Question 7.16: Calculate the slope of elastic line at B and the deflection ...

Let us draw the free-body diagram of the beam as shown in Figure 7.30 to determine the support reactions.

We note from free-body diagram of the beam shown in Figure 7.30(a) that for equilibrium:

\sum M_{ A }=0 \Rightarrow 20 R_{ B }=(5)(200)+15(120) \Rightarrow R_{ B }=140  kN (\uparrow)

and        \sum F_y=0 \Rightarrow R_{ A }=180  kN (\uparrow)

In Figure 7.30(b), we show the equivalent loading diagram, which gives

M_x=(180) x-10 x^2+10\langle x-10\rangle^2-120\langle x-15\rangle  kNm

From flexure equation, we get

(E I) \frac{ d ^2 y}{ d x^2}=-M_x=-180 x+10 x^2-10\langle x-10\rangle^2+120\langle x-15\rangle

Now integrating successively, we obtain:

(E I)\left\lgroup\frac{ d y}{ d x} \right\rgroup =-90 x^2+\frac{10}{3} x^3-\frac{10}{3}\langle x-10\rangle^3+60\langle x-15\rangle^2+C_1           (1)

and        (E I) y=-30 x^3+\frac{10}{12} x^4-\frac{10}{12}\langle x-10\rangle^4+20\langle x-15\rangle^3+C_1 x+C_2             (2)

Noting the boundary conditions of y = 0 at x = 0 m and at x = 20 m, we get from Eq. (2) that:

C_2=0 \quad \text { and } \quad C_1=5625  kN m ^2

Therefore Eqs. (1) and (2) become

(E I) \frac{ d y}{ d x}=-90 x^2+\frac{10}{3} x^3-\frac{10}{3}\langle x-10\rangle^3+60\langle x-15\rangle^2+5625              (3)

and    (E I) y=-30 x^3+\frac{10}{12} x^4-\frac{10}{12}\langle x-10\rangle^4+20\langle x-15\rangle^3+5625 x               (4)

Putting x = 20 m in Eq. (3), we get

\left.\frac{ d y}{ d x}\right|_{x=20 m }=\left.\frac{ d y}{ d x}\right|_{ B }=\tan \theta_{ B } \approx \theta_{ B }=-\frac{5541.67}{E I}=-\frac{5541.67}{(200)\left(10^6\right)(2.5)\left(10^{-12}\right)\left(10^9\right)}

Therefore,

\theta_{ B }=-0.0111  rad ⦨

Again, putting x = 15 m in Eq. (4), we obtain

\text { (EI) }\left.y\right|_{x=15 m }=\left.24791.67 \Rightarrow y\right|_{x=15 m }=\delta_{ D }=\frac{24791.67}{(200)\left(10^6\right)(2.5)\left(10^9\right)\left(10^{-12}\right)}  m

Therefore,