5.Truss Deflection
6. Indeterminate Truss Problems – Method of Consistent Deformations
The truss shown below has 15 members (M=15) and four reaction forces(R=4). The total number of force unknowns is 19. There are nine nodes (N=9). Thus, M+R-2N=1. The problem is statically indeterminate to the first degree. In addition to the 18 equlibrium equations we can establish from the nine nodes, we need to find one more equation in order to solve for the 19 unknowns. This additional equation can be established by considering the consistency of deformations (deflections) in relation to geometrical constraints.
Statically indeterminate truss with one degree of redundancy.
We notice that if the vertical reaction at the central support is known, then the number of force unknowns becomes 18 and the problem can be solved by the 18 equalibrium equation from the nine nodes. The key to solution is then to find the central support reaction, which is called the redundant force. Denoting the vertical reaction of the central support by Rc, the original problem is equivalent to the problem shown below as far as force equlibrium is concerned.
Statically equilvalent problem with the redundant force Rc as unknown.
The truss shown above, with the central support removed, is called the “primary
structure.” Note that the primary structure is statically determinate. The magnitude of Rc
is determined by the condition that the vertical displacement of node c of the primary structure due to (1) the applied load P and (2) the redundan force Rc is zero. This condition is consistent with the geometric constraint imposed by the central support on the original structure. The vertical displacement at node c due to the applied load P can be determined by solving the problem associated with the primary structure as shown below.
P
P Rc
c
Displacement of node c of the primary structure due to the applied load.
The displacement of node c due to the redundant force Rc cannot be computed directly because Rc itself is unknown. We can compute, however, the displacement of node c of the primary structure due to a unit load in the direction of Rc as shown. This
displacement is denoted by δcc , the double subscript ‘cc’ signified displacent at ‘c’ (first subscript) due to a unit load at ‘c’ (second subscript).
Displacement at c due to a unit load at c.
The vertical dispalcement at c due to the redundant force Rc is then Rcδcc , as shown in the figure below.
Displacement at c due to the redundant force Rc.
The condition that the total vertical displacement at node c, ∆c, be zero is expressed as
∆c = ∆’c + Rcδcc = 0 (18)
This is the additional equation needed to solve for the redundant force Rc. Once Rc is obtained, the rest of the force unknowns can be computed from the regular joint equilibrium equations. Eq. 18 is called the condition of compatibility.
c
c
Rc Rcδcc
P ∆’c
c 1 kN δcc
We may summarize the concept behind the above procedures by pointing out that the original problem is solved by replacing the indeterminate truss with a determinate primary structure and superposing the solutions of two problems, each determinate, as shown below.
The superposition of two solutions.
And, the key equation is the condition that the total vertical displacement at node c must be zero, consisdent with the support condition at node c in the original problem. This method of analysis for statically indeterminate structures is called the method of consistent deformations.
Example 17. Find the force in bar 6 of the truss shown, given E=10 GPa, A=100 cm2 for all bars.
Example of an indeterminate truss with one redundant.
1 kN 0.5 kN
1 4
2 3
4 m
3 m 1
2
3
4
5 6
c
P ∆’c
c
Rc Rcδcc
+
Solution. The primary structure is obtained by introducing a cut at bar 6 as shown in the left figure below. The original problem is replaced by that of the left figure and that of the middle figure. The condition of compatibility in this case requires that the total relative displacement across the cut obtained from the superpostion of the two solutions be zero.
∆ = ∆’ + F6δ = 0
Superposition of two solutions.
where ∆’ is the overlap length (opposite of a gap) at the cut due to the applied load and δ is, as defined in the figure below, the overlap length across the cut due to a pair of unit loads applied at the cut.
Overlap displacement at the cut due to the unit-force pair.
The computation needed to find ∆’ and δ is tabulated below.
1 kN
1 4
2 3
1
2
3
4 5
1 4
2 3
1
2
3
4 5
1
1 4
2 3
1
2
3
4 5 F6
δ F6δ
∆
’
1 kN
0.5 kN
1 4
2 3
1
2
3
4
5 6
0.5 kN
+ =
1
Computing for ∆’ and δ
Real Load For ∆’ For δ
Fi EA /L Vι fi fi Vι vi fi vi
Member
(kN) (kN/m) (mm) (kN/kN) (mm) (mm/kN) (mm/kN)
1 -0.33 25,000 -0.013 -0.8 0.010 -0.032 0.026
2 0 33,333 0 -0.6 0 -0.018 0.011
3 0 25,000 0 -0.8 0 -0.032 0.026
4 0.50 33,333 0.015 -0.6 -0.009 -0.018 0.011
5 -0.83 20,000 -0.042 1.0 -0.042 0.050 0.050
6 0 20,000 0 1.0 0 0.050 0.050
Σ -0.040 0.174
Note: Fi = ith member force due to the real applied load.
Vι= Fi / (EA/L)i =ith member elongation due to the real applied load.
fi = ith member force due to the virtual unit load pair at the cut.
vi = fi / (EA/L)i is the ith member elongation due to the virtual unit load pair at the cut.
∆’= -0.040 mm, δ = 0.174 mm/kN From the condition of compatibility:
∆’ + F6δ = 0 F6 = −
174 . 0
040 .
−
0 = 0.23 kN.Example 18. Formulate the conditions of compatibility for the truss problem shown.
Statically indeterminate truss with two degrees of redundancy.
Solution. The primary structure can be obtained by removing the supports at node c and node d. Denoting the reaction at node c and node d as Rc and Rd, respectively, the original
P
c d
problem is equivalent to the superpostion of the three problems as shown in the figure below.
Superposition of three determinate problems.
In the above figure.
∆’c : vertical displacement at node c due to the real applied load,
∆’d : vertical displacement at node d due to the real applied load, δcc : vertical displacement at node c due to a unit load at c, δcd : vertical displacement at node c due to a unit load at d, δdc : vertical displacement at node d due to a unit load at c, and δdd : vertical displacement at node d due to a unit load at d.
The conditions of compatibility are the vertical displacements at nodes c and d be zero:
∆c = ∆’c + Rcδcc + Rdδcd = 0
(19)
∆d = ∆’d + Rc δdc + Rdδdd = 0
c
P ∆’c
c
Rc Rcδcc
c
Rd Rdδcd
d
∆’d
Rcδdc
d
d
Rdδdd
+
+
The above equations can be solved for the two redundant forces Rc and Rd.
Denote
Vi: the ith member elongation due to the real applied load, fic: the ith member force due to the unit load at c.
vic: the ith member elongation due to the unit load at c.
fid: the ith member force due to the unit load at d, and vid: the ith member elongation due to the unit load at d,
we can express the displacements according to the unit load method as
∆’c = Σ fic (Vι)
∆’d = Σ fid (Vι) δcc = Σ fic (vιc) δdc = Σ fic (vιd) δcd = Σ fid (vιc) δdd = Σ fid (vιd)
The member elongation quantities in the above equations are related to the member forces through
Vι =
i i
i i
A E
L F
vιc =
i i
i ic
A E
L f
vιd =
i i
i id
A E
L f
Thus, we need to find only member forces Fi, fic, and fid, corresponding to the real load, a unit load at node c and a unit load at node d , respectively, from the primary structure.