In Example 12.1, it was found that 12-1 3/8 in. A 307 anchor bolts were needed for a vessel with an outside radius R = 7 ft , W1 = 160 kips, M = 1500 ft-kips, and a skirt thickness of 0.375 in. If fc¹ = 3000 psi, determine the actual stress in the concrete and bolts.

Question

In Example 12.1, it was found that [latex]12-1 \frac{3}{8}[/latex] in. A 307 anchor bolts were needed for a vessel with an outside radius R = 7 ft , [latex]W_1[/latex] = 160 kips, M = 1500 ft-kips, and a skirt thickness of 0.375 in. If [latex]f_{c}^{1}[/latex] = 3000 psi, determine the actual stress in the concrete and bolts.

Accepted Answer

By referring to Fig. 12.3 and Table 12.2, for [latex]1 \frac{3}{8}[/latex] in. bolts, the bolt circle can be calculated as

Table 12.2 Bolt Dimensions and Clearances Bolting Data
Bolt Size	Threads		Nut Dimensions		Bolt Spacing B	Radial Distance R	Edge Distance E	Wrench Diameter a
Bolt Size	No. of Threads	Root Area (in.²)	Across Flats	Across Corners	Bolt Spacing B	Radial Distance R	Edge Distance E	Wrench Diameter a
[latex]\frac{1}{2}[/latex]	13	0.126	[latex]\frac{7}{8}[/latex]	0.969	[latex]1 \frac{1}{4}[/latex]	[latex]\frac{13}{16}[/latex]	[latex]\frac{5}{8}[/latex]	[latex]1 \frac{1}{2}[/latex]
[latex]\frac{5}{8}[/latex]	11	0.202	[latex]1 \frac{1}{16}[/latex]	1.175	[latex]1 \frac{1}{2}[/latex]	[latex]\frac{15}{16}[/latex]	[latex]\frac{3}{4}[/latex]	[latex]1 \frac{3}{4}[/latex]
[latex]\frac{3}{4}[/latex]	10	0.302	[latex]1 \frac{1}{4}[/latex]	1.383	[latex]1 \frac{3}{4}[/latex]	[latex]1 \frac{1}{8}[/latex]	[latex]\frac{13}{16}[/latex]	[latex]2 \frac{1}{16}[/latex]
[latex]\frac{7}{8}[/latex]	9	0.419	[latex]1 \frac{7}{16}[/latex]	1.589	[latex]2 \frac{1}{16}[/latex]	[latex]1 \frac{1}{4}[/latex]	[latex]\frac{15}{16}[/latex]	[latex]2 \frac{3}{8}[/latex]
1	8	0.551	[latex]1 \frac{5}{8}[/latex]	1.796	[latex]2 \frac{1}{4}[/latex]	[latex]1 \frac{3}{8}[/latex]	[latex]1 \frac{1}{16}[/latex]	[latex]2 \frac{5}{8}[/latex]
[latex]1 \frac{1}{8}[/latex]	8	0.728	[latex]1 \frac{13}{16}[/latex]	2.002	[latex]2 \frac{1}{2}[/latex]	[latex]1 \frac{1}{2}[/latex]	[latex]1 \frac{1}{8}[/latex]	[latex]2 \frac{7}{8}[/latex]
[latex]1 \frac{1}{4}[/latex]	8	0.929	2	2.209	[latex]2 \frac{13}{16}[/latex]	[latex]1 \frac{3}{4}[/latex]	[latex]1 \frac{1}{4}[/latex]	[latex]3 \frac{1}{4}[/latex]
[latex]1 \frac{3}{8}[/latex]	8	1.155	[latex]2 \frac{3}{16}[/latex]	2.416	[latex]3 \frac{1}{16}[/latex]	[latex]1 \frac{7}{8}[/latex]	[latex]1 \frac{3}{8}[/latex]	[latex]3 \frac{1}{2}[/latex]
[latex]1 \frac{1}{2}[/latex]	8	1.405	[latex]2 \frac{3}{8}[/latex]	2.622	[latex]3 \frac{1}{4}[/latex]	2	[latex]1 \frac{1}{2}[/latex]	[latex]3 \frac{3}{4}[/latex]
[latex]1 \frac{5}{8}[/latex]	8	1.608	[latex]2 \frac{9}{16}[/latex]	2.828	[latex]3 \frac{1}{2}[/latex]	[latex]2 \frac{1}{8}[/latex]	[latex]1 \frac{5}{8}[/latex]	4
[latex]1 \frac{3}{4}[/latex]	8	1.980	[latex]2 \frac{3}{4}[/latex]	3.035	[latex]3 \frac{3}{4}[/latex]	[latex]2 \frac{1}{4}[/latex]	[latex]1 \frac{3}{4}[/latex]	[latex]4 \frac{1}{4}[/latex]
[latex]1 \frac{7}{8}[/latex]	8	2.304	[latex]2 \frac{15}{16}[/latex]	3.242	4	[latex]2 \frac{3}{8}[/latex]	[latex]1 \frac{7}{8}[/latex]	[latex]4 \frac{1}{2}[/latex]
2	8	2.652	[latex]3 \frac{1}{8}[/latex]	3.449	[latex]4 \frac{1}{4}[/latex]	[latex]2 \frac{1}{2}[/latex]	2	[latex]4 \frac{3}{4}[/latex]
[latex]2 \frac{1}{4}[/latex]	8	3.423	[latex]3 \frac{1}{2}[/latex]	3.862	[latex]4 \frac{3}{4}[/latex]	[latex]2 \frac{3}{4}[/latex]	[latex]2 \frac{1}{4}[/latex]	[latex]5 \frac{1}{4}[/latex]
[latex]2 \frac{1}{2}[/latex]	8	4.292	[latex]3 \frac{7}{8}[/latex]	4.275	[latex]5 \frac{1}{4}[/latex]	[latex]3 \frac{1}{16}[/latex]	[latex]2 \frac{3}{8}[/latex]	[latex]5 \frac{7}{8}[/latex]
[latex]2 \frac{3}{4}[/latex]	8	5.259	[latex]4 \frac{1}{4}[/latex]	4.688	[latex]5 \frac{3}{4}[/latex]	[latex]3 \frac{3}{8}[/latex]	[latex]2 \frac{5}{8}[/latex]	[latex]6 \frac{1}{2}[/latex]
3	8	6.324	[latex]4 \frac{5}{8}[/latex]	5.102	[latex]6 \frac{1}{4}[/latex]	[latex]3 \frac{5}{8}[/latex]	[latex]2 \frac{7}{8}[/latex]	7
[latex]3 \frac{1}{4}[/latex]	8	7.487	5	5.515	[latex]6 \frac{3}{4}[/latex]	[latex]3 \frac{7}{8}[/latex]	3	[latex]7 \frac{1}{2}[/latex]
[latex]3 \frac{1}{2}[/latex]	8	8.749	[latex]5 \frac{3}{8}[/latex]	5.928	[latex]7 \frac{1}{8}[/latex]	[latex]4 \frac{1}{8}[/latex]	[latex]3 \frac{1}{4}[/latex]	8
[latex]3 \frac{3}{4}[/latex]	8	10.108	[latex]5 \frac{3}{4}[/latex]	6.341	[latex]7 \frac{5}{8}[/latex]	[latex]4 \frac{3}{8}[/latex]	[latex]3 \frac{1}{2}[/latex]	[latex]8 \frac{1}{2}[/latex]
4	8	11.566	[latex]6 \frac{1}{8}[/latex]	6.755	[latex]8 \frac{1}{8}[/latex]	[latex]4 \frac{5}{8}[/latex]	[latex]3 \frac{5}{8}[/latex]	9

