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36 Analysis of Rigid If the axial thrust is disregarded, the following steel area is required in tension: 180,400 X 12 18,000 X Vs X 46 " 2 " 8q- m- Carry the 1-in. square bars spaced 12 in. from the top of the deck around the corner and add in each interval between these bars one 1^-in. square bar, thus making the total area of tensile reinforcement 2.27 sq. in. per lin. ft. at the corner. No compressive reinforcement will be included in the following stress determination. The depth to the neutral axis in the concrete section equals d (\/2pn + (pri)2 — pri) = 11.5 in. when d = 46 in., A = 2.27 sq. in., n = 10 and E is infinity. Adding the axial thrust will tend to increase the effective depth to, say, 14 in. The section coefficients with the esti mated value of Z = 14 will be: Estimated Section Coefficients Correction Corrected Value A = 12X14+10X2.27 = 190.7 +2.40 193.1 Q = HX12X142 + 22.7 X 46 = 2,220 +2.40X14.10 2,254 I- = HX12X143 + 22.7 X 462 = 59,010 -I-2.40X14.102 59,490 E = 80.5 - 0.5 X 48 = 56.5 Second value of Z is 59,010 + 56.5 X 2,220 _ 184,440 _ 2,220 + 56.5 X 190.7 ~ 12,990 ~ m' The discrepancy in Z is 0.20 in., and the corresponding corrections are made in A, Q and /. Using the corrected values of A, Q and /, determine the final value of Z. 59,490 + 56.5 X 2,254 186,840 / = -ir^-. — , ^„ , = .„ = 14.20 2,254 + 56.5 X 193.1 Now compute 2,254 13,160 9 193 - = 11.67, e = 56.5 + 11.67 = 68.2, Ig = 59,490 - 2,2542 193.1 = 33,180, 26,880 X 68.2 " 33.180 X 14'20 " 785' , W x »>%?n"a X 31.80 - 17,570 Generated on 2015-11-26 09:11 GMT / http://hdl.handle.net/2027/coo.31924003881277 Public Domain, Google-digitized / http://www.hathitrust.org/access_use#pd-google

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36 Analysis of Rigid

If the axial thrust is disregarded, the following steel area is requiredin tension:

180,400 X 12

18,000 X Vs X 46" 2 " 8q- m-

Carry the 1-in. square bars spaced 12 in. from the top of the deckaround the corner and add in each interval between these bars one 1^-in.square bar, thus making the total area of tensile reinforcement 2.27 sq. in.per lin. ft. at the corner. No compressive reinforcement will be includedin the following stress determination.

The depth to the neutral axis in the concrete section equals

d (\/2pn + (pri)2 — pri) = 11.5 in. when d = 46 in., A = 2.27 sq. in.,n = 10 and E is infinity. Adding the axial thrust will tend to increasethe effective depth to, say, 14 in. The section coefficients with the estimated value of Z = 14 will be:

Estimated Section Coefficients CorrectionCorrected

Value

A = 12X14+10X2.27 = 190.7 +2.40 193.1

Q = HX12X142 + 22.7 X 46 = 2,220 +2.40X14.10 2,254

I-= HX12X143 + 22.7 X 462 = 59,010 -I-2.40X14.102 59,490

E = 80.5 - 0.5 X 48 = 56.5

Second value of Z is

59,010 + 56.5 X 2,220 _ 184,440 _2,220 + 56.5 X 190.7

~12,990

~ m'

The discrepancy in Z is 0.20 in., and the corresponding correctionsare made in A, Q and /. Using the corrected values of A, Q and /,determine the final value of Z.

59,490 + 56.5 X 2,254 186,840/ = -ir^-.—, ^„ , = .„ = 14.202,254 + 56.5 X 193.1

Now compute

2,254

13,160

9193

- = 11.67, e = 56.5 + 11.67 = 68.2,

Ig = 59,490 - 2,2542

193.1= 33,180,

26,880 X 68.2"33.180

X 14'20 " 785'

,W x »>%?n"a X 31.80 - 17,570

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