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Chapter 5

Q. 5.3

A plate made of steel 20C8 \left(S_{u t}=440 N / mm ^{2}\right)  in hot rolled and normalised condition is shown in Fig. 5.28. It is subjected to a completely reversed axial load of 30 kN. The notch sensitivity factor q can be taken as 0.8 and the expected reliability is 90%. The size factor is 0.85. The factor of safety is 2. Determine the plate thickness for infinite life.

Step-by-Step

Verified Solution

\text { Given } P=\pm 30 kN \quad S_{u t}=440 N / mm ^{2} \quad(f s)=2 .

R=90 \% \quad q=0.8 \quad K_{b}=0.85 .

Step I Endurance limit stress for plate

S_{e}^{\prime}=0.5 S_{u t}=0.5(440)=220 N / mm ^{2}

\text { From Fig. } 5.24 \text { (hot rolled steel and } S_{u t}= \left.440 N / mm ^{2}\right) .

K_{a}=0.67 .

K_{b}=0.85 .

For 90% reliability,

K_{c}=0.897 .

\frac{d}{w}=\frac{10}{50}=0.2 .

From Fig. 5.2,    K_{t}=2.51 .

From Eq. (5.12),

K_{f}=1+q\left(K_{t}-1\right)                 (5.12).

K_{f}=1+q\left(K_{t}-1\right)=1+0.8(2.51-1)=2.208 .

K_{d}=\frac{1}{K_{f}}=\frac{1}{2.208}=0.4529 .

S_{e}=K_{a} K_{b} K_{c} K_{d} S_{e}^{\prime} .

=0.67(0.85)(0.897)(0.4529)(220) .

=50.9 N / mm ^{2} .

For axial load, (Eq. 5.30)

\left(S_{e}\right)_{a}=0.8 S_{e}             (5.30).

\left(S_{e}\right)_{a}=0.8 S_{e}=0.8(50.9)=40.72 N / mm ^{2} .

Step II Permissible stress amplitude

\sigma_{a}=\frac{\left(S_{e}\right)_{a}}{(f s)}=\frac{40.72}{2}=20.36 N / mm ^{2}               (a).

Step III Plate thickness

\sigma_{a}=\frac{P}{(w-d) t}=\frac{(30)\left(10^{3}\right)}{(50-10) t} N / mm ^{2}             (b).

From (a) and (b),

20.36=\frac{(30)\left(10^{3}\right)}{(50-10) t} .

∴            t = 36.84 mm.