Question 10.32: A horizontal bracket ABCD is loaded at its free end as shown...
A horizontal bracket ABCD is loaded at its free end as shown in Figure 10.66. Calculate the vertical deflection and angle of twist of end D.
We apply a dummy torque T_{ o } at end D and draw the free-body diagram of the assembly as shown in Figure 10.67.
The equilibrium equations are written in Figure 10.68 itself to find the support reactions’ forces and moments at A. We now draw the following free-body diagrams of arbitrary sections considered on members AB, BC and CD of the bracket:
From Figure 10.68, we conclude the following expressions of bending moment and torsional moment at members AB, BC and CD.
For leg AB:
\left.\begin{array}{rl} M_x & =P x-2 P L_1 \\ T_x & =T_{ o }+P L_2 \end{array}\right\}_{ AB } ; \quad 0 \leq x \leq L_1
For leg BC:
\left.\begin{array}{rl} M_x & =P x-\left(T_{ o }+P L_2\right) \\ T_x & =-P L_1+2 P L_1=P L_1 \end{array}\right\}_{ BC } \quad ; \quad 0 \leq x \leq L_2
and for leg CD:
\left.\begin{array}{rl} M_x & =-P x \\ T_x & =T_{ o } \end{array}\right\}_{ CD } \quad ; \quad 0 \leq x \leq L_1
Therefore, angle of twist of end D is
\begin{aligned} \phi_{ D } & =\left.\frac{\partial U}{\partial T_{ o }}\right|_{T_{ o }=0}=\left[\left\lgroup \frac{\partial U}{\partial T_{ o }} \right\rgroup_{ AB }+\left\lgroup \frac{\partial U}{\partial T_{ o }} \right\rgroup_{ BC }+\left\lgroup \frac{\partial U}{\partial T_{ o }} \right\rgroup_{ CD }\right]_{T_{ o }=0} \\ & =\left[\left\{\left\lgroup \frac{1}{E I} \right\rgroup \int_0^{L_1} M_x \frac{\partial M_x}{\partial T_0} d x+\left\lgroup \frac{1}{G J} \right\rgroup \int_0^{L_1} T_x \frac{\partial T_x}{\partial T_0} d x\right\}_{ AB }\right. \end{aligned}
\begin{aligned} & +\left\{\left\lgroup \frac{1}{E I} \right\rgroup \int_0^{L_1} M_x \frac{\partial M_x}{\partial T_0} d x+\left\lgroup \frac{1}{G J} \right\rgroup \int_0^{L_1} T_x \frac{\partial T_x}{\partial T_0} d x\right\}_{ BC } \\ & \left.+\left\{\left\lgroup \frac{1}{E I} \right\rgroup \int_0^{L_1} M_x\left\lgroup \frac{\partial M_x}{\partial T_0} \right\rgroup d x+\left\lgroup \frac{1}{G J} \right\rgroup \int_0^{L_1} T_x \frac{\partial T_x}{\partial T_0} d x\right\}_{ CD }\right]_{T_0=0} \end{aligned}
or
\begin{aligned} \phi_{ D }= & {\left[\left\lgroup\frac{1}{G J}\right\rgroup \int_0^{L_1}\left(P L_2\right) d x+\left\lgroup \frac{1}{E I}\right\rgroup \int_0^{L_2} P\left(x-L_2\right)(-1) d x\right.} \\ & +\left\lgroup\frac{1}{G J}\right\rgroup \int_0^{L_2}\left(P L_1\right)(0) d x+\left\lgroup\frac{1}{E I}\right\rgroup \int_0^{L_1}(-P x)(0) d x \\ & \left.+\left\lgroup\frac{1}{G J}\right\rgroup \int_0^{L_0}(0)(1) d x\right] \end{aligned}
Thus,
