Homework SourseTime 2 hour 30 mins Name J 35 Problem 1 A rectangular concre
Time 2 hour 30 mins. Name: J#: 35 Problem 1: A rectangular concrete beam of width b 16 in. is limited by architectural considerations to a maximum total depth h 22 in. It must carry a total factored load moment Mu = 450 ft-kips. Design the flexural reinforcement for this member. Material strengths are fy-60,000 psi and fc = 4500 psi. Select reinforcement to provide the needed areas, and show a sketch of your final design. 35 Problem 2: A beam of 12 in. width and effective depth of 20 in. carries a factored uniformly distributed load of 5.5 kips/ft, including its own weight, in addition to a central, concentrated factored load of 15 kips. It is simply support with a span of 22 ft. It is reinforced with four No. 9 bars for both positive and negative bending. Material strengths are fy-60,000 psi and fc-4000 psi. Design the web reinforcement. 30 Problem 3: The continuous beam shown in Fig. 1 has been designed to carry a service dead load of 2.25 kips/ft including self-weight and service live load of 3.25 kips/ft. Flexural design has been based on ACI moment coeffici of support and midspan, respectively, resulting in a concrete section with b 14 in. and d - 22 in. Negative reinforcement at the support face is provided by four No. 10 which will be cut off in pairs where no longer required by the ACI Code. Positive bars consist of four No. 8 bars, which will also be cut off in pairs. Specify the exact point of ents of 1/11 and 1/16 at the face cutoff for all negative and positive steel. Check for satisfaction of ACI Code appropriate. Material strengths are fy 60,000 psi and fe 4000 psi. requirements at the point of inflection, and suggest modifications of reinforcement if 41O (No. 32) 41O (No. 32) 0 Figure 1 Longitudinal Reinforcement Details 
Solution
(1)Solution:-
Given
Moment (Mu) = 450 kip feet
b = 16 inch
h = 22 inch
Let cover = 2 inch
Effective depth (d ) = 20 inch
Fy = 60000 psi
Fck = 4500 psi
We know that
Mu = 0.87 fy Ast ( d - fy Ast/( fckb))
450 *103 *12 = 0.87 * 60000 * Ast (20 – 60000*Ast/(4500 * 16))
Ast = 7.54 inch2
Let # 9 rebars are used . Diameter of bars are 1.128 inch
Total number of bars = Ast/(/4 * (1.128)2 )
Total number of bars = 7.54/(/4 * (1.128)2)
Total number of bars = 8 Answer
Provide total 8 number of bars @ 1.128 inch diameter
