Question 2.9: A 6 mm × 75 mm plate, 600 mm long, has a circular hole of 25...
A 6 mm × 75 mm plate, 600 mm long, has a circular hole of 25 mm diameter located at its center. Find the axial tensile force that can be applied to this plate in the longitudinal direction without exceeding an allowable stress of 220 MPa. How does the presence of the hole affect the strength of the plate?
Given: Dimensions of plate, limiting normal stress.
Find: Allowable axial load that can be applied to plate.
Assume: Hole is only feature that causes a stress concentration.
The cross-sectional area normal to an axial load P is A_o = 6 mm × 75 mm = 450 mm² (Figure 2.42). The average normal stress induced by such a load will be
\sigma _{ave}=P/A_oand due to the presence of the hole we must consider the effects of stress concentration:
\sigma _{max}=K\sigma _{ave}=K\frac{\textrm{P}}{\textrm{A}_o} ,
We can find K for this geometry using the graph in Figure 2.27:
\frac{r}{d}=\frac{(25 \textrm{mm})/2}{75 \textrm{mm} -25 \textrm{mm}}= \frac{1}{4} ,
K(\frac{r}{d}=0.25)=2.26so
\sigma _{\max}=K\frac{P}{A_o}=2.26\frac{P}{450 \textrm{mm}^2}=0.005 Pand since we must have
220 MPa ≥ 0.005 P
then
P ≤ 43.8 kN.
Note: If there were no hole in this plate, we would simply have
\sigma _{ave}=\textrm{P/A}_o ,
and we could allow a force
P ≤ 99 kN.
So with the hole, we can permit only 44% of the load we could have allowed without the hole.
