PROBLEM 34. CE Board May 2009
A simply reinforced concrete beam reinforced for tension has a width of 300 mm and a total depth of 600 mm. It is subjected to an external moment Mu = 540 kN-m, f c’ = 28 MPa, f y = 280 MPa, E s = 200 GPa.
540 10 = 0.90 0.90300 30028 280.21 0.21 = 583.21
① Which of the following gives the balance steel
ration in percent.
② Which of the following gives the depth “a” in terms of “d” in percent using ρ = ½ ρ b.
③ Which of the following gives the minimum
effective depth.
Solution:
① Balanced steel ratio: = . + . . = . + = 0.04926% = .% Depth “a” in terms of d in percent:
= ∅ 10.59 10.59 = = 0.04926 0.04926 = 0.0246 = = . = 0.246 10.59 10.59 = 0.246 0.24610.59∗0.246 10.59∗0.246 = ∅ 10.59 10.59
= /2 /2 = 0.85 /2 /2 540 10 = 0.85 0.8528 28300 300583.21/ 2 75630.25 = 583.210.5 583.210.5 0.5 583.2175630.25 = 0 1166.42 1166.42 151260 151260.50 .50 = 0 = .±. = 148.61 . = . = 0.2548 = .% Min. effective depth:
= . . PROBLEM 35. CE Board Nov. 2009
A rectangular beam has a width of 300 mm and an effective depth of 460 mm. The beam is reinforced with 2-28 mm at the top.
∅
= .%
fc’ = 35 MPa, fy = 350 MPa.
① Compute the ratio of the depth of compression
block to the distance of the top fiber fiber to the neutral axis.
② Compute
the balanced steel ratio of the
reinforcement.
③ Compute the max. area of steel permitted.
Max. area of steel permitted:
= = 0.75 460 = 0.75 0.750.0429 0.0429300 300 460 = . .
Solution:
① Ratio of the depth of compression block to the distance of the top fiber to the neutral axis”
PROBLEM 36:
A rectangular beam having a width of 300 mm and an effective depth of 450 mm. It is reinforced with 4-36 mm in diameter bars. fc’ = 28 MPa, MPa.
fy = 270 MPa,
Es = 200000
① Compute the depth of compression block for
a balanced condition.
② Compute the nominal moment capacity of the
beam.
③ a = depth of compression block
If the value of fc’ is increased by 25%, compute the percentage of the increased nominal moment capacity of the beam.
c = distance of top fiber to neutral axis
= =
Solution:
① Depth of compression block for a balanced
= 0.85 85 . −
condition:
= 0.85 85 .− = .. Balanced steel ratio of the reinforcement:
= . +
. . = . +
. = . −
1.350.003 = 0.00135 0.00435 = 1.35 = 310.34 =
③ Percentage of the increased nominal moment capacity of the beam if fc’ is increased by 25%:
= 36 364 = 1296 =
Note:
= 0.85 = 28 = 0.85 0.85310.34 310.34 = .
= 0.85 = = 1.28 1.2828 28 = 35
Nominal moment moment capacity: capacity:
0.85 = 300 = 1296270 0.85 0.8535 35300 = 123.17 =
= 0.85 = 300 = 36 270 0.85 0.8528 28300 364270 = 153.96 = 270 450 . = 36 364270 = 410 10 =
= 1296270450 . = 427 10 = 427 427 Percentage increase in nominal moment:
% = − 100 % = .% PROBLEM 37:
A reinforced concrete beam has a width of 400 mm and an effective depth of 600 mm. It is reinforced for tension with 4-28 mm bars. fc’ fc’ = 20.7 MPa, fy = 414.6 MPa.
∅
① Determine the percent increase in nominal moment if the depth is increased to 700 mm.
② Determine the percent increase in nominal
= 0.00207 >
moment if fc’ is increased to 27.6 MPa.
③ Determine the percent increase in nominal moment if the steel is changed to 4-32 mm ∅. Solution:
① Percent increase in nominal moment if the depth
is
increased
to
700
mm:
Nominal moment moment if d = 600 mm: mm:
= ∅ ∅ 414.6 600 . = 0.90 28 284414.6 = 48475 48475595 59599 = 484 48475 7566 Nominal moment moment if d = 700 mm: mm:
= ∅ ∅
= 0.85 = 400 = 28 414.6 0.85 0.8520.7 20.7400 284414.6 = 145.09 = 145.09 = 0.85 = 170.69 . . = .
= 0.00754 = . = ,
414.6 700 . = 0.90 28 284414.6 = 57666 57666066 06633 = 576 57666 6611 Percent increase in nominal moment:
% = − 100 % = .% Percent increase in nominal moment if fc’ is increased to 27.6 MPa:
= 0.85 = 400 = 28 414.6 0.85 0.8527.6 27.6400 284414.6 = 108.82 = ∅ ∅
= 0.90 284414.6600 . = 501423048.1 = 501423
②
Determine the moment capacity using moment reduction factor of 0.90.
③ Determine
the super-imposed uniform live load it could carry in kPa besides a dead load of 20 kN/m including its own weight if it has a simple span of 6 m and a spacing of 1.8m.
Percentage increase in nominal moment:
% = − 100 % = . % ③ Percent increase in nominal moment if the steel is change to 4-32 mm ∅ : = 0.85 = 0.8520.7400 = 324414.6 = 189.51
Solution:
① Depth of compression block:
= 323 = 2413
= ∅ = 0.90 324414.6600 . = 606490
= =
= 0.014
% = − 100 % = . % PROBLEM 38:
A reinforced concrete beam has a width of 300 mm with an effective depth of 575 mm. It is reinforced with 3-32 mm at the bottom. fc’ = 27.6 MPa, fy = 414 MPa. Balanced steel ratio ρ b = 0.0285.
∅
①Determine the depth of compression block.
= 0.75 = 0.750.02850 = 0.021 > 0.014 : = = 0.85 =
0.8527.6300 = 2413414 = .
