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## Q. 12.1

The gantry shown in Fig. 12.15 is constructed from 100 mm × 50 mm rectangular crosssection and, under service conditions, supports a maximum load P of 20 kN. Determine the maximum distance d at which P can be safely applied if the maximum tensile and compressive stresses for the material used are limited to 30 MN/m² and 100 MN/m² respectively.

How would this value change if the cross-section were circular, but of the same crosssectional area?

## Verified Solution

For the gantry and cross-section of Fig. 12.15 the following values are obtained by inspection:

$R_{c} = 150 mm$    $R_{i} = 100 mm$   $R_{0} = 200 mm$    $b = 50 mm.$

:. From Table 12.1(a)

 Table 12.1. Values of $\int{\frac{dA}{r} }$ for curved bars Cross-section $\int{\frac{dA}{r} }$ (a) Rectangle $b\log _{e}(\frac{R_{0}}{R_{i}} )$ (N.B. The two following cross-sections are simply produced by the addition of terms of this form for each rectangular portion) (b )T-section $b_{1}\log _{e}(\frac{R_{i}+d_{1}}{R_{i}} )+b_{2}\log _{e}(\frac{R_{0}}{R_{i}+d_{1}} )$ (c) I-beam $b_{1}\log _{e}(\frac{R_{i}+d_{1}}{R_{i}} )+b_{2}\log _{e}(\frac{R_{0}-d_{3}}{R_{i}+d_{1}} )+b_{3}\log _{e}(\frac{R_{0}}{R_{0}+d_{3}} )$ (d) trapezoid $[\frac{(b_{1}R_{0}-b_{2}R_{i})}{d} \log _{e}(\frac{R_{0}}{R_{i}} )]-b_{1}+b{2}$ (e ) triangle As above (d) with $b_{2} = 0$ As above (d) with $b_{1} = 0$ $2\pi \left\{(R_{i}+R)-[(R_{i}+R)^{2}-R_{2}]^{1/2}\right\}$

$\int{\frac{dA}{r} } =b\log _{e}(\frac{R_{o}}{R_{i}} )=50 \log _{e}(\frac{200}{100} )$
$=34.6574 mm$
$R_{1}=\frac{A}{\int{\frac{dA}{r} }} =\frac{50\times 100}{34.6574} =144.269 mm$
$h=R_{c}-R_{1}=150-144.269=5.731 mm$

Direct stress (Compressive) due to    $P=\frac{P}{A} =\frac{20\times 10^{3}}{(100\times 50)10^{-6}} =4 MN/m^{2}$

Thus, for maximum tensile stress of 30 MN/m² to be reached at B the bending stress (tensile) must be 30 + 4 = 34 MN/m².

Now  $y_{max}=50+5.731$
$=55.731$   at B

and bending stress at     $B=\frac{My}{hA(R_{1}+y)} =34 MN/m^{2}$
$\frac{(40\times 10^{3} d)\times 55.731\times 10^{-3}}{(5.731\times 10^{-3})(50\times 100\times 10^{-6})(200\times 10^{-3})} =34\times 10^{6}$
$d=174.69 mm$

For maximum compressive stress of 100 MN/m² at A the compressive bending stress must be limited to 100 − 4 = 96 MN/m² in order to account for the additional direct load effect

At A, with  $y_{min}=50-5.731=44.269$

bending stress    $=\frac{(20\times 10^{3})44.269\times 10^{-3}}{(5.731\times 10^{-3})(50\times 100\times 10^{-6})(100\times 10^{-3})} =96\times 10^{6}$

$d = 310.7 mm$.

The critical condition is therefore on the tensile stress at B and the required maximum value of d is 174.69 mm

If a circular section were used of radius R and of equal cross-sectional area to the rectangular section then πR² = 100 × 50 and R = 39.89 mm.
:. From Table 12.1 assuming $R_{c}$ remains at 150 mm

$\int{\frac{dA}{r} }=2\pi \left\{(R_{i}+R)-\sqrt{(R_{1}+R)^{2}-R^{2}} \right\}$
$=2\pi \left\{150-\sqrt{150^{2}-39.894^{2}}\right\}$
$=2\pi 5.4024=33.944 mm$
$R_{1}=\frac{A}{\int{\frac{dA}{r} }}=\frac{50\times 100}{33.944} =147.301 mm$

with  $h = R_{c} – R_{1} = 150 – 147.3 = 2.699 mm.$

For critical tensile stress at B with  y = 39.894 + 2.699 = 42.593.

$\frac{My}{hA(R_{1}+y)} =34 MN/m^{2}$

$\frac{(20\times 10^{3} d)\times(42.593\times 10^{-3})}{2.699\times 10^{-3}(\pi \times 39.894^{2}\times 10^{-6})(150+39.894)} =34\times 10^{6}$
$d=102.3$

i.e. Use of the circular section reduces the limit of d within which the load P can be applied