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.

Figure 7-5(a)

Solution:

According to unit load method (also known as virtual work 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). Tensile forces are considered as positive and compressive forces as negative. L is the length of the member, A is area of cross-section of the member and E is modulus of elasticity of the material of the truss member.

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 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 y-axis 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 7-5

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

       Σ NnL =129.9 kN² m

External virtual work = Internal virtual work

1 kN. δ_E = (Σ NnL)/AE

 δ_E = 129.9/[(250x10^-6)(200x10⁶)]

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