AXIALLY LOADED COLUMNS
(Short Columns)
1. Design a short axially loaded square column for a service dead load of 1080 KN and a
service live load of 990 KN. Use f’c = 34.5 MPa, fy = 414 MPa, ρ g= 2%, 25mm ∅
main bars, 10mm ∅ ties, and 40mm concrete cover.
Solution:
ρ g=
As
= 2%
Ag
As = 0.02Ag
Pu = 1.4 (1080) + 1.7 (990) = 3195 KN
Pu = 0.70 (0.80) [0.85 f’c (Ag – As) + As fy]
3195x103 = 0.70 (0.80) [0.85 (34.5) (Ag – 0.02Ag) + 0.02Ag (414)]
Ag = 154121.78 mm2
Ag = t 2 ; t = √ 154121.78 = 392.58mm
Using 25mm ∅ rebars;
π (25)2
(No. of bars) = As
4
0.02(154121.78)
No. of bars =
= 6.28
π (25)2
4
No. of bars = 8 pcs. 25mm ∅
For lateral ties: (use least value of s)
s = 16 x 25mm (main bar ∅ ¿ = 400mm
s = 48 x 10mm (tie∅) = 480mm
s = t = 392.58mm
s = 390 mm
2. A 450mm round spiral column is reinforced with 6-25mm ∅ bars having fy = 276
MPa. Determine the ultimate strength if f’c = 34 MPa.
Solution:
D = 450mm
As = 6 ¿
As = 2945.24 mm2
Ag = ¿
Ag = 159043.13 mm2
Pu = 0.75 (0.85) [0.85 f’c (Ag – As) + As fy]
Pu = 0.75 (0.85) [0.85 (34) (159043.13 – 2945.24) + 2945.24 (276)]
Pu = 3394.12 KN
3. A 500mm x 500mm tied column is reinforced with 8-28mm ∅ bars. If f’c = 21 MPa
and fy = 414 MPa, find the ultimate axial capacity of the column.
Solution:
As = 8
[
π ( 28 )2
4
]
As = 4926.02 mm2
Ag = 500mm x 500mm
Ag = 250000 mm2
Pu = 0.70 (0.80) [0.85 f’c (Ag – As) + As fy]
Pu = 0.70 (0.80) [0.85 (21) (250000 – 4926.02) + 4926.02 (414)]
Pu = 3591.81 KN
4. Design a spiral column to support an axial dead load of 800 KN and an axial live load
of 1350 KN. Assume that 2% longitudinal steel is desired. Diameter of main bars is
25mm, diameter of spiral ties is 10mm and 40mm concrete cover. Use f’c = 27.6 MPa
and fy = 414 MPa.
Solution:
As = 2% Ag
Pu = 1.4 (800) + 1.7 (1350) = 3415 KN
Pu = 0.75 (0.85) [0.85 f’c (Ag – As) + As fy]
3415x103 = 0.75 (0.85) [0.85 (27.6) (Ag – 0.02Ag) + 0.02 (414)]
Ag = 171305.59 mm2
As = 0.02 (171305.59) = 3426.11 mm2
Using 25mm ∅ main bars:
π ( 25 )2
3426.11 =
(No. of bars)
4
[
]
No. of bars = 6.98
No. of bars = 7 pcs. 25mm ∅
For spacing of spiral:
s=
4 Asties
Dc⍴ s
Ag =
π ( D )2
π ( D )2
; 171305.59 =
; D = 467.03mm
4
4
Dc = D – 2(40mm) = 467.03 – 2(40)
Dc = 387.03mm
π (Dc)2 π (387.03)2
Ac =
=
4
4
Ac = 117646.54mm2
2
Asties = π (10) = 78.54mm2
4
'
171305.59
27.6
⍴s = 0.45 Ag −1 f c = 0.45
−1
117646.54
414
Ac
fy
[
]( )
[
]( )
⍴s = 0.01368
4 (78.54)
s = (387.03)(0.01368)
s = 59.34 mm ≈ 55mm
check specs: 25mm≤ 55mm ≤ 80mm OK.
s = 55 mm
5. A square column is to support a dead load of 1200 KN and a live load of 1500 KN.
Using approximately 3% steel ratio, use f’c = 28 MPa and fy = 415 MPa. Calculate
the required spacing of 10mm diameter ties for a 350mm x 350mm column reinforced
with 4-25mm bars.
Solution:
s = 16 x 25 (main bar ∅ ¿ = 400mm
s = 48 x 10 (tie∅) = 480mm
s = t = 350mm
Use least value of s;
s = 350 mm
6. Determine the axial load capacity of a 350mm diameter spiral column with f’c of 28
MPa and reinforced with 6-25mm bars with fy = 415 MPa.
Solution:
As = 6 ¿ = 2945.24 mm2
Ag = ¿ = ¿ = 96211.28 mm2
Pu = 0.75 (0.85) [0.85 f’c (Ag – As) + As fy]
Pu = 0.75 (0.85) [0.85 (28) (96211.28 – 2945.24) + 2945.24 (415)]
Pu = 2194.28 KN
7. Calculate the minimum spiral reinforcement ratio with the same given of problem #6
and steel covering of 40mm.
Solution:
Dc = D – 2(40) = 350 – 2(40)
Dc = 270mm
π (Dc)2 π (270)2
Ac =
=
4
4
Ac = 57255.53 mm2
'
96211.28
28
⍴s = 0.45 Ag −1 f c = 0.45
−1
57255.53
415
Ac
fy
[
⍴s = 0.02066
]( )
[
]( )
8. A spiral column is to carry a dead load of 1000 KN and a live load of 1227.50 KN, f’c
= 27.50 MPa, fy = 414 MPa. Determine the smallest diameter of spiral column using
approximate steel ratio of 2%.
Solution:
As = 0.02Ag
Pu = 1.4 (1000) + 1.7 (1227.50) = 3486.75 KN
Pu = 0.75 (0.85) [0.85 f’c (Ag – As) + As fy]
3486.75x103 = 0.75 (0.85) [0.85 (27.50) (Ag – 0.02Ag) + 0.02Ag (414)]
Ag = 175371.92 mm2
Ag =
π (D)2
= 175371.92
4
D = 472.54 mm
D = 475 mm
9. A 410mm x 410mm tied column is reinforced with 6-28mm bars with f’c = 27.6 MPa
and fy = 414 MPa. Use steel cover of 75mm, calculate the axial load capacity.
Solution:
Ag = 300 x 500 = 168100 mm2
As = 6
[
π ( 28 )2
= 3694.51 mm2
4
]
Pu = 0.70 (0.80) [0.85 f’c (Ag – As) + As fy]
Pu = 0.70 (0.80) [0.85 (27.6) (168100 – 3694.51) + 3694.51 (414)]
Pu = 3016.43 KN
10. A square tied column supports a dead load of 400 kips and live load of 200 kips.
Design the column assuming a steel ratio of 4%, f’c = 5 ksi and fy = 60 ksi.
Solution:
Pu = 1.4 (400) + 1.7 (200) = 900 kips
As = 0.04Ag
Pu = 0.70 (0.80) [0.85 f’c (Ag – As) + As fy]
900 = 0.70 (0.80) [0.85 (5) (Ag – 0.04Ag) + 0.04 Ag (60)]
Ag = 248.02 ¿2
As = 0.04 (248.02)
As = 9.92 ¿2
WORKING STRESS DESIGN
(Design and Investigation of Singly-Reinforced Beams)
1. A rectangular reinforced concrete beam with width of 250mm and effective depth of
500mm is subjected to a balanced moment capacity of 150 KN-m. The beam is
reinforced with 4-25mm bars. Use n = 9. Determine the maximum stress of concrete
and steel.
Solution:
Mc = Ms
π ( 25 )2
As = 4
= 1963.495 mm2
4
[
]
As = ρ bd
1963.495 = ρ (250) (500) ; ρ=¿0.01571
2
k = √ 2 ρn+ ( ρn)2 – ρn= 2 ( 0.01571 )( 9 ) + [ ( 0.01571 )( 9 ) ] – ( 0.01571 ) ( 9 )
√
k = 0.40886
j=1
−0.408856
= 0.86371
3
Mc =
1
fc k j b d 2
2
150x106 =
1
fc (0.40886) (0.86371) (250) (500 ¿ ¿2
2
fc = 13.59 MPa
Ms = As fs (jd)
150x106 = 1963.495 fs (0.86371) (500)
fs = 176.89 MPa
2. A reinforced concrete beam having a width of 300mm and an effective depth of
520mm is reinforced for tension. Concrete strength fs = 128 MPa, f’c = 27 MPa, fc =
0.45f’c, n = 9. Solve for the balance moment capacity of the beam.
Solution:
9
fs
k = n+
= 9+ 128
= 0.46071
fc
0.45(27)
n
0.46071
= 0.84643
3
j=1–
Mc =
1
fc k j b d 2
2
Mc =
1
(12.15) (0.46071) (0.84643) (300)(520)2
2
Mc = Ms = 192.17 KN-m
3. A reinforced concrete cantilever beam 4m long has a cross-sectional dimension of
400mm by 750mm. It is to carry a superimposed load of 29.05 KN/m including its
own weight. The steel reinforcement has an effective depth of 685mm. Use f’c = 21
MPa, fs = 165 MPa, and n = 9. Determine the required number of 28mm ∅ reinforcing
bars using Working Stress Design Method.
Solution:
9
k = n+ fs = 9+ 165
= 0.34013
fc
0.45(21)
n
j=1–
0.34013
= 0.88662
3
Mc =
1
fc k j b d 2
2
Mc =
1
[0.45(21)] (0.34013) (0.88662) (400) (685 ¿ ¿2
2
Mc = 267439412.7 N-mm = Ms
Ms = As fs (jd)
267.44x106 = As (165) (0.88662) (685)
As = 2668.78
As = ¿ (No. of bars) = 2668.78
No. of bars = 4.33
No. of bars = 5 pcs. 28mm ∅
4. A beam with a width of 350mm and effective depth of 600mm, is reinforced with
1613 mm2 of steel. Assume n = 12, fc = 4.14 MPa, fs = 124 MPa. Find the moment
capacity of the section.
Solution:
As = ⍴bd
1613 = ⍴ (350) (600)
⍴ = 0.00768
2
k = √ 2 ρn+ ( ρn)2 – ρn = 2 ( 0.00768 )( 12 ) + [ ( 0.00768 ) ( 12 ) ] – ( 0.00768 )( 12 )
√
k = 0.34694
j=1–
Mc =
0.34694
= 0.88435
3
1
1
fc k j b d 2 = (4.14) (0.34694) (0.88435) (350) (600 ¿ ¿2
2
2
Mc = 80.02 KN-m
Ms = As fs (jd) = (1613) (124) (0.88435) (600)
Ms = 106.13 KN-m
5. A reinforced concrete beam 300mm wide has an effective depth of 600mm and a total
depth of 650mm. It is reinforced with 4-32mm diameter bars for tension, fc = 9.45
MPa, n = 8. Determine the stress in steel if the balanced moment capacity of the
section is 187 KN-m.
Solution:
As = 4 [
Mc =
π ( 32 )2
] = 3216.99 mm2
4
1
fc k j b d 2
2
187x106 =
1
(9.45) (kj) (300) (600)2
2
kj = 0.36645
j = 1 −¿
k
3
0.36645
k
= 1 −¿ ; k = 0.42732
k
3
j = 0.858
Ms = As fs (jd)
187x106 = (3216.99) (fs) (0.858) (600)
fs = 112.92 MPa
6. In a reinforced concrete beam, b = 300mm, d = 500mm, As = 1500 mm2, n = 8. The
beam is subjected to a bending moment of 70 KN-m, use working stress design
method. Calculate the stress in steel and stress in concrete.
Solution:
As = ρ bd
1500 = ρ (300) (500) ; ρ=¿0.01
2
k = √ 2 ρn+ ( ρn)2 – ρn= 2 ( 0.01 ) ( 8 ) + [ ( 0.01 )( 8 ) ] – ( 0.01 )( 8 )
√
k = 0.328
j = 1 −¿
0.328
= 0.891
3
Ms = As fs (jd)
70x106 = (1500) fs (0.891) (500)
fs = 104.75 MPa
Mc =
1
fc k j b d 2
2
70x106 =
1
fc (0.328) (0.891) (300) (500)2
2
fc = 6.39 MPa
7. Design a reinforced concrete beam that will produce a balanced moment of 140 KNm. Assuming d = 1.5b, fc = 12 MPa, fs =160 MPa and n = 8. Determine also the
width of beam.
Solution:
n
8
fs =
160
k=
n+
8+
fc
12
k = 0.375
j = 1 −¿
Mc =
0.375
= 0.875
3
1
fc k j b d 2
2
140x106 =
1
(12) (0.375) (0.875) (b) (1.5 b)2
2
b = 316.17 mm
Use, b = 320 mm
Ms = As fs (jd)
140x106 = As (160) (0.875) [(1.5) (316.17)]
As = 2108.57 mm2
8. Calculate the number of 16mm ∅rebars to be used for a reinforced concrete beam with
a balanced condition. Use b = 280mm, d = 450mm, fs = 130 MPa, fc = 5 MPa and n =
9.
Solution:
n
9
fs =
130
k=
n+
9+
fc
5
k = 0.257
j = 1 −¿
0.257
= 0.914
3
Mc =
1
fc k j b d 2
2
Mc =
1
(5) (0.257) (0.914) (280) (450 ¿ ¿ 2
2
Mc = 33296791.5 N-mm
33296791.5 = As (130) (0.914) (450)
As = 622.73 mm2
As = (No. of bars) ¿ = 622.73
No. of bars = 3.097
No. of bars = 4 pcs. 16mm ∅
9. A rectangular beam is 300mm wide and have 550mm effective depth. It is reinforced
with 4 bars of 14mm diameter. Determine the moment of resistance using working
stress method of analysis. Use fc = 10 MPa, fs = 135 MPa and n = 12.
Solution:
As = 4 ¿ = 615.75 mm2
As = ρ bd
615.75 = ρ (300) (550) ; ρ= 0.0037
2
k = √ 2 ρn+ ( ρn)2 – ρn= 2 ( 0.0037 ) ( 12 )+ [ ( 0.0037 ) (12 ) ] – ( 0.0037 )( 12 )
√
k = 0.257
j = 1 −¿
0.257
= 0.914
3
Mc =
1
fc k j b d 2
2
Mc =
1
(10) (0.257) (0.914) (300) (550 ¿ ¿2
2
Mc = 106.58 KN-m
Ms = As fs (jd)
Ms = (615.75) (135) (0.914) (550)
Ms = 41.79 KN-m
10. Determine the area of steel and number of 28mm ∅bars required for a rectangular
beam having a width of 400mm, d = 700mm, fc = 12 MPa, fs = 160 MPa, n = 10. The
beam is to produce a moment in balance condition.
Solution:
n
10
fs =
160
k=
n+
10+
fc
12
k = 0.429
j = 1 −¿
Mc =
0.429
= 0.857
3
1
1
fc k j b d 2 = (12) (0.429) (0.857) (400) (700)2
2
2
Mc = 432.36KN-m
Ms = As fs (jd)
432.36x106 = As (160) (0.857) (700)
As = 4504.5 mm2
As = (No. of bars) ¿ = 4504.5
No. of bars = 7.32
No. of bars = 8 pcs. 28mm ∅