Design of One-way Slab using Trial-and-Error Method: Problem 2) Design a solid one-way slab for flexure using trial-and-error method assuming the following properties and conditions: Span, L= 16 ft (simply supported) Live load = 120 psf Dead load = 40 psf + self weight of the slab. (use density of RC = 150 pcf) fy 60 ksi (Gr. 60 steel) Pc = 4 ksi (Concrete) No. 4 bars for temperature and shrinkage, spacing = ? ANo. 4 bars at 8" on-center (a) Estimate the minimum thickness for the slab if deflections are not computed (i.e. thickness rules). (Table 4.1]. Select a thickness, h (round to ½" increment). hmin =. in, h = in (b) Design the slab using minimum thickness from part (a) and determine the total factored load per linear foot (Ibs/ft) and also the maximum factored moment (use 150 Ibs/ft³ for the density of the slab, and design using a 12"-strip). Uniform load due to self-weight of slab, DLslab =, Ib/ft Factored uniform load, Wy = Ib/ft, M, =, ft*kip (c) Determine the minimum cover (cs ) for the slab steel if the concrete is not exposed to weather or ground contact. Then calculate the depth ( d ) to the primary steel assuming #4 bars will be used. Hint, d = h-cs- . Minimum cover, CS = in., d = in.
Design of One-way Slab using Trial-and-Error Method: Problem 2) Design a solid one-way slab for flexure using trial-and-error method assuming the following properties and conditions: Span, L= 16 ft (simply supported) Live load = 120 psf Dead load = 40 psf + self weight of the slab. (use density of RC = 150 pcf) fy 60 ksi (Gr. 60 steel) Pc = 4 ksi (Concrete) No. 4 bars for temperature and shrinkage, spacing = ? ANo. 4 bars at 8" on-center (a) Estimate the minimum thickness for the slab if deflections are not computed (i.e. thickness rules). (Table 4.1]. Select a thickness, h (round to ½" increment). hmin =. in, h = in (b) Design the slab using minimum thickness from part (a) and determine the total factored load per linear foot (Ibs/ft) and also the maximum factored moment (use 150 Ibs/ft³ for the density of the slab, and design using a 12"-strip). Uniform load due to self-weight of slab, DLslab =, Ib/ft Factored uniform load, Wy = Ib/ft, M, =, ft*kip (c) Determine the minimum cover (cs ) for the slab steel if the concrete is not exposed to weather or ground contact. Then calculate the depth ( d ) to the primary steel assuming #4 bars will be used. Hint, d = h-cs- . Minimum cover, CS = in., d = in.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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please answer a,b,c,d,e
![Design of One-way Slab using Trial-and-Error Method:
Problem 2) Design a solid one-way slab for flexure using trial-and-error method assuming the following
properties and conditions:
Span, L = 16 ft (simply supported)
Live load = 120 psf
Dead load = 40 psf + self weight of the slab. (use density of RC = 150 pcf)
fy 60 ksi (Gr. 60 steel)
f'c = 4 ksi (Concrete)
No. 4 bars for temperature and shrinkage,
spacing = ?
No. 4 bars at 8" on-center
L
(a) Estimate the minimum thickness for the slab if deflections are not computed (i.e. thickness rules).
[Table 4.1]. Select a thickness, h (round to ½" increment). hmin =_
in, h =
in
(b) Design the slab using minimum thickness from part (a) and determine the total factored load per linear
foot (Ibs/ft) and also the maximum factored moment (use 150 Ibs/ft³ for the density of the slab, and
design using a 12"-strip).
Uniform load due to self-weight of slab, DLslab =
Ib/ft
Factored uniform load, Wy =
Ib/ft, My =
ft*kip
(c) Determine the minimum cover (cs ) for the slab steel if the concrete is not exposed to weather or
ground contact. Then calculate the depth ( d ) to the primary steel assuming #4 bars will be used. Hint,
d = h-cs-. Minimum cover, CS =
in., d =
in.
(d) Determine the flexural steel required ( A, reg) per foot using trial-and-error approach. Let your initial
req
trial moment arm (d -4) equal to 0.9d. Assume reduction factor ( ø ) equal to 0.9. 'a' converges at
about 2nd or 3rd iteration. Asreq =
in.?
(e) Determine the maximum spacing for the slab including the flexural steel requirement from part (d).
3h =
in.
18in
max
(12in) =.
in.
A reg](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2867fa33-d354-46d8-b33d-713eb0a8e48f%2F0602a783-897f-4390-98e1-5bcd88793ee4%2F9vq1scf_processed.png&w=3840&q=75)
Transcribed Image Text:Design of One-way Slab using Trial-and-Error Method:
Problem 2) Design a solid one-way slab for flexure using trial-and-error method assuming the following
properties and conditions:
Span, L = 16 ft (simply supported)
Live load = 120 psf
Dead load = 40 psf + self weight of the slab. (use density of RC = 150 pcf)
fy 60 ksi (Gr. 60 steel)
f'c = 4 ksi (Concrete)
No. 4 bars for temperature and shrinkage,
spacing = ?
No. 4 bars at 8" on-center
L
(a) Estimate the minimum thickness for the slab if deflections are not computed (i.e. thickness rules).
[Table 4.1]. Select a thickness, h (round to ½" increment). hmin =_
in, h =
in
(b) Design the slab using minimum thickness from part (a) and determine the total factored load per linear
foot (Ibs/ft) and also the maximum factored moment (use 150 Ibs/ft³ for the density of the slab, and
design using a 12"-strip).
Uniform load due to self-weight of slab, DLslab =
Ib/ft
Factored uniform load, Wy =
Ib/ft, My =
ft*kip
(c) Determine the minimum cover (cs ) for the slab steel if the concrete is not exposed to weather or
ground contact. Then calculate the depth ( d ) to the primary steel assuming #4 bars will be used. Hint,
d = h-cs-. Minimum cover, CS =
in., d =
in.
(d) Determine the flexural steel required ( A, reg) per foot using trial-and-error approach. Let your initial
req
trial moment arm (d -4) equal to 0.9d. Assume reduction factor ( ø ) equal to 0.9. 'a' converges at
about 2nd or 3rd iteration. Asreq =
in.?
(e) Determine the maximum spacing for the slab including the flexural steel requirement from part (d).
3h =
in.
18in
max
(12in) =.
in.
A reg
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