Question 4.21: The C-frame of a 100 kN capacity press is shown in Fig. 4.68...

\text { Given } P=100 kN \quad S_{u t}=200 N / mm ^{2} .

(fs) = 3.

Step I Calculation of permissible tensile stress

\sigma_{\max }=\frac{S_{u t}}{(f s)}=\frac{200}{3}=66.67 N / mm ^{2} .

Step II Calculation of eccentricity (e)
Using notations of Eq. (4.66) and Fig. [4.65(e)],

R_{N}=\frac{t_{i}\left(b_{i}-t\right)+t h}{\left(b_{i}-t\right) \log _{e}\left(\frac{R_{i}+t_{i}}{R_{i}}\right)+t \log _{e}\left(\frac{R_{o}}{R_{i}}\right)}             (4.66).

b_{i}=3 t \quad h=3 t \quad R_{i}=2 t .

R_{o}=5 t \quad t_{i}=t \quad t=0.75 t .

From Eq. (4.66),

R_{N}=\frac{t_{i}\left(b_{i}-t\right)+t h}{\left(b_{i}-t\right) \log _{e}\left(\frac{R_{i}+t_{i}}{R_{i}}\right)+t \log _{e}\left(\frac{R_{o}}{R_{i}}\right)}             (4.66).

R_{N}=\frac{t_{i}\left(b_{i}-t\right)+t h}{\left(b_{i}-t\right) \log _{e}\left(\frac{R_{i}+t_{i}}{R_{i}}\right)+t \log _{e}\left(\frac{R_{o}}{R_{i}}\right)} .

=\frac{t(3 t-0.75 t)+0.75 t(3 t)}{(3 t-0.75 t) \log _{e}\left(\frac{2 t+t}{2 t}\right)+0.75 t \log _{e}\left(\frac{5 t}{2 t}\right)} .

= 2.8134t.

From Eq. (4.67),

R=R_{i}+\frac{\frac{1}{2} t h^{2}+\frac{1}{2} t_{i}^{2}\left(b_{i}-t\right)}{t h+t_{i}\left(b_{i}-t\right)}             (4.67).

R=R_{i}+\frac{\frac{1}{2} t h^{2}+\frac{1}{2} t_{i}^{2}\left(b_{i}-t\right)}{t h+t_{i}\left(b_{i}-t\right)} .

=2 t+\frac{\frac{1}{2}(0.75 t)(3 t)^{2}+\frac{1}{2} t^{2}(3 t-0.75 t)}{(0.75 t)(3 t)+t(3 t-0.75 t)}=3 t .

e=R-R_{N}=3 t-2.8134 t=0.1866 t .

Step III Calculation of bending stress

h_{i}=R_{N}-R_{i}=2.8134 t-2 t=0.8134 t .

A=(3 t)(t)+(0.75 t)(2 t)=\left(4.5 t^{2}\right) mm ^{2} .

M_{b}=100 \times 10^{3}(1000+R) .

=100 \times 10^{3}(1000+3 t) N – mm .

From Eq. (4.56), the bending stress at the inner fibre is given by,

\sigma_{b i}=\frac{M_{b} h_{i}}{A e R_{i}}=\frac{100 \times 10^{3}(1000+3 t)(0.8134 t)}{\left(4.5 t^{2}\right)(0.1866 t)(2 t)} .

=\frac{100 \times 10^{3}(1000+3 t)(2.1795)}{\left(4.5 t^{2}\right)} N / mm ^{2} .

Step IV Calculation of direct tensile stress

\sigma_{t}=\frac{P}{A}=\frac{100 \times 10^{3}}{\left(4.5 t^{2}\right)} N / mm ^{2} .

Step V Calculation of dimensions of cross-section Adding the two stresses and equating the resultant stress to permissible stress,

\sigma_{b i}+\sigma_{t}=\sigma_{\max } .

\frac{100 \times 10^{3}(1000+3 t)(2.1795)}{\left(4.5 t^{2}\right)}+\frac{100 \times 10^{3}}{\left(4.5 t^{2}\right)}=66.67 .

t^{3}-2512.83 t-726500=0 .

Solving the above cubic equation by trial and error method,
t = 99.2 mm or t = 100 mm.