The rods AD and CE shown in Fig. 4–17a each have a diameter of 10 mm. The second area moment of beam ABC is I = 62.5\left(10^3\right) \ mm^4 . The modulus of elasticity of the material used for the rods and beam is E = 200 GPa .
The threads at the ends of the rods are single-threaded with a pitch of 1.5 mm. The nuts are first snugly fit with bar ABC horizontal. Next the nut at A is tightened one full turn. Determine the resulting tension in each rod and the deflections of points A and C.
There is a lot going on in this problem; a rod shortens, the rods stretch in tension, and the beam bends. Let’s try the procedure!
1-The free-body diagram of the beam is shown in Fig. 4–17b. Summing forces, and moments about B, gives
F_B-F_A-F_C=0 (1)
4F_A-3F_C=0 (2)
2-Using singularity functions, we find the moment equation for the beam is
M=-F_Ax+F_B\left\langle x-0.2\right\rangle ^1
where x is in meters. Integration yields
EI\frac{dy}{dx} =-\frac{F_A}{2}x^2+\frac{F_B}{2}\left\langle x-0.2\right\rangle ^2 +C_1
EIy =-\frac{F_A}{6}x^3+\frac{F_B}{6}\left\langle x-0.2\right\rangle ^3 +C_1x+C_2x (3)
The term EI = 200 \left(10^9 \right) 62.5 \left(10^{−9}\right) = 1.25 \left(10^4 \right) N·m^2 .
3 -The upward deflection of point A is
\left({Fl}/{AE}\right) _{AD}-Np , where the first term is the elastic stretch of AD, N is the number of turns of the nut, and p is the pitch of the thread. Thus, the deflection of A in meters is
y_A = \frac{F_A\left (0.6\right) } {\frac{\pi } {4}\left(0.010\right)^2 \left(200\right) \left(10^9\right) } – \left(1\right) \left(0.0015\right) \\= 3.8197 \left(10^{-8}\right) F_A-1.5 \left(10^{-3}\right) (4)
The upward deflection of point C is \left({Fl}/{AE}\right) _{CE} , or
y_C=\frac{F_C\left(0.8\right) }{\frac{\pi }{4}\left(0.010\right)^2\left(200\right)\left(10^9\right) }=5.093 \left(10^{-8}\right) F_C (5)
Equations (4) and (5) will now serve as the boundary conditions for Eq. (3). At
x = 0 \ , \ y = y_A. Substituting Eq. (4) into (3) with x = 0 \ and \ EI = 1.25 \left(10^4\right) , noting that the singularity function is zero for x = 0 , gives
-4.7746\left(10^{-4}\right) F_A+C_2=-18.75 (6)
At x = 0.2 \ m, \ y = 0, and Eq. (3) yields
-1.3333\left(10^{-3}\right)F_A+0.2C_1+C_2=0 (7)
At x = 0.35 \ m, \ y = y_C. Substituting Eq. (5) into (3) with x = 0.35 \ m \ and \ EI =1.25\left(10^4\right) gives
−7.1458 \left(10^{−3}\right) F_A + 5.625 \left(10^{−4}\right) F_B − 6.3662 \left(10^{−4}\right) F_C + 0.35 C_1 + C2 = 0 (8)
Equations (1), (2), (6), (7), and (8) are five equations in F_A\ , \ F_B , \ F_C, \ C_1, \ and \ C_2 . Written in matrix form, they are
\left [ \begin{matrix} -1 & 1 & -1 & 0 & 0 \\ 4 & 0 & -3 & 0 & 0 \\ -4.7746\left(10^{-4}\right) & 0 & 0 & 0 & 1 \\ -1.3333\left(10^{-3}\right) & 0 & 0 & 0.2 & 1 \\ -7.1458\left(10^{-3}\right) & 5.625\left(10^{-4}\right) & -6.3662\left(10^{-4}\right) & 0.35 & 1 \end{matrix} \right ] \left [ \begin{matrix} F_A \\ F_B \\ F_C \\ C_1 \\ C_2 \end{matrix} \right ] =\left [ \begin{matrix} 0 \\ 0 \\ -18.75 \\ 0 \\ 0 \end{matrix} \right ]
Solving these equations yields
F_A=2988 \ N F_B=6971 \ N F_C=3983 \ N
C_1=06.54 \ N.m^2 C_2=-17.324 \ N.m^3
Equation (3) can be reduced to
y=-\left(39.84 x^3-92.95\left\langle x-0.2\right\rangle^3-8.523x+1.386 \right) \left(10^{-3}\right)at x=0 \ , \ y=y_A=-1.386\left(10^{-3}\right) \ m =-1.386 \ mm
at x=0.35 \ m, \ y=y_C=-\left[39.84 \left(0.35\right)^3 – 92.95 \left(0.35-0.2\right)^3-8.523\left(0.35\right)+1.386\right]\left(10^{-3}\right) =0.203\left(10^{-3}\right) \ m=0.203 \ mm
