Question 7.8: Determine the deflection at point D of the beam shown in Fig...

Determine the deflection at point D of the beam shown in Fig. 7.13(a) by the virtual work method.

7.13a
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The real and virtual systems are shown in Fig. 7.13(b) and (c), respectively. It can be seen from Fig. 7.13(a) that the flexural rigidity EI of the beam changes abruptly at points B and D. Also, Fig. 7.13(b) and (c) indicates that the real and virtual loadings are discontinuous at points C and D, respectively. Consequently, the variation of the quantity (M_vM/EI) will be discontinuous at points B, C, and D. Thus, the beam must be divided into four segments, AB, BC, CD, and DE; in each segment the quantity (M_vM/EI) will be continuous and, therefore, can be integrated.

The x coordinates selected for determining the bending moment equations are shown in Fig. 7.13(b) and (c). Note that in any particular segment of the beam, the same x coordinate must be used to write both equations—that is, the equation for the real bending moment (M) and the equation for the virtual bending moment (M_v). The equations for M and M_v for the four segments of the beam, determined by using the method of sections, are tabulated in Table 7.6. The deflection at D can now be computed by applying the virtual work expression given by Eq. (7.30).

1(Δ) = \int_{0}^{L}{\frac{M_vM}{EI} dx}                 (7.30)

1(Δ_D) = \int_{0}^{L}{\frac{M_vM}{EI} dx}

 

1(Δ_D) = \frac{1}{EI}[\int_{0}^{3}{(\frac{x}{4})(75x) dx} + \frac{1}{2}\int_{3}^{6}{(\frac{x}{4})(75x) dx}

 

+\frac{1}{2}\int_{6}^{9}{(\frac{x}{4})(-75x + 900) dx} + \int_{0}^{3}{(\frac{3}{4}x)(75x) dx}]

 

(1 kN)Δ_D = \frac{2,193.75 kN^2-m^3}{EI}

Therefore,

Δ_D = \frac{2,193.75 kN-m^3}{EI} = \frac{2,193.75}{200(300)} = 0.0366 m = 36.6 mm

 

Δ_D = 36.6 mm ↓
TABLE 7.6
Segment x Coordinate EI
(I = 300 × 10^6 mm^4)
M
(kN-m)
M_v
(kN-m)
Origin Limits(m)
AB A 0-3 EI 75x \frac{x}{4}
BC A 3-6 2EI 75x \frac{x}{4}
CD A 6-9 2EI 75x – 150(x – 6) \frac{x}{4}
ED E 0-3 EI 75x \frac{3}{4}x
7.13b
7.13c

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