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                                        Deflection of beam by Unit Load method -- solved example  

Problem 7-3

Use unit load method to find the deflection at the center of the beam shown in figure 7-3(a). Take E= 200 GPa and I=400x106 mm4

Figure 7-3(a)

Solution:

In the case of unit load method the deflection at a point of beam is given as

 

To write the equations of bending moment for different parts of the beam we have to first calculate the support reactions by applying the equations of equilibrium

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(i) reactions due to actual loading

Ay = 13.75 kN;     Dy = 31.25 kN

write the equations of bending moment for different part of the beam (origin of axes at A).

Mx = 13.75x ;                              (0 ≤ x ≤ 2)

Mx = 13.75x - 5(x-2) ;                 (2 ≤ x ≤ 4)

Mx = 13.75x - 5(x-2) -10(x-4)(x-4)/2;     (4 ≤ x ≤ 8)

(ii) reactions due to virtual unit load at C

ay = 0.5 kN;   by = 0.5 kN

mx = 0.5x                                 (0 ≤ x ≤ 4)

mx = 0.5x - 1(x-4)                     (4 ≤ x ≤ 8)

Depending upon the validity of different bending moment equations, the deflection as point C can be written as follows;

substituting the appropriate bending moment equations for Mx and mx

 δc = 302.58 kN.m3/EI

substituting the values of E I in the above expression we get;

  δc = (302.58 kN.m3)/{(200*106kN/m2)(400*10-6m4)}

  δc = 0.0038 m = 3.8 mm

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Last updated on Thursday January 31, 2013