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The simply supported beam shown spans 20 ft and, in addition to its own weight carries...
A rectangular beam having b=16 in and d=26 in spans 28 ft face to face of...
A rectangular beam having b=16 in and d=26 in spans 28 ft face
to face of simple supports. It is reinforced for flexure with 6#11
bars that continue uninterrupted to the ends of the span. It is to
carry service dead load 2.0 kips/ft (including self weight) and
service live load 3.6 kips/ft, both uniformly distributed along the
span. Design the shear reinforcement using vertical U stirrups.
Economize the spacing of stirrups in appropriate number of bands.
Material strengths are...
A rectangular beam having b-10 in. and d= 17.5 in. spans 15 ft face to face...
A rectangular beam having b-10 in. and d= 17.5 in. spans 15 ft face to face of simple supports. It is reinforced for flexure with three No. 9 (No. 29) bars that continue uninterrupted to the ends of the span. It is to carry service dead load D 1.27 kips/ft (including self-weight) and service live load L -3.70 kips/ft, both uniformly distributed along the span. Design the shear reinforcement, using No. 3 (No. 10) vertical U stirrups. The more approximate...
A 26.0 ft. simply supported beam is subjected to factored uniformly distributed load of wa equal ...
A 26.0 ft. simply supported beam is subjected to factored uniformly distributed load of wa equal to 6.0 kips/t. Beam properties are as follows: b-14 in, d-24 in, 4,000 psi (normal-weight concrete) and fy -60,000 psi. Design shear reinforcement for the entire beam (you can use symmetry if applicable). Use two different spacings over the beam length to come up with economical design. Perform all necessary checks and calculations. Draw final design for shear ma reinforcement indicating distribution (i.e, lengths,...
Figure 2 shows a simply supported beam and the cross section at mid span. The beam...
Figure 2 shows a simply supported beam and the cross section at mid span. The beam supports a uniform service (unfactored) dead load consisting of its own weight plus 1.4 kips/ft and a uniform service (unfactored) live load of 1.5 kips/ft. The concrete strength is 3500 psi, and the yield strength of the reinforcement is 60,000 psi. The concrete is normal-weight concrete. For the midspan section shown in Figure 2, compute фМп and show that it exceeds Mu. WD 1.4...
For the simply supported beam shown, the span is 40 ft, dead load including own weight...
For the simply supported beam shown, the span is 40 ft, dead load including own weight is 5 k/ft and live load is 4 k/ft. Assume fy-60,000 psi and fc 4000 psi, find: Size of the longitudinal reinforcement at the critical section (5 points) Spacing of U stirrups # 4 at the critical section (support width is 12 in.) (5 points) a. b. 48" 26.5" As 16
A rectangular reinforced concrete beam of span 16 ft supports a concentrated load of 40 kips...
A rectangular reinforced concrete beam of span 16 ft supports a concentrated load of 40 kips at mid-point of the beam and a uniformly-distributed load of 4 kips/ft over the entire span. Given: (a) The breadth of the beam is 18 inches. (b) Concrete compressive strength, fc, is 3500 psi. (c) Rebar is Grade 40 steel i) Determine the maximum moment and maximum shear acting on the beam. ii) Determine the minimum effective depth 'd' of the beam section (rounded...
A rectangular reinforced concrete beam of span 16 ft supports a concentrated load of 40 kips...
A rectangular reinforced concrete beam of span 16 ft supports a concentrated load of 40 kips at mid-point of the beam and a uniformly-distributed load of 4 kips/ft over the entire span. Given: (a) The breadth of the beam is 18 inches. (b) Concrete compressive strength, f 'c, is 3500 psi. (c) Rebar is Grade 40 steel i) Determine the maximum moment and maximum shear acting on the beam. ii) Determine the minimum effective depth 'd' of the beam section...
(40 pts) The simply supported beam shown in the figure below is being designed for shear. #3 stirrups (Grade 40) are to be used. 2. Wu = 4 kips/ft (includes self weight) wDL-1 kips/ft, wu...
(40 pts) The simply supported beam shown in the figure below is being designed for shear. #3 stirrups (Grade 40) are to be used. 2. Wu = 4 kips/ft (includes self weight) wDL-1 kips/ft, wu 1.75 kips/ft h= 18" d = 15" L=20ft- fe-4000 psi Yoon- 150 lbf/t E-29,000 ksi fy60 ksi 2 IS Draw the factored shear envelope for Vu c. d. Determine the resistance to shear provided by the concrete. Are there any regions where no shear stirrups...
WD= 1 k/ft WL 2.0 k/ft 17in 20 in 4 # 9 - 20 ft 12 in. 21 in. The simply-supported beam shown above is comprised of cla...
WD= 1 k/ft WL 2.0 k/ft 17in 20 in 4 # 9 - 20 ft 12 in. 21 in. The simply-supported beam shown above is comprised of class 5000 concrete and grade 60 rebar. Assume that the beam uses #3 stirrups (not shown) and has 1.5" of concrete cover (to the stirrups) on all sides (a) (20%) Determine the maximum instantaneous deflection for the unfactored (i.e. "service") dead load
WD= 1 k/ft WL 2.0 k/ft 17in 20 in 4 #...
