
Problem 75
A pinjointed truss is shown in
figure 75(a). Determine the vertical displacement of joint E
by using
unit load method.
All the members have crosssectional area of 250 mm^{2} and same modulus
of elasticity 200GPa.
Figure 75(a)
Solution:
According to
unit load method the
deflection of a joint of truss is given by the following formula
δ_{E} = (Σ
NnL)/AE
Here we have to find the member forces
two times. First we will calculate member forces "N" due to the real loading and then
"n" due to unit virtual
load applied at the point of required deflection (in this case joint
E).
Step 1: The member forces
"N" due to
real load are calculated in
Example 31 .So, we take those values. 
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Step 2: Calculation of member
forces "n" due to unit virtual load applied at E as shown in
figure 75(b).
As the unit load is applied at center
of the truss, the support reactions at A and C will be 0.5 kN each.
Considering the equilibrium of joint D,
we get F_{DE} =0 and F_{DC}
=0;
Similarly the equilibrium conditions at
joint F gives;
F_{FE} =0 and F_{FA}
=0.
Consider the equilibrium of joint B
along yaxis we get F_{BE} =0.
Now consider the equilibrium of
joint A.
Σ F_{y} = 0;
F_{AE} sin45 + A_{y}
 F_{AF} = 0
(i)
F_{AE}
sin45 + 0.5  0 =0
Therefore, F_{AE}
= 0.5/sin45 = 0.707 kN
Σ F_{x} = 0;
F_{AE}
cos45 + F_{AB} = 0
(ii)
Therefore, F_{AB} =
F_{AE}
cos45 =0.5 kN
As the loading is symmetric,
the equilibrium of joint C will also yield the forces
similar to the member forces at A
Therefore, F_{CE} =
0.707 kN, F_{CB} = 0.5 kN
All the calculated values are entered in Table 75 

Σ
NnL =129.9 kN^{2} m
External virtual work = Internal
virtual work
1 kN. δ_{E} = (Σ
NnL)/AE
δ_{E}
= 129.9/[(250x10^{6})(200x10^{6})]
vertical deflection
at E =0.0026
m = 2.6 mm (Ans)
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