Deflection of Truss by unit load method

Problem 7-5

A pin-jointed truss is shown in figure 7-5(a).   Determine the vertical displacement of joint E by using unit load method.  All the members have cross-sectional area of 250 mm2 and same modulus of elasticity 200GPa.

pin-jointed plane truss

Figure 7-5(a)

Solution:

According to unit load method the deflection of a joint of truss is given by the following formula

     δE = (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 3-1 .So, we take those values.

Figure 7-5(b)

Step 2: Calculation of member forces "n" due to unit virtual load applied at E  as shown in figure 7-5(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 FDE =0 and FDC =0;

Similarly the equilibrium conditions at joint F gives; 

   FFE =0   and   FFA =0.

Consider the equilibrium of joint B along y-axis we get FBE =0.

Now consider the equilibrium of joint A.

Σ Fy = 0;

FAE sin45 + Ay - FAF = 0              (i)

FAE sin45 + 0.5 - 0 =0

Therefore, FAE = -0.5/sin45 = -0.707 kN

Σ Fx = 0;  

FAE cos45 + FAB = 0                  (ii)

Therefore, FAB = -FAE 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, FCE = -0.707 kN, FCB = 0.5 kN

All the calculated values are entered in Table 7-5

(In case of mobile device please scroll horizontally to view the full width of table)

Table 7-5 Result of Member Forces Calculations
Member N (kN) n (kN) L (m) NnL (kN2 m)
AB 22.5 0.5 2 22.5
AF -25 0 2 0
AE -31.82 -0.707 2.83 63.67
BC 22.5 0.5 2 22.5
BE 20 0 2 0
CD 0 0 2 0
CE -10.61 -0.707 2.83 21.23
DE -15 0 2 0
EF 0 0 2 0

       Σ NnL =129.9 kN2 m

External virtual work = Internal virtual work

1 kN. δE = (Σ NnL)/AE

 δE = 129.9/[(250x10-6)(200x106)]

  Vertical displacement of Joint E =0.0026 m = 2.6 mm      (Ans)

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