Q.1 A crane having an approximate trapezoidal c/s is shown in fig. It is made of plain carbon steel 45C8 $(Syt=380 \ N/mm^2$) and the FOS is 3.5. Determine the load carrying capacity of hook.

1. $\sigma_{max}$ =$\frac{Syt}{FOS}$ = $\frac{380}{3.5} = 108.57 \ N/mm^2$

2.

$\;$(I) PSG 6.3

$r_{n}$ = $\frac{(\frac{b_{i}+b_{o}}{2})\;h}{(\frac{b_{i}.r_{o}-b_{o}.r_{i}}{h})\;ln\;(\frac{r_{o}}{r_{i}})-(b_{i}-b_{o})}$

$\therefore r_{n}$ = $\frac{(\frac{90+30}{2})\;120}{(\frac{90 \times170-30 \times 50}{120})\;ln\;(\frac{170}{50})-(90-30)}$

$\therefore r_{n}$ = $\frac{7200}{(115 \times 1.22377)-60}$

$\therefore r_{n} = 89.1815 \ mm = \text{radius of neutral axis.}$

(II) $R = r_{i}$ + $\frac{h[b_i + 2b_o]}{3[b_i + b_o]}$ ......... PSG 6.3

$\therefore R = 50 + \frac{120[90 + 2\;\times\;30]}{3[90 + 30]}$

$\therefore R = 100 \ mm$ ... Radius of centroidal axis.

(III) $e = R - r_{n} = 100 - 89.1815$

$\therefore e = 10.8185 \ mm = eccentricity$

1. (I) $h_i = r_{n} - r_{i} = 89.1815 - 50$

$\therefore h_i = 39.1815 \ mm$ ...Dist. of innermost fibre from N.A

(II) $A = (b_{i} + b_{o})\;\frac{h}{2}$...c/s area

$\therefore A = (90+30) \;\frac{120}{2}$

$\therefore A = 7200\ mm^2$

(III) $M_{b} = P.R = P(100) = 100\ P$.....Bending moment wrt centroidal axis.

(IV) Inside fibre, from PSG 6.2

$(\sigma_{b})_{i}$ = $\frac{M_{b} . h_i}{A.e.r_{i}}$ = $\frac{(100P)(39.1815)}{7200 \times 10.8185 \times 50}$

$\therefore$ $(\sigma_{b})_{i}$ = 1.006 $\times 10^{-3} P$

For symmetric section, maximum loading moment occurs at inside fibre,

$\therefore$ $(\sigma_{b})_{i}$ = $(\sigma_{b})_{max} = 1.006 \;\times\; 10^{-3} P$

(V) $\sigma_{t}$ = $\frac{P}{A}$ = $\frac{P}{7200}$ = 1.3888 $\;\times\; 10^{-4} P$....Direct tensile stress.

(VI) $(\sigma_{b})_{max}$ + $\sigma_{t}$ = $\sigma_{max}$

$\therefore (1.006 \;\times\;10^{-3} P) + (1.3888 \;\times\; 10^{-4} P)$ = 108.57

$\therefore P = 94.83 \;\times\;10^3 N = 94.83\ kN$.

Q.2 A curved link of the mechanism made from a round steel bar is shown in fig. The material of the link is plain carbon steel 30C8(Syt = 400 N/mm$^2$) and the factor of safety is 3.5. Determine the dimensions of the link.

1. $\sigma_{max}$ = $\frac{Syt}{FOS}$ = $\frac{400}{3.5} = 114.285 \ N/mm^2$

2. At section X-X,

R = 4D...given

(I) From PSG 6.3,

$r_{i} = R - 0.5D = 4D - 0.5D$

$\therefore r_{i} = 3.5D$

$r_{o} = R + 0.5D = 4D + 0.5D$

$\therefore r_{o} = 4.5D$

r$_{n}$ = $\frac{(\sqrt{r_{0}} + \sqrt{r_{i}})^2}{4}$

$\therefore$ r$_{n}$ = $\frac{(\sqrt{4.5D} + \sqrt{3.5D})^2}{4}$

$\therefore r_{n} = 3.9843D$

(II) $e = R - r_{n} = 4D - 3.9843D = 0.0157D$

1. (I) $h_i = r_{n} - r_{i} = 3.9843D - 3.5D = 0.4843D$

(II) $A = \frac{\pi}{4}\;D^2 = 0.7854 D^2$

(III) $M_{b} = P . R = 1 \times 10^3 \times 4D = 4 \times 10^3D$

(IV) Inside fibre

$(\sigma_{b})_{i}$ = $\frac{M_{b} . hi}{A.e.r_{i}}$ = $\frac{4 \times 10^3 D \times 0.4843D}{0.7854D^2 \times 0.0157D \times 3.5D}$

