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(103) mm4. 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.

Question

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[latex] I = 62.5\left(10^3\right) \ mm^4[/latex] . The modulus of elasticity of the material used for the rods and beam is [latex]E = 200 GPa[/latex] .

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.

Accepted Answer

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

[latex]F_B-F_A-F_C=0[/latex] (1)

[latex]4F_A-3F_C=0[/latex] (2)

2-Using singularity functions, we find the moment equation for the beam is

[latex]M=-F_Ax+F_B\left\langle x-0.2 ight angle ^1[/latex]

where x is in meters. Integration yields

[latex]EI\frac{dy}{dx} =-\frac{F_A}{2}x^2+\frac{F_B}{2}\left\langle x-0.2 ight angle ^2 +C_1[/latex]

[latex]EIy =-\frac{F_A}{6}x^3+\frac{F_B}{6}\left\langle x-0.2 ight angle ^3 +C_1x+C_2x[/latex] (3)

The term [latex]EI = 200 \left(10^9 ight) 62.5 \left(10^{−9} ight) = 1.25 \left(10^4 ight) N·m^2[/latex] .

3 -The upward deflection of point A is

[latex]\left({Fl}/{AE} ight) _{AD}-Np[/latex] , 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

[latex]y_A = \frac{F_A\left (0.6 ight) } {\frac{\pi } {4}\left(0.010 ight)^2 \left(200 ight) \left(10^9 ight) } - \left(1 ight) \left(0.0015 ight) \= 3.8197 \left(10^{-8} ight) F_A-1.5 \left(10^{-3} ight) [/latex] (4)

The upward deflection of point C is [latex]\left({Fl}/{AE} ight) _{CE}[/latex] , or

[latex]y_C=\frac{F_C\left(0.8 ight) }{\frac{\pi }{4}\left(0.010 ight)^2\left(200 ight)\left(10^9 ight) }=5.093 \left(10^{-8} ight) F_C[/latex] (5)

Equations (4) and (5) will now serve as the boundary conditions for Eq. (3). At
[latex]x = 0 \ , \ y = y_A[/latex]. Substituting Eq. (4) into (3) with [latex] x = 0 \ and \ EI = 1.25 \left(10^4 ight)[/latex] , noting that the singularity function is zero for[latex] x = 0 [/latex], gives

[latex]-4.7746\left(10^{-4} ight) F_A+C_2=-18.75[/latex] (6)

At [latex]x = 0.2 \ m, \ y = 0[/latex], and Eq. (3) yields

[latex]-1.3333\left(10^{-3} ight)F_A+0.2C_1+C_2=0 [/latex] (7)

At [latex] x = 0.35 \ m, \ y = y_C[/latex]. Substituting Eq. (5) into (3) with [latex]x = 0.35 \ m \ and \ EI =1.25\left(10^4 ight) [/latex] gives

[latex]−7.1458 \left(10^{−3} ight) F_A + 5.625 \left(10^{−4} ight) F_B − 6.3662 \left(10^{−4} ight) F_C + 0.35 C_1 + C2 = 0[/latex] (8)

Equations (1), (2), (6), (7), and (8) are five equations in [latex] F_A\ , \ F_B , \ F_C, \ C_1, \ and \ C_2 [/latex] . Written in matrix form, they are

[latex]\left [ \begin{matrix} -1 & 1 & -1 & 0 & 0 \ 4 & 0 & -3 & 0 & 0 \ -4.7746\left(10^{-4} ight) & 0 & 0 & 0 & 1 \ -1.3333\left(10^{-3} ight) & 0 & 0 & 0.2 & 1 \ -7.1458\left(10^{-3} ight) & 5.625\left(10^{-4} ight) & -6.3662\left(10^{-4} ight) & 0.35 & 1 \end{matrix} ight ] \left [ \begin{matrix} F_A \ F_B \ F_C \ C_1 \ C_2 \end{matrix} ight ] =\left [ \begin{matrix} 0 \ 0 \ -18.75 \ 0 \ 0 \end{matrix} ight ] [/latex]

Solving these equations yields

[latex]F_A=2988 \ N[/latex] [latex]F_B=6971 \ N[/latex] [latex]F_C=3983 \ N[/latex]

[latex]C_1=06.54 \ N.m^2[/latex] [latex]C_2=-17.324 \ N.m^3[/latex]

Equation (3) can be reduced to

[latex]y=-\left(39.84 x^3-92.95\left\langle x-0.2 ight angle^3-8.523x+1.386 ight) \left(10^{-3} ight) [/latex]

at [latex] x=0 \ , \ y=y_A=-1.386\left(10^{-3} ight) \ m =-1.386 \ mm[/latex]

at [latex] x=0.35 \ m, \ y=y_C=-\left[39.84 \left(0.35 ight)^3 - 92.95 \left(0.35-0.2 ight)^3-8.523\left(0.35 ight)+1.386 ight]\left(10^{-3} ight) =0.203\left(10^{-3} ight) \ m=0.203 \ mm[/latex]

Question 1.4.15: The rods AD and CE shown in Fig. 4–17a each have a diameter ...

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