\phi_{ D }=\left\lgroup \frac{1}{G J}\right\rgroup P L_2 L_1+\left\lgroup \frac{1}{E I}\right\rgroup P\left[L_2 x-\frac{x^2}{2}\right]_0^{L_2}
=\frac{P L_1 L_2}{G J}+\frac{P L_2^2}{2 E I}=P L_2\left\lgroup\frac{L_1}{G J}+\frac{L_2}{2 E I} \right\rgroup
or \phi_{ D }=P L_2\left\lgroup\frac{L_2}{2 E I}+\frac{L_1}{G J} \right\rgroup ⦨
Now, for determining \left(\delta_v\right)_{ D } , we do not require any T_0 . Therefore,
\left(\delta_{ v }\right)_{ D }=\frac{\partial U}{\partial P}=\left[\left\lgroup \frac{\partial U}{\partial P} \right\rgroup_{ AB }+\left\lgroup \frac{\partial U}{\partial P} \right\rgroup_{ BC }+\left\lgroup \frac{\partial U}{\partial P} \right\rgroup_{ CD }\right]
or
\begin{aligned} \left(\delta_{ v }\right)_{ D }= & {\left[\left\{\left\lgroup \frac{1}{E I} \right\rgroup \int_0^{L_1} M_x \frac{\partial M_x}{\partial P} d x+\left\lgroup \frac{1}{G J} \right\rgroup \int_0^{L_1} T_x \frac{\partial T_x}{\partial P} d x\right\}_{ AB }\right.} \\ & +\left\{\left\lgroup \frac{1}{E I} \right\rgroup \int_0^{L_1} M_x \frac{\partial M_x}{\partial P} d x+\left\lgroup \frac{1}{G J} \right\rgroup \int_0^{L_2} T_x\left\lgroup \frac{\partial T_x}{\partial P} \right\rgroup d x\right\}_{ BC } \\ & \left.+\left\{\left\lgroup \frac{1}{E I} \right\rgroup \int_0^{L_1} M_x \frac{\partial M_x}{\partial P} d x+\left\lgroup \frac{1}{G J} \right\rgroup \int_0^{L_1} T_x\left\lgroup \frac{\partial T_x}{\partial P} \right\rgroup d x\right\}_{ CD }\right] \end{aligned}
or
\begin{aligned} \left(\delta_{ v }\right)_{ D }= & {\left[\left\{\left\lgroup \frac{1}{E I} \right\rgroup \int_0^{L_1} P\left(x-2 L_1\right)^2 d x+\left\lgroup \frac{1}{G J} \right\rgroup \int_0^{L_1}\left(P L_2^2\right) d x\right\}\right.} \\ & +\left\{\left\lgroup \frac{1}{E I} \right\rgroup \int_0^{L_2} P\left(x-L_2\right)^2 d x+\left\lgroup \frac{1}{G J} \right\rgroup \int_0^{L_2}\left(P L_1^2\right) d x\right\} \\ & \left.+\left\{\left\lgroup \frac{1}{E I} \right\rgroup \int_0^{L_1}(-P x)(-x) d x\right\}\right] \end{aligned}
So,
\begin{aligned} \left(\delta_{ v }\right)_{ D } & =\left\lgroup \frac{P}{3 E I} \right\rgroup\left[\left(x-2 L_1\right)^3\right]_0^{L_1}+\frac{P L_2^2 L_1}{G J}+\left\lgroup \frac{P}{3 E I} \right\rgroup\left[\left(x-L_2\right)^3\right]_0^{L_2}+\left\lgroup \frac{P L_1^2 L_2}{G J} \right\rgroup +\frac{P L_1^3}{3 E I} \\ & =\left\lgroup \frac{P}{3 E I} \right\rgroup\left(-L_1^3+8 L_1^3\right)+\frac{P L_2^2 L_1}{G J}+\left\lgroup \frac{P L_2^3}{3 E I} \right\rgroup+\frac{P L_1^2 L_2}{G J}+\frac{P L_1^3}{3 E I} \\ & =\left\lgroup \frac{8 P L_1^3}{3 E I}+\frac{P L_2^3}{3 E I} \right\rgroup +\frac{P L_1 L_2}{G J}\left(L_1+L_2\right) \end{aligned}
Therefore,
\left(\delta_{ v }\right)_{ D }=\left\lgroup \frac{P}{3 E I} \right\rgroup\left(8 L_1^3+L_2^3\right)+\frac{P L_1 L_2}{G J}\left(L_1+L_2\right)
