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34 Analysis of Rigid which is equal to the value of Z determined by Step II. It is also advisable to make a check by determining the algebraic sum of all the stresses in the section — a sum which is equal to the axial load, P. Problem 9 — Stresses at Crown and Corner The frame in Problem 1 is subject to dead load, live load and change in length of deck. A summary of these loads and the moments and thrusts they create is given in Fig. 11 on page 23. Choose tensile and compressive reinforcement and ascertain that the corresponding unit stresses do not exceed the allowable working stresses, which will be chosen conservatively as /„ = 18,000 and /„ = 1,000 p. s. i. Moments and thrusts at the midpoint of the deck are Axial Moment Thrust Dead Load +24,500 +6,740 Live Load +19,300 +3,120 Change in Deck Length +13,400 -690 Total -^7,200 +9,170 Eccentricity with respect to centerline: 57,200/9,170 = 6.24 ft., say, 75.0 inches. The tensile steel area for this moment and thrust must be somewhat less than that required when the axial thrust is disregarded; namely 57,200 X 12 18,000 X% X 19 " 2 29 Sq- m- A tensile reinforcement of 1-in. square bars spaced 6 in. (A' = 2.00 sq. in.) will be chosen. With this reinforcement — and axial thrust still disregarded — the extreme fiber stress in concrete is less than 900 p. s. i. This stress will be raised by the addition of axial thrust, and compressive reinforcement equal to 1-in. square bars spaced 12 in. will be chosen. The depth to the neutral axis in the concrete section equals d ( 2pn + (jm)2 ~ pn) = 6.5 in. when d = 19 in., A' = 2.00 sq. in., n = 10, and E is infinity. Adding the axial thrust will tend to increase the effective depth to, say, 7 in. The section coefficients with the estimated value of Z = 7 will be: Corrected Value Estimated Section Coefficients Correction A = 12X7+(10-1)X1.00 + 10X2.00 = 113 -3.84 109.2 Q= 3^X12 X 72+ 9 X 2 + 20X19 = 692 -3.84X6.84 666 / = HX12X73+9X22+20X192 = 8630 -3.84X6.842 8,450 E = 75.0 - 0.5 X 21 = 64.5 Generated on 2015-11-26 09:10 GMT / http://hdl.handle.net/2027/coo.31924003881277 Public Domain, Google-digitized / http://www.hathitrust.org/access_use#pd-google

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34 Analysis of Rigid

which is equal to the value of Z determined by Step II. It is also advisableto make a check by determining the algebraic sum of all the stresses in

the section —a sum which is equal to the axial load, P.

Problem 9—Stresses at Crown and Corner

The frame in Problem 1 is subject to dead load, live load and changein length of deck. A summary of these loads and the moments andthrusts they create is given in Fig. 11 on page 23. Choose tensile and

compressive reinforcement and ascertain that the corresponding unitstresses do not exceed the allowable working stresses, which will be chosen

conservatively as /„ = 18,000 and /„ = 1,000 p. s. i.Moments and thrusts at the midpoint of the deck are

AxialMoment Thrust

Dead Load +24,500 +6,740

Live Load +19,300 +3,120

Change in Deck Length +13,400 -690Total -^7,200 +9,170

Eccentricity with respect to centerline: 57,200/9,170 = 6.24 ft.,say, 75.0 inches.

The tensile steel area for this moment and thrust must be somewhatless than that required when the axial thrust is disregarded; namely

57,200 X 12

18,000 X % X 19" 2 29 Sq- m-

A tensile reinforcement of 1-in. square bars spaced 6 in. (A' = 2.00

sq. in.) will be chosen. With this reinforcement—and axial thrust stilldisregarded — the extreme fiber stress in concrete is less than 900 p. s. i.This stress will be raised by the addition of axial thrust, and compressivereinforcement equal to 1-in. square bars spaced 12 in. will be chosen.

The depth to the neutral axis in the concrete section equals

d ( 2pn + (jm)2 ~ pn) = 6.5 in.when d = 19 in., A' = 2.00 sq. in., n = 10, and E is infinity. Adding theaxial thrust will tend to increase the effective depth to, say, 7 in. Thesection coefficients with the estimated value of Z = 7 will be:

CorrectedValueEstimated Section Coefficients Correction

A = 12X7+(10-1)X1.00 + 10X2.00 = 113 -3.84 109.2

Q= 3^X12 X 72+ 9 X 2 + 20X19 = 692 -3.84X6.84 666

/ = HX12X73+9X22+20X192 = 8630 -3.84X6.842 8,450

E = 75.0 - 0.5 X 21 = 64.5

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