The ring girder shown in Fig. 12.14 is supported at eight points. If W = 200 kips, find the forces in the ring at the supports and at the point of maximum torsional moment.

Question

Accepted Answer

[latex]w=\frac{ W }{\pi d}=\frac{200}{(\pi)(100)}=0.637 ext{ k - in.}[/latex]

from Table 12.5, with N = 8,

Table 12.5 Ring Girder Coefficients
Number of Supports	Angle Between Supports (degrees)	[latex]K_{3}[/latex]	[latex]K_{4}[/latex]	[latex]K_{5}[/latex]	[latex]K_{6}[/latex]	Angle of Maximum Torsion from Support (degrees)
2	180	-1.0000	1.5707	-0.5707	[latex]-3.307 imes 10^{-1}[/latex]	39.55
3	120	-0.3954	1.0471	-0.2091	[latex]-8.278 imes 10^{-2}[/latex]	25.80
4	90	-0.2146	0.7853	-0.1107	[latex]-3.313 imes 10^{-2}[/latex]	19.21
5	72	-0.1351	0.6283	-0.0690	[latex]-1.654 imes 10^{-2}[/latex]	15.30
6	60	-0.0931	0.5235	-0.0471	[latex]-9.471 imes 10^{-3}[/latex]	12.74
8	45	-0.0519	0.3926	-0.0262	[latex]-3.940 imes 10^{-3}[/latex]	9.53
10	36	-0.0331	0.3141	-0.0166	[latex]-2.007 imes 10^{-3}[/latex]	7.62
12	30	-0.0229	0.2617	-0.0115	[latex]-1.154 imes 10^{-3}[/latex]	6.34
16	22.5	-0.0128	0.1963	-0.0065	[latex]-3.722 imes 10^{-3}[/latex]	4.72
20	18	-0.0082	0.1570	-0.0042	[latex]-2.469 imes 10^{-4}[/latex]	3.79

[latex]K_{3}=-0.0519[/latex] [latex]K_{4}=0.3926[/latex]

[latex]K_{5}=-0.0262[/latex] [latex]K_{6}=3.940 imes 10^{-3}[/latex]

Maximum torsion occurs at 9.53° from support.

The forces given by Eqs. 12.16, 12.19, and 12.20 are determined in Table 12.6 and illustrated in Fig. 12.15 at the supports and at the point of maximum torsion.

[latex]\begin{array}{ll}M_{s}=K_{3} w r^{2} & M_{m}=K_{5} w r^{2} \V_{s}=K_{4} w r & V_{m}=0 \T_{s}=0 & T_{m}=0\end{array}[/latex] (12.16)

where [latex]M_{s}, V_{s}, T_{s}[/latex] = support moment, shear, and torsion, respectively. Positive direction is shown in Fig. 12.11.

[latex]M_{m}, V_{m}, T_{m}[/latex] = midspan moment, shear and torsion, respectively

[latex]K_{3}=K_{5}[/latex] = constants obtained from Table 12.5

w = uniform load

r = radius

[latex]\sigma=\frac{m r y}{I_{x}}[/latex]

or

[latex]\sigma=\frac{w b^{2} d^{2} h r y}{4 I_{x}^{2}}[/latex] (12.19)

where [latex]\sigma[/latex] = stress

b = width of flange

d = distance between flanges

h = flange thickness

[latex]I_{x}[/latex] = moment of inertia of girder

r = radius of vessel

t = shell thickness

w = uniform applied load

[latex]\begin{aligned}&M_{l}=\frac{H r}{2}\left(\cos heta \cot \frac{\alpha}{2}-\sin heta-\frac{2}{\alpha} ight) \quad-\frac{\alpha}{2}< heta<\frac{\alpha}{2} \&F_{l}=\frac{H}{2} \frac{\cos \left( heta+\frac{\alpha}{2} ight)}{\sin \frac{\alpha}{2}} \&V_{l}=\frac{-H}{2} \frac{\sin \left( heta+\frac{\alpha}{2} ight)}{\sin \frac{\alpha}{2}}\end{aligned}[/latex] (12.20)

Table 12.6
	At Support	At Point of Maximum Torsion
Eq. 12.16
[latex]M_s , M_m[/latex]	-82.65 k-in.	0 k-in.
[latex]V_s , V_m[/latex]	12.50 k	7.20 k
[latex]T_s , T_m[/latex]	0	6.27 k-in.
Eq. 12.19
M	-60.05 k-in.	-60.05 k-in.
Eq. 12.20[latex]^a[/latex]
[latex]{M_l}^b[/latex]	+40.54 k-in.	-101.50 k-in.
[latex]F_l[/latex]	14.80 k	13.58 k
[latex]V_l[/latex]	-6.13 k	-8.50 k
[latex]^a[/latex] These equations apply at points B and D, points A and C have opposite signs. [latex]\begin{aligned}{ }^{b} H &=\frac{W}{N}\frac{\left(e+e^{\prime} ight)}{d} \&=\frac{200}{8} \frac{1.89+4}{12} \&=12.26^{k}\end{aligned}[/latex] [latex]α = \frac{360}{8}=45°[/latex]

Question 12.5: The ring girder shown in Fig. 12.14 is supported at eight po...

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