Test1Spr07 - Closed book exam No notes allowed CE 3213...

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Closed book exam. No notes allowed. CE 3213 REINFORCED CONCRETE DESIGN EXAM I – SPRING 2007 STUDENT NAME I. What is the purpose of the value “a” when designing reinforced concrete beams? The purpose of the “a: in the design of reinforced concrete beams is to replace the existing stress distribution with and equivalent rectangular stress distribution. II. In the design of a reinforced concrete beam you decided to use the minimum amount of reinforcement. Is the resulting cross-section the smallest or the largest cross-section.? (2 points) The largest cross-section III. In reinforced concrete what is an admixture? (2 points) It is a substance other than the basic components of water, aggregate and hydraulic cement that is added to the concrete mix to modify its properties. IV. A steel manufacturer has delivered two packages of #6 bars. One package is A706 (low alloy bars) and the second package is A615 (billet steel bars). Which package will you use if the bars are going to be welded? (2 points) For welding purposes A706 (low alloy bars) should be used. V. Explain the term creep deformation in reinforced concrete (2 points) It is the shortening that occurs after the initial deformation when the concrete is subjected to a sustained load.
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1. For the beam shown, determine the nominal and design moments. Assume f’ c = 5500 psi , f y = 60,000 psi and #4 stirrups. (20 points) Solution: 1: Define , ' , , , , , y c s t f f A d d b and s E 16 = b in, 500 , 5 ' = c f psi , 000 , 60 = y f psi , 6 10 29 × = s E psi 00 . 6 9 # 6 = = s A 2 in For this case, d t = h – cover - diameter of stirrup - ½ diameter of bar 44 . 25 ) 128 . 1 ( 5 . 0 8 4 5 . 1 28 = = t d in and, d = h –cover – diameter of stirrup - diameter of main bar – 0.5 . 37 . 24 5 . 0 ) 128 . 1 ( 8 4 5 . 1 28 in d = = You could also find d as h - d s , where d s is the centroid of the main steel. In this case () . 628 . 3 00 . 3 00 . 3 )) 128 . 1 )( 2 / 1 ( 1 128 . 1 8 / 4 5 . 1 ( 00 . 3 ) 128 . 1 )( 2 / 1 ( 8 / 4 5 . 1 00 . 3 in d s = + + + + + + + + = d = 28 -3.628 = 24.37 in. 3: Check min s A d b f d b f f A w y w y c s 200 3 ' min = 2 ' 45 . 1 ) 37 . 24 )( 16 ( 000 , 60 500 , 5 3 3 in d b f f w y c = = , 2 30 . 1 ) 37 . 24 )( 16 ( 000 , 60 200 200 in d b f w y = = 2 min 45 . 1 in A s = min 2 00 . 6 s s A in A > = O.K.
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Test1Spr07 - Closed book exam No notes allowed CE 3213...

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