Question 15.6: Determine the slope and deflection at D for the beam and loa...

The slope and deflection at any point of the beam can be obtained  by superposing the slopes and deflections caused respectively by the concentrated load and by the distributed load (Fig. 15.28).
Since the concentrated load in Fig. 15.28b is applied at quarter span, we can use the results obtained for the beam and loading of Example 15.3 and write

(θ_{D})_{P} = – \frac{PL^{2}}{32EI}=  – \frac{(150 × 10^{3}) (8)^{2}}{32(100 × 10^{6})}= -3 × 10^{-3}  rad

(y_{D})_{P} = – \frac{3PL^{3}}{256EI}= – \frac{3(150 × 10^{3}) (8)^{3}}{256(100 × 10^{6})} = -9 × 10^{-3}  m

= -9 mm

On the other hand, recalling the equation of the elastic curve obtained for a uniformly distributed load in Example 15.2, we express the deflection in Fig. 15.28c as

y = \frac{w}{24EI}(-x^{4} + 2L x^{3} – L^{3}x)                          (15.44)

nd, differentiating with respect to x,

θ = \frac{dy}{dx} = \frac{w}{24EI} (-4x^{3} + 6L x^{2} – L^{3})                          (15.45)

Making w = 20 kN/m, x = 2 m, and L = 8 m in Eqs. (15.45) and (15.44), we obtain

(θ_{D})_{w} = \frac{20 × 10^{3}}{24(100 × 10^{6})} (-352) = -2.93 × 10^{-3}  rad

(y_{D})_{w} = \frac{20  × 10^{3}}{24(100 × 10^{6})} (-912) = -7.60 × 10^{-3}  m

= -7.60 mm

Combining the slopes and deflections produced by the concentrated and the distributed loads, we have

θ_{D} = (θ_{D})_{P} + (θ_{D})_{w} = -3 × 10^{-3} – 2.93 × 10^{-3}

= -5.93 × 10^{-3}  rad

y_{D} = (y_{D})_{P} + (y_{D})_{w} = – 9 mm  – 7.60  mm = -16.60  mm