Deflections of a Three-Bar Device The rigid member BC is attached by the 12 mm diameter rod AB and the 10 mm diameter rod CD (Figure 4.3a). Each rod is made of cold-rolled yellow brass with yield strength Sy and elastic modulus of elasticity E. What is the displacement of point E of the bar caused

Question

Deflections of a Three-Bar Device

The rigid member [latex]BC[/latex] is attached by the 12 mm diameter rod [latex]AB[/latex] and the 10 mm diameter rod [latex]CD[/latex] (Figure 4.3a). Each rod is made of cold-rolled yellow brass with yield strength [latex]S_{y}[/latex] and elastic modulus of elasticity [latex]E[/latex]. What is the displacement of point [latex]E[/latex] of the bar caused by a vertical load [latex]P[/latex] applied at this point?

Given: [latex]S_{y}[/latex] = 435 MPa, [latex]P[/latex] = 60 kN, [latex]E[/latex] = 105 GPa (by Table B.1)

Assumptions: Failure will not occur at the pin-connected joints. Both rods deform
uniformly.

Accepted Answer

The cross-sectional areas of the bars equal

[latex] A_{A B}=\frac{\pi}{4}(12)^2=113.1 mm ^2 \quad A_{C D}=\frac{\pi}{4}(10)^2=78.5 mm ^2 . [/latex]

Free body of bar [latex]BC[/latex]. Applying equations of equilibrium to Figure 4.3b, the tensile forces in each bar are

[latex] \begin{array}{c} \sum M_C=0: \quad-F_{A B}(0.6)+60(0.4)=0 \quad F_{A B}=40 kN \ \sum F_y=0: \quad-60+40+F_{C D}=0 \quad F_{C D}=20 kN \end{array} [/latex]

Displacements of [latex]B[/latex] and [latex]C[/latex]. The deflections of the bottom of the links, respectively, are

[latex] \begin{array}{l} \delta_B=\frac{F_{A B} L_{A B}}{A_{A B} E}=\frac{40\left(10^3 ight)(0.6)}{(113.1)(105)\left(10^3 ight)}=2.02 mm \ \delta_C=\frac{F_{C D} L_{C D}}{A_{C D} E}=\frac{20\left(10^3 ight)(0.6)}{(78.5)(105)\left(10^3 ight)}=1.46 mm \end{array} [/latex]

Displacement of [latex]E[/latex]. A diagram showing the centerline displacements of points [latex]B, C,[/latex] and [latex]E[/latex] on the rigid bar is represented in Figure 4.3b. It follows that, by proportion of the shaded triangle, the displacement of point [latex]E[/latex] is found as

[latex] \delta_E=\delta_C+\left(\delta_B-\delta_C ight)\left(\frac{0.4}{0.6} ight)=\left[1.46+(0.56)\left(\frac{2}{3} ight) ight] [/latex]

or

[latex]\delta _{E}[/latex] = 1.83 mm

Comments: A positive sign means downward displacement. The largest axial stress is in rod [latex]AB,[/latex] [latex] \sigma_{A B}=40\left(10^3 ight) / 113.1 imes 10^{-6}=354 MPa <435 MPa .[/latex] Therefore, the bar will not deform permanently.

Question 4.4: Deflections of a Three-Bar Device The rigid member BC is att...

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