# Test1Ret07 - Closed book exam. No notes allowed. CE 3213...

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Closed book exam. No notes allowed. CE 3213 REINFORCED CONCRETE DESIGN RETEST – Fall 2007 STUDENT NAME I. What is the meaning of balanced condition in beam design ? (2 points) Balanced condition means that the concrete reaches crushing conditions ( ε c = 0.003) at the same time the tension steel reaches yielding ( ε y ) II. What minimum strength a mechanical anchor must have to be used as splice? (2 points) Mechanical anchors must develop at least 25% more strength than the yield value of the reinforcing steel being spliced (1.25 f y ). III. When would you specify a retarder admixture for concrete? (2 points) To obtain better workability and finishes and to delay setting during hot-weather concrete placing. IV. List three items of good hot weather concrete placing practices. - Place concrete at night or early in the morning - Avoid prolonged mixing - Use water for curing - Start curing process early - Cool concrete trucks by sprinkling water V. What is the main reason to place part of the negative reinforcement in the flange area of T-beams? Part of the negative reinforcement in T-beams is spread over the effective flange to control flexural cracking in the flange.

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1. For the beam shown determine the spacing of #4 stirrups. Sketch your solution. The loads given below do not include the beam weight ( assume b= 30 in., h = 20 in., and d = 17.5 in. ). f’ c = 3000 psi f y = 60,000 psi w L = 2 k/ft, w D = 1k/ft, P L = 30 kips a) Sketch the shear diagram and find the design shear (5points) b) Determine if stirrups are needed. If they are needed also Find the length stirrups are required (5points) c) Obtain S max , and S crit (5points) d) Select spacing of stirrups (minimum two different spacing values not counting the 2” or 3 “ spacing) (15 points) e) Sketch the shear provided superimpose over the shear diagram (5 points). Solution: Beam weight: 20(30) (0.150) 0.625k/ft 144 = a) u1 u2 W 1.4(1+0.625) 2.275 kips/ft 1.2(1+0.625)+1.6(2) 5.15 kips/ft 1.2(0) 1.6(30) 48 kips == =+ = u P W u = 5.15 kips/ft Assuming that both dead load and live load are present on the full span then b) Since the beam is loaded on the top and supported at the bottom, V u is critical at a distance d = 17.5 in. from the support: 5.85(17.5) 127 119.5 12 =− kips 40 ft w D + w L P L W u = 3.87 k/ft 5.15(40) 48 127 22 2 2 += u u P WL kips 127 k φ V s V c 43.1 k 118.5kips 17.5 in. z l = 240 in. ½ φ V c = 21.6 k 17.5 = at uu f a c e u VV V w d ud P u = 48 k 24 k
c) Check if stirrups are required: φ V c = 2 d b f w c ' ACI 11.3.1.1 = 0.75 [ 2 3,000 (30)(17.5) ] 1000 1 = 43.1 kips 2 1 V c = 2 1 (43.1) = 21.6 kips Since V u 118.5 kips > 21.6 kips, stirrups are required (ACI 11.5.5.1) Since the minimum shear is 24 kips > 21.6 kips stirrups are needed for the full length of the beam.

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## This note was uploaded on 02/28/2009 for the course CE 3213 taught by Professor Diaz during the Spring '09 term at The University of Texas at San Antonio- San Antonio.

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Test1Ret07 - Closed book exam. No notes allowed. CE 3213...

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