is applied at Joint C. we may introduce dummy load Q. as shown below
Using castigliano's theorem and denoting by the force Fi in a given member i caused by the combined loading of P and Q. we have
Free body diagram : The free body diagram is as shown below
Force in Members:
Considering in sequence, the equilibrium of joints E, C, B and D, we may determine the force in each member caused by load Q.
Joint E: FCE = FDE = 0
Joint C: FAC = 0; FCD = -Q
Joint B: FAB = 0; FBD = -3/4Q
The total force in each member under the combined action of Q and P is
Member
|
Fi
|
¶ Fi / ¶ Q
|
Li ,m
|
Ai ,m2
|  |
|---|---|---|---|---|---|
AB
AC
AD
BD
CD
CE
DE
|
0
+15P/8
+5P/4+5O/4
-21P/8-3Q/4
-Q
15P/8
-17P/8
|
0
0
5/4
-3/4
-1
0
0
|
0.8
0.6
1.0
0.6
0.8
1.5
1.7
|
5000x10-6
5000x10-6
5000x10-6
1000x10-6
1000x10-6
500x10-6
1000x10-6
|
0
0
3125P+3125Q
1181P+338Q
+800Q
0
0
|
P = 60 KN
Sub-(2) in (1)
Deflection of C.
Since the load Q is not the part of loading therefore putting Q = 0
3. For the beam and loading shown, determine the deflection at point D. Take E = 200Gpa, I = 28.9x106 mm4
Solution:
Castigliano's Theorem :
Since the given loading does not include a vertical load at point D, we introduce the dummy load Q as shown below. Using Castigliano's Theorem and noting that E.I is constant, we write.
The integration is performed seperatly for portion AD and DB
Reactions
Using F.B.D of the entire beam
Portion AD of Beam :
From Using the F.B.D.we find
Portion DB of Beam :
From Using the F.B.D shown below we find the bending moment at a distance V from end B is
Deflection at point D:
Recalling eq (1) . (2) and (3) we have
4. For the uniform loaded beam with following supports. Determine the reactions at the supports
Solution:
