d) Calculate the first story column axial load capacity using 8-#11. Verify that the column size and reinforcement satisfy the design strength requirement: Pu<ØPn. Material Properties: fc'= 5 ksi, (concrete compressive strength) fy= 60 ksi (steel yield strength) e) Determine the number and size of minimum required vertical bars at third story. Note that reinforcing bar cages are typically fabricated two stories tall. That is, the first and second stories have the same reinforcing bars, and the third and fourth stories have the same reinforcing bars. Since the same size column is used at all stories, for construction economy, the vertical reinforcing bars may be reduced as the factored load demand reduces at the upper stories.

Structural Analysis
6th Edition
ISBN:9781337630931
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Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
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8.9 The following table shows the design dead and live loads for the lowest four stories of a 18" square reinforced concrete column.
a) Calculate the column factored loads. (Note: these loads are cumulative and included the loads from the levels above)
Table Column design loads
Story
4
3
2
1
Dead
(kip)
250
350
450
550
Reduced Live
(kip)
110
160
200
250
Factored load
(kip)
b) Assume 8-#11vertical bars are used at the first story. Draw column cross-section at 1"=1'-0" scale; show column
overall dimensions, transverse ties, and vertical reinforcing bars.
c) Verify that 8-#11 vertical bars meet the minimum and maximum steel area requirements for columns.
0.01 Ag<As < 0.08 Ag
Transcribed Image Text:8.9 The following table shows the design dead and live loads for the lowest four stories of a 18" square reinforced concrete column. a) Calculate the column factored loads. (Note: these loads are cumulative and included the loads from the levels above) Table Column design loads Story 4 3 2 1 Dead (kip) 250 350 450 550 Reduced Live (kip) 110 160 200 250 Factored load (kip) b) Assume 8-#11vertical bars are used at the first story. Draw column cross-section at 1"=1'-0" scale; show column overall dimensions, transverse ties, and vertical reinforcing bars. c) Verify that 8-#11 vertical bars meet the minimum and maximum steel area requirements for columns. 0.01 Ag<As < 0.08 Ag
d) Calculate the first story column axial load capacity using 8-#11. Verify that the column size and reinforcement satisfy
the design strength requirement: Pu<ØPn.
Material Properties:
fc'= 5 ksi, (concrete compressive strength)
fy= 60 ksi (steel yield strength)
e) Determine the number and size of minimum required vertical bars at third story. Note that reinforcing bar cages are
typically fabricated two stories tall. That is, the first and second stories have the same reinforcing bars, and the third
and fourth stories have the same reinforcing bars. Since the same size column is used at all stories, for construction
economy, the vertical reinforcing bars may be reduced as the factored load demand reduces at the upper stories.
Transcribed Image Text:d) Calculate the first story column axial load capacity using 8-#11. Verify that the column size and reinforcement satisfy the design strength requirement: Pu<ØPn. Material Properties: fc'= 5 ksi, (concrete compressive strength) fy= 60 ksi (steel yield strength) e) Determine the number and size of minimum required vertical bars at third story. Note that reinforcing bar cages are typically fabricated two stories tall. That is, the first and second stories have the same reinforcing bars, and the third and fourth stories have the same reinforcing bars. Since the same size column is used at all stories, for construction economy, the vertical reinforcing bars may be reduced as the factored load demand reduces at the upper stories.
Expert Solution
Step 1

Answer

(d) Calculation of axial load capacity of square column

We know that Axial load capacity is given by ϕ.Pn = ϕ.0.80*(0.85*f'c *(Ag-Ast)+fy*Ast)

ϕ= strength reduction factor

ϕ= 0.65 for tied column

Ag = gross section area

Ag = 18*18

Ag = 324 in2 

Ast = area of steel for 8-#11

Ast = 12.50 in2     (from areas of groups of standard bars)

f'c = compressive strength of concrete

f'c = 5 ksi

fy = yield strength of steel

fy = 60 ksi

Substitute the values to get " ϕ.Pn"

ϕ.Pn = ϕ.0.80*(0.85*f'c *(Ag-Ast)+fy*Ast)

ϕ.Pn = 0.65.0.80*(0.85*5 *(324-12.50)+60*12.50)

ϕ.Pn = 1078.415 in2 

" Axial load capacity of first storey column = 1078.415 in2 "

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