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Chapter 10

Q. 10.33

The structure shown consists of two steel columns AB and CD and a beam BC under uniformly distributed load intensity w_o (Figure 10.69). Find the horizontal component of the pin reaction at D. Assume modulus of elasticity = E.

10.69

Step-by-Step

Verified Solution

We identify the given problem to be statically indeterminate one. Let us proceed by drawing the free-body diagram of the frame as shown in Figure 10.70.

In Figure 10.70, preliminary equations of statical equilibrium are written. Let us now draw the bending moment diagrams of the members AB, BC and CD as shown in Figure 10.71:

From Figure 10.71, we get

For member AB [refer Figure 10.71(a)]:

M_x=-x R_{ A _x} ;  0     \leq x \leq h

For member BC [refer Figure 10.71(b)]:

M_x=\left\lgroup \frac{w_{ o } L}{2} \right\rgroup x-\frac{w_{ o } x^2}{2}-h R_{ A _x} ; \quad 0 \leq x \leq L

For member CD [refer Figure 10.71(c)]:

M_x=-x R_{ D _x} ; \quad 0 \leq x \leq h

The bending moment equations are rewritten as follows (using equations of equilibrium as shown in Figure 10.70).

M_x=x R_{ D _x} ; \quad 0 \leq x \leq h \text { for } AB

=\left\lgroup \frac{w_{ o } L}{2} \right\rgroup x-\frac{w_{ o } x^2}{2}+h R_{ D _x} ; \quad 0 \leq x \leq L \text { for } BC

=-x  R_{ D _x} ;  \quad  0 \leq x \leq h  \text {  for  }  CD

Now,              \left(\delta_x\right)_{ D }=\frac{\partial U}{\partial R_{D_x}}

Therefore,

\left(\delta_x\right)_{ D }=\left\lgroup \frac{\partial U}{\partial R_{ D _x}} \right\rgroup_{ AB }+\left\lgroup \frac{\partial U}{\partial R_{ D _x}} \right\rgroup_{ BC }+\left\lgroup \frac{\partial U}{\partial R_{ D _x}} \right\rgroup_{ CD }

or

\left(\delta_x\right)_{ D }=\left[\left.\left\lgroup \frac{1}{E I_1} \right\rgroup \int_0^h M_x\left\lgroup \frac{\partial M_x}{\partial R_{ D _x}} \right\rgroup\right|_{ AB }+\left.\left\lgroup \frac{1}{E I_2} \right\rgroup \int_0^L M_x\left\lgroup \frac{\partial M_x}{\partial R_{ D _x}} \right\rgroup d x\right|_{ BC }+\left.\left\lgroup \frac{1}{E I_1} \right\rgroup \int_0^h M_x\left\lgroup \frac{\partial M_x}{\partial R_{ D _x}} \right\rgroup d x\right|_{ CD }\right]

=\left\lgroup \frac{1}{E I_1} \right\rgroup \int_0^h R_{ D _x} x^2 d x+\left\lgroup \frac{1}{E I_2} \right\rgroup \left\{\int_0^L\left[\left\lgroup \frac{w_{ o } h L}{2} \right\rgroup x-\left\lgroup \frac{w_{ o } h}{2} \right\rgroup x^2+h^2 R_{ D _x}\right] d x\right\} +\left\lgroup \frac{1}{E I_1} \right\rgroup \int_0^h R_{ D _x} x^2 d x

=\left\lgroup \frac{2}{E I_1} \right\rgroup R_{ D _x} \frac{h^3}{3}+\left\lgroup \frac{1}{E I_2} \right\rgroup \left[\frac{w_{ o } h L^3}{4}-\frac{w_{ o } h L^3}{6}+h^2 L R_{ D _x}\right]

But as point D is horizontally constrained evidently, \left(\delta_x\right)_{ D }=0 . Therefore,

\left(R_{ D _x}\right)\left[\frac{2 h^3}{3 E I_1}+\frac{h^2 L}{E I_2}\right]+\frac{w_{ o } h L^3}{12 E I_2}=0

or              R_{ D _x}=-\frac{\frac{w_{ o } L^3}{12 E I_2}}{\left\lgroup \frac{2 h^2}{3 E I_1}+\frac{L h}{E I_2} \right\rgroup }

=-\frac{\frac{w_{ o } L^3}{12 I_2}}{\left\lgroup \frac{2}{3} \frac{h^2}{I_1}+\frac{h}{I_2} \right\rgroup }=-\frac{w_{ o } L^3 I_1}{8 h^2 I_2+12 h I_1}

Therefore, horizontal reaction at pin D is

\frac{w_0 L^3 I_1}{8 h^2 I_2+12 h I_1}(\leftarrow)

10.70
10.71