d = 2(84 + 0.25 + 1.875) = 172.25 in.

Also

[latex]t_{c}=2(0.25+1.875+1.375)+0.375=7.375 ext { in. }[/latex]

From Example 12.1,

[latex]f_{s}=18.2 \ ksi[/latex]

From Eq. 12.2

[latex]\sigma=\frac{-W}{2 \pi R t} \pm \frac{M}{\pi R^{2} t}[/latex] (12.2)

where σ = axial stress in skirt

W = weight of vessel

M = moment due to wind or earthquake forces

R = radius of skirt

t = thickness of skirt

[latex]\begin{aligned}t_{s} &=\frac{12(1.23)}{\pi(172.25)} \&=0.0273 \ in.\end{aligned}[/latex]

From Table 12.3

Table 12.3 Concrete Properties
Compressive Stress (psi)	Allowable Compressive Stress (psi)	Modulus of Elasticity (psi)	n
[latex]f_{c}^{\prime}[/latex]	[latex]f_{c}=0.45 f_{c}^{\prime}[/latex]	[latex]E_{c}=57,000 \sqrt{f_{c}^{\prime}}[/latex]	[latex]E_{s}^{a} / E_{c}[/latex]
2500	1,125	2,850,000	11
3000	1,350	3,120,000	10
3500	1,575	3,370,000	9
4000	1,800	3,610,000	8
[latex]{ }^{a} E_{s}=30 imes 10^{6} psi[/latex].

[latex]f_{c}=1350 \ psi[/latex]

n = 10

and Eq. 12.5 gives K = 0.43.