= 47.98
② Moment
= . . = .
capacity using moment reduction factor of 0.90:
= 141.44 = 0.85 = 166.99 ∅ = 0.90: = 0.003 408.01 166.99 = 0.0073 > 0.005 ∅ = 0.90 Check whether we could use
= ∅ = 0.90 2413414575 . = 453.4 Super-imposed uniform live load it could carry in kPa besides a dead load of 20 kN/m including its own weight if it has a simple span of 6 m and a spacing of 1.8m:
= 453.4 = = 100.76
= 1.61.2 100.76 = 1.6 1.220
PROBLEM 39:
A fixed ended rectangular beam must support a uniform service dead and live loads of 220 kN/m and 182.6 kN/m respectively. It has a span of 6m. fc’ = 27.6 MPa, fy = 414.7 MPa, ρ b = 0.028.
① Effective depth of the beam using ω = 0.18. ② Compute the flexural reinforcements at the support.
③ Compute for the flexural reinforcements at the midspan.
Solution:
① Eff ective depth of the beam using ω = 0.18:
= 1.21.6 = 1.2220 1.6182.6 = 556.2 / Max. moment occurs at the fixed supports:
=
= .
= 1668.60
<
Moment at midspan:
=
= . = 834.3 Effective depth of beam:
= ∅ 10.59 1668.6 10 = 0.9027.6600 0.18[1 0.590.18] = 834.2 850 = Flexural reinforcements at the supports:
= 0.18 = = .. . = 0.01198 = . = 0.00338
= 0.75 = 0.750.028 = 0.01198 = 0.021
= = 0.01198600850 = Flexural reinforcements at the midspan:
= 834.3 = ∅ 10.59 843.3 10 = 0.9027.6600850[10.59] [10.59] = 0.0774 1.6949 0.1313 = 0 = 0.0814 =
= .. . = 0.00541 = . = 0.00338
< 0.00541 < 0.021 = = 0.00541600850 =
PROBLEM 40:
A rectangular concrete beam has a width of 250 mm and a total depth of 450 mm. It is reinforced with a total steel area of 1875 mm 2 placed at an effective depth of 375 mm. fc’ = 27.6 MPa, fy = 414.7 MPa.
① Determine the depth of compression block. ② Determine the moment capacity reduction factor.
③ Determine the safe live load that the beam
could carry in addition to a dead load of 20 kN/m if it has a span of 6m.
Solution:
① Depth of compression block:
. . = .
= 0.0042125 > 0.002 < 0.005 = . = , = 0.0020735 < 0.0042125 Steel yields:
Assuming steel yields:
= 0.85 = 0.8527.6250 = 1875414.7 = . ② Moment capacity reduction factor: = 132.58 = 0.85 = 155.98
Since t is between 0.002 and 0.005, this value is within the transition range between compression controlled section and tension controlled section.
∅ = 0.65 0.002 ∅ = 0.65 0.00421250.002 ∅ = . ③ Live load it could carry: = ∅
= 0.834 1875414.7375 . = 200.2 10 = 200.2
=
200.2 = = 44.48 /
Wu = 1.2 (10.5) + 1.6 (22) Wu = 47.8 kN/m
= = . = . . 2. Approximate flexural resistance factor: Mu = ø R b d2 d = 400 – 62.5 = 337.50 215.1 x 106 = 0.90 R (300) (337.5)2
= 1.21.6 44.48 = 1.220 1.6 = .
R = 6.99
3. Number of 32 mm ø bars:
(1 – 0.59 ) 6.99 = 27.6 (1 – 0.59 ) 0.59 – 27.6 + 6.99 = 0 – 46.78 + 11.85 = 0 R = f c’
PROBLEM 41:
Architectural considerations limit the height of a 6 m. long simple span beam to 400 mm and width of 300 mm. The following loads are material properties are given: Use 62.5 mm ø as covering from center of reinforcing bars. DL = 10.5 kN/m
LL = 22 kN/m
f c’ = 34.6 MPa
f y = 414.7 MPa
= . ±. = 0.25 = 0.25 = . . = 0.0209
1. Determine the factored moment carried
As = b d
by the beam. 2. Determine
the
approximate
flexural
As = 0.0209 (300) (337.50) As = 2112 mm2
resistance factor. Assume ø = 0.90 3. Determine the number of 32 mm ø bars needed for the beam.
= 32 = 2112
= 2.63 3 Solution:
1. Factored moment carried by the beam. Wu = 1.2 DL + 1.6 LL
Use 3 – 32 mm ø bars As =
= 32 3 = 2413
= 0.75 = . + = 0.85 . − = 0.85 . .−
= 0.817
.. = .. +. = 0.03426 = 0.75 = 0.750.03426 = 0.0257
– 0.002)
ø = 0.65 + (0.00426 – 0.002)
ø = 0.838 (moment capacity reduction factor)
Mu = ø As f y (d - ) Mu = 0.838 (2413) (414.7) (337.50 -
.)
Mu = 235.46 kN.m > 215.10 kN.m (safe)
PROBLEM 42:
A reinforced rectangular concrete beam has a width of 250 mm and an effective depth of 360
= = . = 0.0238 < 0.0257
Use ø = 0.65 + (
mm. It is reinforced for tension only at the bottom with a total tension steel area of 600 mm 2. f c’ = 40 ok
MPa, f y = 400 MPa.
1. Determine the tension reinforcement index for this beam. 2. Determine the distance of the neutral axis below the compression surface. 3. Determine the ultimate flexural strength of the beam.
Solution:
C=T 0.85 f c’ ab = As f y 0.85 (34.6) (a) (300) = 2413 (414.7) a = 113.42
a= c 113.42 = 0.817 c c = 138.82
= . . . = 0.00426 < 0.005
1. Tension reinforcement index for this beam.
= = = = 0.0067 = = . = . 2. Distance of the neutral axis below the compression surface.
Mu = ø As f y (d - )
= . . . = 0.0026983
= 0.85 . − = 0.85 . − = 0.78 Assume steel yields:
= = = 0.002
0.85 (40) (a) (250) = 600 (400)
> steel yields (under reinforced) = = 0.0026983 < 0.005
a = 28.23 mm
ø = moment capacity reduction factor
C=T 0.85 f c’ ab = A s f y
a= c
ø = 0.70 + (
– 0.002)
28.23 = 0.78 c c = 36.20 mm (neutral axis below compression surface)
3. Ultimate flexural strength of the beam.
Use ø = 0.65 + (
– 0.002)
ø = 0.65 + (0.0026983 – 0.002)
ø = 0.71
Steel does not yield: f s ≠ f y Using Hookes Law, solve for the actual fs:
. = − = −. = −. = − fs =
Mu = ø As f y (d - ) Mu = 0.71 (600) (400) (360 -
.)
Mu = 58.93 kN.m
PROBLEM 43:
A rectangular beam has a width of 280 mm and an effective depth of 500 mm. It is reinforced with 4 – 36 mm ø bars at the tension side of the beam placed 65 mm above the bottom of the beam.
T=C As f y = 0.85 f c’ ab
4072 600 − = 0.85250.85280 500 – c = 0.00207 c 2
1. Which of the following will give the
c2 + 483.09 c – 241546 = 0 c = 306.08 mm
location of the neutral axis from the top of the beam. 2. Which of the following give the stress of steel. 3. Which of the following will give the ultimate capacity of the beam.
2. Stress of steel.
a= c a = 0.85 (306.08) a = 260.17 mm
Solution:
1. Location of the neutral axis from the top of the beam .
= − = −. . = . < 3. Ultimate moment capacity of the beam
Mu = ø As f y (d - ) Mu = 0.90 (4072) (380.14) (500 Mu = 515.3 kN.m
.)
PROBLEM 44:
A rectangular beam has a width of 280 mm and an effective depth of 500 mm and is reinforced with steel area in tension equal to 4000 mm 2. f c’ = 25 MPa, f y = 400 MPa. 1. Compute the depth of compression strees
= −. = = −. = −. = −
block. 2. Compute the ultimate moment capacity
0.85 f c’ ab = As f s
of the beam. 3. What is the correct description of the beam? a.
C=T
0.85 f c’ cb = 40000 f s 0.85 (25) (0.85) c (280) =
Under reinforced
c2 + 474.54c – 237271.4 = 0
b. Over reinforced c.
−
c = 304.55
Balanced condition
d. Reduction compression
in zone
depth results
of in
decrease in steel strain at failure.
Solution:
= − = −. . = 385.06 < = 400 ⸫ Steel does not yield
1. Depth of compression stress block
ok as assumed
Assume steel does not yield
a= c a = 0.85 (304.55) a =
258.87
mm
depth
of
compression block
2. Ultimate moment capacity of the beam
Mu = ø As f y (d - ) f s ≠ f y
. = − fs =
Mu = 0.90 (4000) (385.06) (500 Mu = 513.7 kN.m
3. Description of beam
.)
= −. = −.. . = 0.00193
1. Depth of compression block for a balanced condition.
= = = 0.002 < 0.00193 < 0.002 The beam is Over Reinforced.
PROBLEM 45:
A beam has a width of 300 mm and an effective
= = = 0.002 . = . − c = 300 mm
depth of 500 mm. f c’ = 28 MPa, f y =414 MPa, Es = 200,000 MPa 1. Determine the depth of compression
a = 0.85 (300)
block for a balanced condition. 2. Determine
the
balanced
steel
a= c
a = 255 mm
area
required. 3. Determine the moment capacity for maximum steel area in a balanced condition.
2. Balanced steel area required. C=T 0.85 f c’ ab = As f y 0.85 (28) (255) (300) = Asb (414)
Solution:
Asb = 4398 mm2
3. Moment capacity for maximum steel area in a balanced condition. As = 0.75 Asb (max. steel area for balanced condition) As = 0.75 (4398) As = 3298.5 mm2
1. Total depth of the beam for a balanced condition.
= = 0.002 < 0.005 Use ø = 0.65
Mu = ø As f y (d - ) Mu = 0.65 (3298.5) (414) (500 -
)
Mu = 330.64 kN.m
= = = 0.0021
a= c 255 = 0.85 (c) c = 300 mm
PROBLEM 46:
The width of a rectangular beam is 300 mm. The depth of compression block for a balanced
By ratio and proportion:
condition is 255 mm. f c’ = 28 MPa, f y =414 MPa,
. = . −
Es = 200,000 MPa. Use 70 mm as steel covering.
d = 300 + 210
Unit weight of concrete is 24kN/m 3.
d = 510 mm
1. Determine the total depth of the beam for a balanced condition.
Total depth = 510 + 70 = 580 mm
2. Determine the total area of reinforcement for a balanced condition. 3. Determine the factored support imposed
2. Area of reinforcement for a balanced condition.
uniform load that a 6-m. simple span beam could support for a balanced
T=C
condition.
Asb f y = 0.85 f c’ ab Asb (414) = 0.85 (28) (255) (300)
Solution:
Asb = 4398 mm2
3. Factored super imposed uniform load that a 6-m. simple span beam could support for a a balanced condition.
Solution:
1. Depth of compression block for a balanced condition.
Mu = ø As f y (d - ) Mu = 0.90 (4398) (414) (500 -
)
Mu = 626.8 kN.m
Mu =
626.8 = Wu = 139.29 kN/m
Wt. of concrete = 0.3 (0.58) (24) Wt. of concrete = 4.18 kN/m
Wu = 4.18 (1.2) + W s
= = = 0.0020750 . = . − c = 295.57 mm
139.29 = 4.18 (1.2) + W s Ws = 134.27 kN/m
a= c PROBLEM 47:
A reinforced concrete beam has a width of 250
a = 0.85 (295.57) a = 251.23 mm
mm and an effective depth of 500 mm. The compression strength of concrete is 28 MPa and the yield strength of steel is f y = 415 MPa.
1. Determine the depth of compression block for a balanced condition. 2. Determine the steel area required for a balanced condition. 3. Determine the ultimate moment capacity to ensure that concrete fails in a ductile manner.
2. Steel area required for a balanced condition.
Check for ø: C=T 0.85 f c’ ab = A sb f y 0.85 (28) (251.53) (250) = Asb (415) Asb = 3606 mm2 (balanced steel area)
3. Ultimate moment capacity to ensure that concrete fails in a ductile manner.
Use As = 0.75 Asb
a= c
As = 0.75 (3606)
188.63 = 0.85 (c)
As = 2704.50 mm
2
c = 221.92 mm
. . = . = 0.00376 < 0.005 ø = 0.65 + (0.00376 – 0.002)
Use ø = 0.65 + ( – 0.002)
ø = 0.80
Mu = ø As f y (d - ) C=T
Mu = 0.80 (2704.05) (415) (500 -
0.85 f c’ ab = A sb f y
Mu = 364.26 kN.m
.)
0.85 (28) (a) (250) = 2704.50 (415) a = 188.63
PROBLEM 48:
A rectangular reinforced concrete beam has a
Mu = ø As f y (d - )
width b = 300 mm and an effective depth d = 400 mm. If f c’ = 28 MPa, f y = 280 MPa, Es = 200,000 MPa.
1. Which of the following gives the nearest
2. Compressive force of concrete:
vaule of the distance of the N.A. from the top of the beam so that the strain in concrete
= 0.003 will be attained at
the same time with the yield strain of steel
.
2. Which of the following gives the nearest value of the total compressive force of concrete. 3. Which of the following gives the nearest value of the balanced steel ratio.
a= c Solution:
a = 0.85 (272.73)
1. Distance from N.A. to the top of the
a = 231.82 mm
beam:
= = = 0.0014 . = . − c = 272.73 mm
C = 0.85 f c’ ab C = 0.85 (28) (231.82) (300) C = 1655194.8 N C = 1655.2 kN
3. Balanced steel ratio:
= . + . = .+ = 0.0493 = . % PROBLEM 49:
The beam has a cross section as shown in the figure. It carries an ultimate moment of 156 kN.m. Using f c’ = 20.7 MPa, f y = 414 MPa, E s = 200,000 MPa.
T=C 509804 = 0.85 f c’ Ac 509804 = 0.85 (20.7) Ac Ac = 28974 mm2 Ac = 100 (a) (2) + 100(a – 75) 28974 = 200a + 100a – 7500 300a = 36474 a = 121.58 mm
a= c 121.58 = 0.85 (c) 1. Compute the location of the neutral axis
c = 143.04 mm
measured from the top of the beam. 2. Compute the number of 20 mm ø bars needed.
2. Number of 20 mm ø A1 = 100 (121.58) (2)
3. Compute the actual strain of the steel reinforcements used.
A1 = 24316 A2 = 100 (46.58) A2 = 4658
Solution:
1. Location of N.A.
̅
A = A1y1 + A2y2
̅ ̅= 66.82
(24316 + 4658) = 24316 (60.79) + 4658 (98.29)
z = 400 – 66.82 = 333.18 mm
Mu = ø T z 156 x 106 = 0.90 As (414) (333.18) As = 1256.62 mm2 Approximate
z = 0.85 d z = 0.85 (400) z = 340
Mu = ø T z 6
156 x 10 = 0.90 T (340) T = 509804
20 = 1256.62 N = 4 bars Use 4 – 20 mm ø bars
3. Actual strain of steel bars
Solution:
= . . . = . (actual strain) = = = 0.00207
1. Location of neutral axis from the top of the beam for a balance condition.
> steel yields = PROBLEM 50:
A symmetrical cross-section of a reinforced concrete shown has a value of f c’ = 24.13 MPa, f y = 482.7 MPa, E s = 200,000 MPa.
= . = = 0.00241 . = . .− c = 377.77 mm
2. Balanced steel area A sb
1. Which of the following gives the location of neutral axis from the top of the beam for a balance condition. 2. Which of the following gives the balanced steel area A sb. 3. Which of the following gives the max. area permitted by the code.
3. Which of the following gives the max.
a= c a = 0.85 (377.77)
area permitted by the code.
a = 321.10 mm Solution:
C = 0.85 f c’ ab C = 0.85 (24.13) [375(125) + 125(321.10)]
1. Location of neutral axis from the top of the beam for a balance condition.
C = 1784670 N
T=C Asb (482.7) = 1784670 Asb = 3697 mm2
3. Max. area permitted by the code Asmax = 0.75 (3697) Asmax = 2773 mm2
PROBLEM 51:
The hallow box beam in the figure must carry a factored moment of 540 kN.m. f c’ = 28 MPa, f y =
= = = 0.0017
345 MPa, E s = 200,000 MPa.
. = . − c = 462.77 mm
2. Balanced steel area A sb
a= c a = 0.85 (462.77) a = 393.35 mm
C = 0.85 f c’ A 1. Which of the following gives the location of neutral axis from the top of the beam
A = 125 (393.35)(2) + 250(150) A = 135,837.5 mm2
for a balance condition. 2. Which of the following gives the balanced steel area A sb.
C = 0.85 (28) (135837.5) C = 3232933 N
T=C
T=C
Asb (345) = 3232933
Asb f y = 0.85 f c’ Ac
Asb = 9371 mm2
12 (3) (414) = 0.85 (20.7) (255) A
c
Ac = 7983.3 mm2 3. Max. area permitted by the code Asmax = 0.75 (9371) Asmax = 7028.25 mm2
= 7983.3 = .
PROBLEM 52:
A triangular beam having a base width of 300
By ration and proportion
mm. has a total depth of 600 mm. It is reinforced
=
with 3 – 12 mm ø bars placed at 70 mm above the
= 0.5
bottom of the beam. f c’ = 20.7 MPa, f y = 414 MPa, Es = 200,000 MPa.
1. Compute the neutral axis of the beam from the apex of the section. 2. Compute the ultimate strength capacity of the beam. 3. What would be the steel area required for
0.5 = . = 178.70
a= c 178.70 = 0.85 (c) c = 210.24 mm
a balanced condition? 2. Ultimate strength capacity Solution:
1. Neutral axis of the beam
z = 530 – 2/3 (178.7) z = 410.87
Mu = ø T z Mu = 0.90
12 (3) (414) (410.87)
Mu = 51.9 kN.m
3. Steel area required for a balanced condition.
As = (755) As = Asb
= = = 0.00207 . = . −
As = 566.25 mm2
PROBLEM 53:
1.59 – 0.003c = 0.00207 c = 313.61
A triangular beam has an effective depth of 687.50 mm and a base of 750 mm. The beam carries an ultimate moment of 197 kN.m. f c’ = 27.5 MPa, f y = 414 MPa
1. Compute the neutral axis of the beam 2. Compute the value of total compressive force of concrete. 3. Compute the streel area required.
Solution:
1. Neutral axis of the beam.
a= c a = 0.85 (313.61) a = 266.57 mm
T=C Asb f y = 0.85 f c’ Ac
. =
=
x = 133.29
a=b Asb f y = 0.85 f c’ Ac Asb (414) = 0.85 (20.7) Asb = 755 mm2
..
C=T
= A f s y .. = 414.7 0.85 f c’
As = 0.0282 a 2
PROBLEM 54:
Mu = ø T z
The beam has a cross section shown in the figure.
197 x 106 = 0.90 As (414.7) (687.50 - a)
f c’ = 20.7 MPa, f y = 414.7 MPa
197 x 106 = 0.90 (0.282) a 2 (414.7) (687.50 – 0.667a) 18717186.73 = 687.50a2 – 0.667a3
Solve for “a” by trial and error a = 181.8
a= c 181.8 = 0.85 (c) c = 213.88 mm
2. Compressive force of concrete
. ...
C = 0.85 f c’
=
C = 386286 N
1. Compute
the
minimum
steel
area
permitted by the NSCP Specs. 2. Compute the flexural design strength ø Mn if it is reinforced with minimum steel area.
C = 386.3 kN
3. Compute the maximum area of flexural steel that can be used in reinforced the
3. Steel area required
section.
C=T 386286 = As f y
Solution:
386286 = As (414.7) As = 931.5 mm
1. Min. steel permitted by the NSCP Specs. 2
= . . = . = 0.00338
= = 0.00338 1252437.5 = . 2. Flexural strength ø M n
C=T 0.85 f c’ (750) a = As f y 0.85 (20.7) (750) a = 369.2 (414.7) a = 11.60 mm
. ø M = 0.90 (369.2) (414.7) 437.5 ø Mn = 0.90 T
n
ø Mn = 59.5 x 10 6 N.mm ø Mn = 59.5 kN.m 1. Determine the neutral axis of the section 3. Max. steel area that can be used
= . + .. = . .+.
from the top of the beam. 2. Determine the flexural capacity of the cross section. 3. Determine the strain in the steel at failure.
= 0.0213 Solution:
= 0.75 = 0.75 0.0213 = 0.016
1. Neutral axis of the beam.
= = 0.016 1252437.5 = PROBLEM 55:
Each leg of the cross section is reinforced with a 28 mm ø bar.
C=T 0.85 f c’ Ac = As f y 0.85 (24.8) (500) a =
28(2) (414.7)
a = 48.45 mm
a= c 48.45 = 0.85 (c) c = 57 mm
2. Flexural capacity ø M n ø Mn = 0.90 T ø Mn = 0.90
28 (2) (414.7) 375 .
ø Mn = 161.2 x 10 6 N.mm ø Mn = 161.2 kN.m
3. Strain in steel at failure:
= 1963.5
= . = . PROBLEM 56: CE BOARD MAY 2010
A 12 m simply supported beam is provided by an
= . = = 0.017
additional support at midspan. The beam has a width of b = 300 mm and a total depth h = 450 mm. It is reinforced with 4 – 25 mm ø at the tension side and 2 – 25 mm ø at the compression side
with
70
mm
cover
to
centroid
If only tension bars are needed:
= 0.75 = 0.023 > 0.017
of
reinforcement. f c’ = 30 MPa, f y = 415 MPa. Use
Therefore, the beam needs only tension
0.75 = 0.023
bars a specified in the problem.
1. Determine the depth of the rectangular stress block. 2. Determine the nominal bending moment, Mn.
C=T 0.85 f c’ ab = As fy 0.85 (30) (a)(300) =
254(415)
a = 106.52 mm
3. Determine the total factored uniform load including the beam weight considering moment capacity reduction of 0.90. Solution:
1. Depth of the rectangular stress block:
2. Nominal bending moment:
. Mn = 25 4(415) 380 Mn = As fy
Mn = 266.2 x 10 6 N.mm Mn = 266.2 kN.m Check if compression bars is needed.
= 254
3. Total factored uniform load including beam weight:
PROBLEM 57:
A rectangular beam has a width of 300 mm and an effective depth of 537.50 mm to the centroid of tension steel bars. Tension reinforcement consists of 6 – 28 mm ø in two rows, compression reinforcement consists of 2 – 22 mm ø. f c’ = 27.6 MPa, f y = 414.7 MPa. Assume steel covering is 60 mm for compression bars.
= = = =
1. Compute the depth of compression block. 2. Compute the factored moment capacity of beam. 3. Compute the maximum total tension steel
2R + P = wL 2R = wL 2R =
allowed by specifications.
R =
Solution:
1. Depth of compression block.
= =
=
Mu = 0.90 M n Mu = 0.90 (266.2) Mu = 239.58 kN.m
239.58 = 239.58 = w = 53.24 kN/m
= 286 = 3695 ′ = 222 = 760 =
= . = 0.0229 ′ = ′ = . = 0.00471
As1 = As – As2 As2 = As’ when compression bars will yield As1 = As – As’
0.85 f c’ ab = (As – As’) fy 0.85 (27.6) a (300) = (3695 – 760) 414.7 a = 172.94 mm
Check if compression bar are really needed:
2. Factored moment capacity of the beam.
= . +
a= c
.. = . .+.
172.94 = 0.85 (c) c = 203.46 mm
= 0.0284
= . . . = 0.00511 > 0.005
When,
> compression bars are needed = 0.75 0.0229 > 0.750.0284 0.0229 > 0.0213
Use capacity reduction factor: ø = 0.90
Check: if compression bars will yield
> . − = 0.0229 0.00471 = 0.0182 .. = . ..−.
= 0.0173
M1 = ø As1 fy
As2 = As’ As1 = As - As 2 As1 = As – As’
0.0182 > 0.0173 (compression bars will yield)
. M = 0.90 (3695 – 760) 414.7 537.50 M1 = ø (As – As’) fy
Check:
< 0.0182 < 0.0213
1
M1 = 494.07 x 10 6 N.mm
′ = ø As’ fy ′
M2 = ø As2 fy C1 = T1 0.85 f c’ ab = As 1 fy
M2
M2 = 0.90 (760) (414.7) (537.5 – 60)
630 = 1.8 b
M2 = 135.45 x 10 6 N.mm
b = 350 mm
Mu = M1 + M2
2. Reinforcement for compression
= . +
Mu = 494.07 + 135.45 Mu = 629.52 kN.m
3. Maximum total tension steel allowed by
.. = ..+. = 0.0285
specifications. 0.85 f c’ ab = As1 f y
Max. As = 0.75 Max. As = 0.75 Max. As = 300537.50.02130.00471 Max. As = > Max. As =
PROBLEM 58:
A doubly reinforced concrete beam has a max.
Assume
effective depth of 630 mm and is subjected to a
. = ..
total factored moment of 1062 kN.m including its own weight. f c’ = 27.58 MPa, f y = 413.4 MPa. Use 62.5 mm steel covering.
1. Determine the width of the beam 2. Determine
the
reinforcement
= . ... . . = 0.018 As1 = As – As2
for
compression 3. Determine the total reinforcement for
As2 = As’ if compression steel will yield As1 =
bd
As1 = 0.018 (350) (630)
tension
As1 = 3969 mm2
Solution:
M1 = ø As1 fy
1. Width of the beam Approximate proportion of b and d d = 1.5 to 2 b Try d = 1.80 b
M1 = 0.90 (3969) 413.4
630 .
M1 = 782.7 kN.mm Mu = M1 + M2
Use As = 5292 mm2
1062 = 494.07 + M 2 M2 = 279.3 kN.m
PROBLEM 59:
Check whether compression bars will yield:
> . − As = As1 + As2
A rectangular beam has a width of 300 mm and an effective depth to the centroid of the tension reinforcement
of
600
mm.
The
tension
reinforcement has an area of 4762 mm 2 and the
As2 = As’
area of compression reinforcement placed 62.50
As = As1 + As’
bd = bd + ′ bd = 0.018 > .... .−. 0.018 > 0.0154
mm from the compression face to the beam is 987.5 mm2. f c’ = 34.56 MPa, f y = 414.6 MPa. Balanced steel ratio is 0.034. Assume that steel yields.
1. Determine the depth of compression Therefore, compression bars will yield.
block 2. Determine the design strength using 0.90
fs’ = fy
as the reduction factor
As’ fy = As2 fy As’ = As2 M2 = ø As’ fy
(ok as assumed)
′
279.3 x 106 mm = 0.90 As’ (413.4) (630 – 62.5) As’= 1323 mm2
3. Total reinforcement for tension. As = As1 + As2 As = 3969 + 1323 As = 5292 mm2
′ = ′ = = 0.006 Max. As = bd ( 0.75 ) Max. As = 350(630) [0.75(0.0285) + 0.00669)] Max. As = 6036 mm2 > 5292 mm2 ok
3. Determine the concentrated live loads at the midspan in addition to a dead load of 20 kN/m including the weight of the beam if it has a span of 6 m.
Solution:
1. Depth of compression block
= . +
= 0.85 . − = 0.85 . .− = 0.803 .. = ..+. = 0.034 = = = 0.0264 . ′ = = ′ = 0.0054
Since, compression bars will yield: fs’ = fy T = C1 + C2 As fy = 0.85 f c’ ab + As’ fy
=
Check the beam first as a singly reinforced beam to see if the compression bars can be disregarded.
= 0.75 = 0.75 0.034 = 0.0255 = 0.0264 > 0.0255
=
− . −.. . .
= . 2. Design strength using 0.90 as the reduction factor.
Therefore, the beam must be analyzed as doubly reinforced
beam.
Check
if
the
bars
in
compression will really yield, by computing the steel ratio that will ensure yielding of the compression bar at failure.
= 0.0264
> . − ... 0.0264 0.0054 > .. −. 0.021 > 0.0195 (therefore, compression
bars will yield)
C1 = T1 0.85 f c’ ab = As1 fy 0.85 (34.56) (177.57) (300) = As 1 (414.6) As1 = 3774.50 mm2
As = As1 + As2 As2 = As – As1
PROBLEM 60:
As2 = 4762 – 3774.50
A reinforced concrete beam has a width of 375
As2 = 987.50 mm2 ok
mm and a total depth of 775 mm. Steel covering for both compression and tension bars is 75 mm.
Mu = ø (M1 + M2)
Area of compression bars is 1290 mm 2 while that
M = As fy . M = 3774.5 (414.6) 600
of the tension bars it is 6529 mm 2. f c’ = 27.6 MPa,
M1 = T1 1
f y = 414.6 MPa.
1
1
M1 = 800 x 10 6 N.mm
1. Determine the depth of compression block. 2. Determine the capacity reduction factor
M2 = ø As2 fy
′
M2 = 987.50 (414.6) (600 – 62.5) M2 = 220 x 10 6 N.mm
Mu = ø (M1 + M2) Mu = 0.90 (800 + 220)
for moment. 3. Determine the ultimate moment capacity of the beam.
Solution:
1. Depth of compression block.
Mu = 918 kN.m
3. Concentrated live loads it could support at its midspan.
. = − = .− (1.2) Mu = (1.6) (1.2) 918 = (1.6) P = 337.5 kN
= ′ = .− = −
T = C1 + C2 As fy = 0.85 f c’ ab + As’ fs’ 6529 (414.6) = 0.85(27.6) (0.85) c (375) +
− 1932923.4 c = 7477.88c2 – 58050000 = 0 c2 – 258.49 c – 7762.90 = 0 c = 285.66 mm
= − = .− . = 442.47 > 414.6
Note: When
< 0.005
Value of ø = 0.65 + (
– 0.002)
ø = 0.65 + (0.00435 – 0.002) Compression steel yield:
ø = 0.846
Use fs’ = fy = 414.6 MPa
3. Ultimate moment capacity of the beam.
a= c a = 0.85 (285.66) c = 242.81 mm
2. Capacity reduction factor for moment.
= . − = .−
=
.−. .
= 0.00435 < 0.005
C2 = T2 As’ fy = As2 fy As’ = As2 As2 = 1290 mm2
As1 = As – As2 As1 = 6529 – 1290 As1 = 5239 mm2
= ∅ = ∅ = 0.8465239414.6700 . 1290414.670075 = 1346 10 . = . PROBLEM 61:
A rectangular concrete beam has a width of 350 mm and a total depth of 675 mm. It is reinforced for tension at the bottom with 4 – 36 mm ø bars at an effective depth of 537.5 mm and two 28 mm ø bars at the top placed at 62.5 mm from the top of the beam. f c’ = 20.7 MPa, f y = 414.6 MPa.
1. Determine the minimum tensile steel ratio that will ensure yielding of the compression steel at failure. 2. Determine the total compressive force of
Solution:
1. Minimum tensile steel ratio that will ensure yielding of the compression steel at failure.
′ = . + A = = 28 2 = 1232 = . + s
... = . .. +. . = 0.020 Check:
= A = = 36 4 = 4071.5 . = . = 0.0216 > 0.020 = . s
2. Total compressive force of concrete.
concrete. 3. Determine the design moment capacity of the beam.
= 36 4 = 4071.5 As’ = = 28 2 = 1232 . = . As =
= 0.0216 ′ = . ′ = 0.0065
C1 = 0.85 (20.7) (188) (350) When:
C1 = 1157751 N
(compression > . −
bars will
yield)
C2 = As’ fs’ C2 = 1232 (414.6)
= 0.0216 0.0065 = 0.0151
C2 = 510787 N
... 0.0151 > . ..−. 0.0151 > 0.0136
M = 1157751 537.5 M1 = C1 1
M1 = 513.46 x 10 6 N.mm
Therefore, compression steel yields. M2 = C2
fs’ = fy = 414.6 MPa
′
M2 = 510787 (537.5 – 62.5)
C1 = 0.85 fc’ ab
M2 = 242.62 x 10 6 N.mm
a= c a = 0.85 (221.18)
Mu = ø (M1 + M2)
a = 188 mm
Mu = 0.90 (513.46 + 242.62)
C1 = 0.85 (20.7) (188) (350)
Mu = 635.11 kN.m
C1 = 1157751 N C1 = 1157.75 kN
PROBLEM
3. Design moment capacity of the beam.
62:
CE
Board
Nov.
2010,
Nov.2012
A simply supported beam is reinforced with 4 – 28 mm ø at the bottom and 2 – 28 mm ø at the top
. . = . = 0.00429 <
of the beam. Steel covering to centroid of reinforcement is 70 mm at the top and bottom of
0.005
the beam. The beam has a total depth of 400 mm and a width of 300 mm. f c’ = 30 MPa, f y = 415
Use ø = 0.65 + (
MPa. Balanced steel ratio
– 0.002) ø = 0.65 + (0.00429 – 0.002) ø = 0.84
= 0.031.
1. Determine the depth of compression block. 2. Determine the design strength using 0.90
Mu = ø (M1 + M2) C1 = 0.85 f c’ ab
as the reduction factor.
3. Determine the live load at the mid-span in addition to a DL = 20 kN/m including
Therefore, compression bars will not yield.
the weight of the beam if it has a span of
fs’ ≠ fy
6 m.
. = − = .− = ′
Solution:
1. Depth of compression block:
= 28 4 = 2463 As’ = = 28 2 = 1231.5 As =
= = = 0.0248
T = C1 + C2 As fy = 0.85 f c’ ab + As’ fs’
= 0.75 = 0.75 0.031 = 0.023
2463(415)
=
(30)
(a)
(300)
+
.−
>
− − 1022145 = 7650 (0.85c) +
Therefore, reinforcement for compression is
157.19 c = c2 + 113.63 (c – 70)
needed. To ensure that compression bars will
c2 + 43.56 c -7654.10 = 0
yield.
= . ±.
> . − . > . − > 0.036 ′ = . ′ = ′ = 0.0124 > 0.036 0.02480.0124 < 0.036 0.0124 < 0.036
1022145 = 7650a +
c = 113.59
a= c a = 0.85 (113.59) a = 96.55 mm
1231.5
2. Design strength using 0.90 as reduction factor.
Mu = ø (M1 + M2) Mu = 0.90 (208 + 73.7) Mu = 253.53 kN.m
3. Concentrated live load at mid-span:
(1.2) (1.2) 253.53 = (1.6) Mu = (1.6)
P = 60.64 kN C1 = 0.85 f c’ ab C1 = 0.85 (30) (96.55) (300) C1 = 738607.50 N
PROBLEM 63:
A reinforced concrete beam has a width of 350 mm and an effective depth of 562.5 mm. It is
C2 = As’ fs’
reinforced for tension at the bottom of the section
= ′ = .− = ..− . = 0.0011512 = 0.0011512 200000 = 230.25 < 415
having an area of 4896 mm 2 and for compression at the top of the beam 62.5 mm below the extreme compression fibers of the beam, having an area of 1530 mm2. f c’ = 34.6 MPa, f y = 414.7 MPa.
1. Determine the depth of compression block. 2. Determine the ultimate moment capacity.
C2 = As’ fs’ C2 = 1231.5 (230.25)
3. Determine the maximum total tension area that could be used in this section.
C2 = 283553 Solution:
. M = 738607.5 330 M1 = C1 1
M1 = 208 x 10 6 N.mm M2 = C2
′
1. Depth of compression block.
= = . = 0.0249
M2 = 283553 (330 – 70) M2 = 73.7 x 10 6 N.mm
′ = ′ = .
′ = 0.0078
C1 + C2 = T1 + T2 C1 + C2 = T + As fy
= 0.0249 0.0078 = 0.0171 When,
8234.8 c +
918000 −. = 4896 (414.7)
c2 + 111.48 (c – 62.5) = 246.56 c c2 – 135.08 c – 6967.5 = 0
> . −
c = 174.91 mm
(compression bar did not yield)
a= c a = 0.85 (174.91) a = 139.93 mm
2. Ultimate moment capacity. C2 = As’ fs’
. = − = .− = ′
) 200,000 = .(− = −.
= .−. . = 385.60 = .−. . = 589974 = 589.97 C1 = 8234.8 c C1 = 8234.8 (174.91)
C1 = 0.85 f c’ ab
C1 = 1440349 N
C1 = 0.85 f c’ cb C1 = 0.85 (34.6) (0.8) c (350) C1 = 8234.8 c
. M = 1440349 562.5 M1 = C1 1
M1 = 709.42 x 10 6 N.mm
C2 = As’ fs’
−. −. C2 = 918000 C2 =
M2 = C2
′
M2 = 589974 (562.5 – 62.5) M2 = 294.984 x 106 N.mm
Mu = ø (M1 + M2)
= = = 0.0253
Mu = 0.90 (709.42 + 294.98) Mu = 903.96 kN.m
3. Maximum total tension area that could be used in this section. Max. As = bd (
0.75 )
Max. As = 350(562.5) [0.75(0.03346) +
. .] . Max. As = 6369 mm2
PROBLEM 64:
A rectangular concrete beam has a width of 375 mm and an effective depth of 500 mm. compression bars has an area of 1968 mm 2
′ = ′ = ′ = 0.010496 = 0.75
= . + .. = ..+. = 0.02844
located at 100 mm from the compression face of the beam. The tension bars have an area of 4744 mm2. f c’ = 27.6 MPa, f y = 414.6 MPa.
