Question 8.2: Maximum Contact Pressure between a Cylindrical Rod and a Bea......
Maximum Contact Pressure between a Cylindrical Rod and a Beam
A concentrated load F at the center of a narrow, deep beam is applied through a rod of diameter d laid across the beam width of width b. Determine
a. The contact area between rod and beam surface.
b. The maximum contact stress.
c. The maximum value of the subsurface shear stress.
Given: F=4 kN, d=12 mm, L=125 mm.
Assumptions: Both the beam and the rod are made of steel having E=200 GPa and \nu=0.3.
We use the equations on the third column of case A in Table 8.4.
a. Since E_1=E_2=E \text { or } \Delta=2 / E \text {, } the half-width of the contact area is
\begin{aligned} a & =1.076 \sqrt{\frac{F}{L} r_1 \Delta} \\ & =1.076 \sqrt{\frac{4\left(10^3\right)}{0.125} \frac{(0.006) 2}{200\left(10^9\right)}}=0.0471 mm \end{aligned}
The rectangular contact area equals
2 a L=2(0.0471)(125)=11.775 mm ^2
b. The maximum contact pressure is therefore
p_o=\frac{2}{\pi} \frac{F}{a L}=\frac{2}{\pi} \frac{4\left(10^3\right)}{5.888\left(10^{-6}\right)}=432.5 MPa
c. Observe from Figure 8.9b that the largest value of the shear stress is at approximately z=0.75a for which
\frac{\tau_{y z, \max }}{p_o}=0.3 \quad \text { or } \quad \tau_{y z, \max }=0.3(432.5)=129.8 MPa
This stress occurs at a depth z = 0.75(0.0471)=0.0353 mm below the surface.
| TABLE 8.4 Maximum Pressure p_o , and Deflection \delta of Two Bodies at the Point of Contact |
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| Configuration | \text { Spheres: } p_o=1.5 \frac{F}{\pi a^2} | \text { Cylinders: } p_o=\frac{2}{\pi} \frac{F}{a L} |
| \quad\quad\quad\quad\quad\quad\quad\quad
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\text { Sphere on a flat surface } \\ \\ \begin{aligned} & a=0.880 \sqrt[3]{F r_1 \Delta} \\ & \delta=0.775 \sqrt[3]{F^2 \frac{\Delta^2}{r_1}} \end{aligned} | \text { Cylinder on a flat surface } \\ \\ \begin{aligned} & a=1.076 \sqrt{\frac{F}{L} r_1 \Delta} \\ & \text { For } E _1= E _2= E \\ & \delta=\frac{0.579 F}{E L}\left(\frac{1}{3}+\ln \frac{2 r_1}{a}\right) \end{aligned} |
| \quad\quad\quad\quad\quad\quad\quad\quad
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\text { Two spherical balls } \\ \\ \begin{aligned} & a=0.880 \sqrt[3]{F \frac{\Delta}{m}} \\ & \delta=0.775 \sqrt[3]{F^2 \Delta^2 m} \end{aligned} | \text { Two cylindrical rollers } \\ \\ a=1.076 \sqrt{\frac{F \Delta}{L m}} |
| \quad\quad\quad\quad\quad\quad\quad\quad
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\text { Sphere on a spherical seat } \\ \\ \begin{aligned} & a=0.880 \sqrt[3]{F \frac{\Delta}{n}} \\ & \delta=0.775 \sqrt[3]{F^2 \Delta^2 n} \end{aligned} | \text { Cylinder on a cylindrical seat } \\ \\ a=1.076 \sqrt{\frac{F \Delta}{L n}} |
| Source: [9]. Notes: \Delta=\frac{1}{E_1}+\frac{1}{E_2}, m=\frac{1}{r_1}+\frac{1}{r_2}, n=\frac{1}{r_1}-\frac{1}{r_2} , where the modulus of elasticity (E ) and radius (r ) are for the contacting members, 1 and 2. The L represents the length of the cylinder (Figure 8.7). The total force pressing two spheres or cylinder is F . Poisson’s ratio \nu in the formulas is taken as 0.3. |
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