• A curved beam is defined as a beam in which the neutral axis in unloaded condition is curved instead of straight.

• Following assumptions are made in the stress analysis of curved beams:

  1. Plane sections perpendicular to the axis of the beam remain plane after bending.
  2. The moduli of elasticity in tension and compression are equal.
  3. The material is homogeneous and obeys Hooke’s law.

• There are two factors, which distinguish the analysis of straight and curved beams:

1. The neutral and centroidal axes of the straight beam are coincident. However, in curved beam the neutral axis is shifted towards the center of curvature.
2. The bending stresses in straight beam vary linarty with the distance from the neutral axis. In curved beams, the stress distribution is hyperbolic.

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

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

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

$\therefore t = 99.18496 \ mm$