[latex]k=\frac{1}{1+f_{s} / n f_{c}}[/latex] (12.5)

From Table 12.4 with K = 0.47,

Table 12.4
k	[latex]\gamma[/latex]	[latex]2 l_{1} / d[/latex]	[latex]2 l_{2} / d[/latex]	[latex]2 l_{3} / d[/latex]	[latex]K_{1}[/latex]	[latex]K_{2}[/latex]
0.01	78.52	0.98	1.489	0.016	3.113	0.267
0.02	73.74	0.96	1.477	0.032	3.085	0.378
0.03	70.05	0.94	1.465	0.048	3.059	0.463
0.04	66.93	0.92	1.452	0.064	3.033	0.535
0.05	64.16	0.90	1.439	0.080	3.008	0.599
0.06	61.64	0.88	1.426	0.096	2.983	0.657
0.08	57.14	0.84	1.400	0.128	2.935	0.760
0.10	53.13	0.80	1.373	0.160	2.887	0.852
0.15	44.43	0.70	1.304	0.239	2.772	1.049
0.20	36.87	0.60	1.233	0.318	2.661	1.218
0.25	30.00	0.50	1.161	0.397	2.551	1.370
0.30	23.58	0.40	1.087	0.475	2.442	1.509
0.35	17.46	0.30	1.013	0.553	2.333	1.640
0.40	11.54	0.20	0.938	0.631	2.224	1.765
0.45	5.74	0.10	0.862	0.709	2.113	1.884
0.50	0.00	0.00	0.785	0.785	2.000	2.000
0.55	-5.74	-0.10	0.709	0.862	1.884	2.113
0.60	-11.54	-0.20	0.631	0.938	1.765	2.224
0.65	-17.46	-0.30	0.553	1.013	1.640	2.333
0.70	-23.58	-0.40	0.475	1.087	1.509	2.442
0.75	-30.00	-0.50	0.397	1.161	1.370	2.551
0.80	-36.87	-0.60	0.318	1.233	1.218	2.661
0.85	-44.43	-0.70	0.239	1.304	1.049	2.772

[latex]\frac{2 l_{1}}{d}=0.140[/latex]

[latex]\frac{2 l_{2}}{d}=0.892[/latex]

[latex]\frac{2 l_{3}}{d}=0.678[/latex]

[latex]K_{1}=2.157[/latex]

[latex]K_{2}=1.836[/latex]

The magnitude of T is obtained from Eq. 12.10 as

[latex]T=\frac{M-W\left(l_{1}+l_{3} ight)}{l_{2}+l_{3}}[/latex] (12.10)

[latex]\begin{aligned}T &=\frac{1,500,000 imes 12-160,000(0.140+0.678)(172.25)}{(0.892+0.678)(172.25)} \&=49,750 \ lb\end{aligned}[/latex]

The value of [latex]f_{s}[/latex] is determined from Eq. 12.7 as

[latex]T=f_{s} t_{s}\left(\frac{d}{2} ight) K_{1}[/latex] (12.7)

[latex]\begin{aligned}f_{s} &=\frac{49,750}{(0.0273)(172.25 / 2)(2.157)} \&=9810 psi\end{aligned}[/latex]

From Eq. 12.11

C = T + W (12.11)

C = 49,750 + 160,000

= 209,750 lb

Equation 12.9 gives

[latex]C=\left(t_{s}+n t_{c} ight)\left(\frac{d}{2} ight) f_{c}\left[\frac{2\left(\cos \gamma+\gamma \sin \gamma-\frac{\pi}{2} \sin \gamma ight)}{1-\sin \gamma} ight][/latex] (12.9)

or

[latex]C=\left(t_{s}+n t_{c} ight)\left(\frac{d}{2} ight) f_{c} K_{2}[/latex]

[latex]\begin{aligned}f_{c} &=\frac{209,750}{[0.0273+(10)(7.375)](172.25 / 2)(1.836)} \&=18 \ psi\end{aligned}[/latex]

The calculated values of [latex]f_{s}^{1}[/latex] and [latex]f_{c}[/latex] result in a K value of

[latex]\begin{aligned}K &=\frac{1}{1+9810 /(10)(18)} \&=0.02\end{aligned}[/latex]

which is considerably lower than the assumed value of K = 0.43. Hence another trial is needed with a K value of 0.02. After recalculating values of T , [latex]f_{s}[/latex], C, and [latex]f_{c}[/latex], a new value of K is obtained and compared with the assumed one. If both values are approximately the same, the analysis is completed. If they are not, a new analysis is performed. Thus, in this example after a few trials, for K = 0.075, the following values were obtained:

[latex]\frac{2 l_{1}}{d}=0.85[/latex]

[latex]\frac{2 l_{2}}{d}=1.407[/latex]

[latex]\frac{2 l_{3}}{d}=0.120[/latex]

[latex]K_{1}=2.947[/latex]

[latex]K_{2}=0.734[/latex]

[latex]\begin{aligned}T_{1} &=\frac{15 imes 10^{5} imes 12-16 imes 10^{4}(0.85+0.120)(172.25)}{(1.407+0.120)(172.25)} \&=35,230\end{aligned}[/latex]

[latex]\begin{aligned}f_{s} &=\frac{35,230}{(0.0273)(172.25 / 2)(2.947)} \&=5085 \ psi\end{aligned}[/latex]

C=195,230

[latex]f_{c}=42 \ psi[/latex]

and

[latex]K=\frac{1}{1+(5085 / 10+42)}=0.076[/latex]

which is approximately the same as the assumed value. Hence [latex]f_{s}[/latex] = 5085 psi and [latex]f_{c}[/latex] =42 psi is the answer.

Question 12.2: In Example 12.1, it was found that 12-1 3/8 in. A 307 anchor...

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