Question 12.1: The gantry shown in Fig. 12.15 is constructed from 100 mm × ...

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{(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