The T-beam shown in Figure 1 supports the un-factored dead load of 1.4 kips/ft and live...
The T-beam shown in Figure 1 supports the un-factored dead load
of 1.4 kips/ft and live load of 1.5 kips/ft. The dead load does not
include the self-weight of the beam. The material properties are as
follows: fc’=3000 psi; fy=60,000 psi. Design the shear
reinforcement (stirrups). Plot the stirrups distribution along the
span of the beam.
DL= 1.4 kips/ft ; L2=1.5 kips/Ft * 75 Sz=7 X * b=3616. hr-6in k ) انا امه hw-lain + * bw=12 in
A rectangular beam having b=16 in and d=26 in spans 28 ft face
to face of simple supports. It is reinforced for flexure with 6#11
bars that continue uninterrupted to the ends of the span. It is to
carry service dead load 2.0 kips/ft (including self weight) and
service live load 3.6 kips/ft, both uniformly distributed along the
span. Design the shear reinforcement using vertical U stirrups.
Economize the spacing of stirrups in appropriate number of bands.
Material strengths are...
A rectangular beam having b-10 in. and d= 17.5 in. spans 15 ft face to face of simple supports. It is reinforced for flexure with three No. 9 (No. 29) bars that continue uninterrupted to the ends of the span. It is to carry service dead load D 1.27 kips/ft (including self-weight) and service live load L -3.70 kips/ft, both uniformly distributed along the span. Design the shear reinforcement, using No. 3 (No. 10) vertical U stirrups. The more approximate...
A 26.0 ft. simply supported beam is subjected to factored uniformly distributed load of wa equal to 6.0 kips/t. Beam properties are as follows: b-14 in, d-24 in, 4,000 psi (normal-weight concrete) and fy -60,000 psi. Design shear reinforcement for the entire beam (you can use symmetry if applicable). Use two different spacings over the beam length to come up with economical design. Perform all necessary checks and calculations. Draw final design for shear ma reinforcement indicating distribution (i.e, lengths,...
Figure 2 shows a simply supported beam and the cross section at mid span. The beam supports a uniform service (unfactored) dead load consisting of its own weight plus 1.4 kips/ft and a uniform service (unfactored) live load of 1.5 kips/ft. The concrete strength is 3500 psi, and the yield strength of the reinforcement is 60,000 psi. The concrete is normal-weight concrete. For the midspan section shown in Figure 2, compute фМп and show that it exceeds Mu. WD 1.4...
For the simply supported beam shown, the span is 40 ft, dead load including own weight is 5 k/ft and live load is 4 k/ft. Assume fy-60,000 psi and fc 4000 psi, find: Size of the longitudinal reinforcement at the critical section (5 points) Spacing of U stirrups # 4 at the critical section (support width is 12 in.) (5 points) a. b. 48" 26.5" As 16
A rectangular reinforced concrete beam of span 16 ft supports a concentrated load of 40 kips at mid-point of the beam and a uniformly-distributed load of 4 kips/ft over the entire span. Given: (a) The breadth of the beam is 18 inches. (b) Concrete compressive strength, fc, is 3500 psi. (c) Rebar is Grade 40 steel i) Determine the maximum moment and maximum shear acting on the beam. ii) Determine the minimum effective depth 'd' of the beam section (rounded...
(40 pts) The simply supported beam shown in the figure below is being designed for shear. #3 stirrups (Grade 40) are to be used. 2. Wu = 4 kips/ft (includes self weight) wDL-1 kips/ft, wu 1.75 kips/ft h= 18" d = 15" L=20ft- fe-4000 psi Yoon- 150 lbf/t E-29,000 ksi fy60 ksi 2 IS Draw the factored shear envelope for Vu c. d. Determine the resistance to shear provided by the concrete. Are there any regions where no shear stirrups...
WD= 1 k/ft WL 2.0 k/ft 17in 20 in 4 # 9 - 20 ft 12 in. 21 in. The simply-supported beam shown above is comprised of class 5000 concrete and grade 60 rebar. Assume that the beam uses #3 stirrups (not shown) and has 1.5" of concrete cover (to the stirrups) on all sides (a) (20%) Determine the maximum instantaneous deflection for the unfactored (i.e. "service") dead load
WD= 1 k/ft WL 2.0 k/ft 17in 20 in 4 #...
The T-beam shown in Figure 1 supports the un-factored dead load
of 1.4 kips/ft and live load of 1.5 kips/ft. The dead load does not
include the self-weight of the beam. The material properties are as
follows: fc’=3000 psi; fy=60,000 psi. Design the shear
reinforcement (stirrups). Plot the stirrups distribution along the
span of the beam.
DL= 1.4 kips/ft ; L2=1.5 kips/Ft * 75 Sz=7 X * b=3616. hr-6in k ) انا امه hw-lain + * bw=12 in
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