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Chapter 4

Q. 4.21

The C-frame of a 100 kN capacity press is shown in Fig. 4.68(a). The material of the frame is grey cast iron FG 200 and the factor of safety is 3. Determine the dimensions of the frame.


Verified Solution

\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.