1. Determine the depth of the compression block. 2. Determine the max. steel ratio.
= 0.750.02844 = 0.02133 0.0253 > 0.02133 Therefore, compression bars are really needed.
3. Determine the ultimate moment capacity of the beam.
Check if compression bars will yield or not.
Solution:
1. Depth of the compression block.
> . −
compression bars will
not yield
> ... . −. 0.0253 – 0.010496 < 0.03113 0.0148 < 0.03113 not yield fs’ ≠ fy
) = (− C1 + C2 = T1 + T2
compression bars will
As = As1 + As2
3. Ultimate moment capacity of the beam.
T1 + T2 = T As1 fy + As2 fy = T (As1 + As2) fy = T T = As fy
C1 + C2 = T As’ fs’ = As2 fy
0.85 fc’ ab + As’ fs’ = As fy 0.85 (27.6) (0.85) c (375) + 4744 (414.6)
−
1968 (282.15) = As2 (414.6) =
As2 = 1339 mm2
c2 – 105.12 c – 15790.57 = 0
As = As1 + As2
c = 188.77
As = 4744 + 1339
a= c
As = 3405 mm2
a = 0.85 (188.77) a = 160.45 mm
2. Max. steel ratio.
= 0.75 ′ ) = (− = .− . = 282.15 = 0.028440.010496 . . = 0.02847 0.0253 < 002847
<
ok
. M = 3405 (414.6) 500 M1 = As1 fy 1
M1 = 704.44 x 10 6 N.mm
M2 = As2 fy
′
M2 = 1339 (414.6) (562.5 – 62.5) M2 = 222.06 x 10 6 N.mm
Mu = ø (M1 + M2)
Check if we could use ø = 0.90
. . = . = 0.005
ok
= 32 6 = 4825 = =
Use ø = 0.90
As =
Mu = 0.90 (704.44 + 222.06) Mu = 833.85 kN.m
= 0.0268
PROBLEM 65:
A rectangular beam has a width of 300 mm and
= 0.75 = . +
an effective depth to the centroid of the tension reinforcement
of
600
mm.
The
tension
reinforcement consists of 6 – 32 mm ø bars placed in
two
rows.
Compression
= 0.85 . − = 0.85 . .− = 0.80 .. = ..+. = 0.0335
reinforcement
consisting of two 25 mm ø bars is placed 62.5 mm from the compression face of the beam. f c’ = 34.6 MPa, f y = 414.7 MPa.
1. Determine the depth of compression block.
= 0.750.0335 = 0.0252
2. Determine the maximum steel ratio. 3. Determine the design moment capacity of the beam.
Compression bars is needed.
Solution:
>
1. Depth of compression block. Check whether compression bars is really
0.0268 > 0.0252
needed.
Check if compression bars will yield or not.
> . −
′ =
′ = ′ = 0.00545 > .... . −. 0.0268 – 0.00545 > 0.01954 0.02135 > 0.01954 fs’ = fy (steel compression yields)
As’ =
= 25 2 = 981.75
a= c 180.64 = 0.85 (c) c = 221.10 mm
T1 + T2 = C1 + C2
. . = . = 0.0051 >
T = C1 + C2 As fy = 0.85 fc’ ab + As’ fy 4825 (414.7) = 0.85 (34.6) (a) (300) + 981.75
fs = fy Use ø = 0.90
+ C ′ = 0.85 fc’ ab + As’ fy ′ Mn = C1
Mn
2
C1 = 0.85 fc’ ab C1 = 0.85 (34.6) (180.64) (300) C1 = 1593787 N (414.7) a = 180.64 mm
C2 = As’ fy C2 = 981.75 (414.7) C2 = 407132 N
2. Maximum steel ratio
= 0.75 ′ = 0.75 0.0335 0.00545 = .
+ C ′ . + 407132 600 M = 1593787 600 62.5 Mn = C1
2
n
3. Design moment capacity of the beam.
Mn = 1031 x 106 N.mm
Mu = ø Mn Mu = 0.90 (1031) Mu = 927.9 kN.m
PROBLEM 66:
A rectangular beam reinforced for both tension
T = C1 + C2
and compression bars has an area of 1250 mm 2
As fy = 0.85 f c’ ab + As’ fs’
and 4032 mm2 for tension bars. The tension bars
4032 (414.6) = 0.85(20.7) (0.85) c (350)
are placed at a distance of 75 mm from the bottom of the beam while the compression bars are placed 62.5 mm from the top of the beam. f c’ = 20.7 MPa, f y = 414.60 MPa. Width of beam is 350
+
−.
1671667.2 c = 5234.5c 2 – 750000 c - 46875000 c2 – 176.08 c – 8955 = 0 c = 217.29 mm
mm with a total depth of 675 mm.
1. Determine the depth of compression block. 2. Determine the ultimate moment of
= .−. = ..−. . = 0.00214
capacity of the beam. 3. Determine the safe live concentrated load that the beam could support at its midspan if it has a span of 6-m. Assume weight of concrete to be 23.5 kN/m.
Solution:
1. Depth of the compression block.
= . = = 0.00207 > > a = c a = 0.85 (217.29) a = 184.70 mm
2. Ultimate moment capacity of the beam
. = −. = .−.
= ′ = .−. = −.
. = − = .− = .−. . =
0.00529 > 0.005
Note:
< 0.005 Use ø = 0.65 + ( – 0.002) When
= 0.90 2782414.6600 . 1250414.660062.5 = 777.7 10 . = . . 3. Safe concentrated live load at mid span of the beam:
(1.4) Mu = (1.7) Mu =
WDL = 23.5 (0.35) (0.675) WDL = 5.55 kN/m Use ø = 0.90
(1.4) . (1.4) 777.7 = (1.7) Mu = (1.7)
P = 291.27 kN
As’ fy = As2 fy As’ = As2 As = As1 + As2 As1 = As – As2 As1 = As – As’ As1 = 4032 – 1250 As1 = 2782 mm2
= ∅ = 0.90