$\therefore$ $(\sigma_{b})_{i}$ = $\frac{44.8865 \times 10^3}{D^2}$ = $(\sigma_{b})_{max}$

(V) $\sigma_{t}$ = $\frac{P}{A}$ = $\frac{1 \times 10^3}{0.7854D^2}$ = $\frac{1273.2365}{D^2}$

(VI) $(\sigma_{b})_{max}$ + $\sigma_{t}$ = $\sigma_{max}$

$\therefore$ $\frac{44.8865 \times 10^3}{D^2}$ + $\frac{1273.2365}{D^2} = 114.285$

$\therefore D = 20.0972 \ mm$

$D \approx 21 \ mm$

Q.3 The C frame of a $100\ KN$ capacity press is shown in fig. The material of the frame is grey cast iron FG 200 and factor of safety is 3. Determine dimensions of frame.

1. $\sigma_{max}$ = $\frac{Syt}{FOS}$ = $\frac{200}{3} = 66.6667 N/mm^2$

   1. (I)

$\begin{aligned} r_{n} & = \frac{(b_{i} - t) t_{i} + t_{h}}{(b_{i} - t)\;ln\;\frac{r_{i} + t_{i}}{r_{i}} + t\;ln\frac{r_{o}}{r_{i}}} \\ \therefore r_{n} & = \frac{(3t - 0.75t) t + (0.75t)(3t)}{(3t - 0.75t)\;ln\;\frac{2t + t}{2t} + 0.75t\;ln(\frac{5t}{2t})} \\ & = \frac{t^2(3 - 0.75) + t^2(0.75\;\times\;3)}{t(3 - 0.75)\;ln\;(\frac{3}{2}) + t\;[0.75\;ln\;(\frac{5}{2})]} \\ & = \frac{t^2[(3 - 0.75)\;+\;(0.75\;\times\;3)]}{t[(3 - 0.75)\;ln\;(\frac{3}{2})\;+\;(0.75\;ln\;(\frac{5}{2}))]} \\ \therefore r_{n} & = 2.8134t. \end{aligned}$

(II)
$\begin{aligned} R & = r_{i} + [\frac{[\frac{h^2 t}{2} + \frac{t_{i}^2(b_{i} - t)}{2}]}{ht + (b_{i} - t)t_{i}}] \\ & = 2t + [\frac{[\frac{(3t)^2(0.75t)}{2} + \frac{t^2(3t - 0.75t)}{2}]}{(3t)(0.75t) + (3t - 0.75t)t}] \\ & = 2t + [\frac{[\frac{(t^3(9 \times 0.75)}{2} + \frac{t^3(3 - 0.75)}{2}]}{t^2(3 \times 0.75) + t^2(3 - 0.75)}] \\ & = 2t + 1t \\ R & = 3t. \\ \end{aligned}$

(III) $e = R - r_{n} = 3t - 2.8134t = 0.1866t$

1. (I) $h_{i} = r_{n} - r_{i} = 2.8134t = 2t = 0.8134t$

(II) $A = (3t)(t) + (2t)(0.75t) = 4.5t^2$

(III) $M_{b} = P \times (1000 + R) = 100 \times 10^3 \times (1000 + 3t)$

(IV) $(\sigma_{b})_{i}$ = $\frac{M_{b} . hi}{A.e.r_{i}}$

= $\frac{100 \times 10^3 (1000 + 3t)\; \times 0.8134t}{4.5t^2 \times 0.1866t \times 2t}$

= $\frac{(1000 + 3t)\;\times 81.34\;\times\;10^3 t}{1.6794.t^4}$

= $(\sigma_{b})_{max}$

(V) $\sigma_{t}$ = $\frac{P}{A}$ = $\frac{100 \times 10^3}{4.5t^2}$ = $\frac{22.2222\;\times\;10^3}{t^2}$

(VI) $(\sigma_{b})_{max}$ + $\sigma_{t}$ = $\sigma_{max}$

$\therefore 66.6667 = \frac{(1000 + 3t)\;\times 81.34\;\times\;10^3 t}{1.6794t^4} + \frac{22.2222\;\times\;10^3}{t^2}$

$\therefore 66.6667 = \frac{(1000 + 3t)\;48.434\;\times\;10^3}{t^3} + \frac{22.2222\;\times\;10^3}{t^2}$

$\therefore t = 99.18496 \ mm$