1. Reinforced Concrete Design Module 1
Subject: CE 74B- Reinforced Concrete Design
1. Title of the Module
Introduction and Singly Reinforced Beam
2. Introduction
In this module, the students will be given a clear background on Working Stress
Design, the specifications regarding the same, types of beams, design of rectangular
beams reinforced with tension bars only and investigation of beams both rectangular
and beams with irregular sections.
3. Learning Outcomes
At the end of the course, the students should be able to:
a. be familiar with WSD specifications and apply the same in solving problems with
speed and accuracy
b. be able to design and investigate beams reinforced with tension bars only
c.. investigate beams with irregular cross sections
4. Learning Content
It contains readings, discussions, questions and sets of activities that students can work
on individually or by group.
Topic 1. Introduction with WSD Specifications
Topic 2. Diagonal Tension and Bond Stress
Topic 3. Design and Investigation of Rectangular Beams
Topic 4. Investigation of Beams with Irregular Section
2. Working Stress Design (ACI 318-63)
PHILIPPINE STANDARD REINFORCING BARS
Philippine Near Nominal Unit
Standard ASTM Sectional Weight
Designation Designation Area (mm2
) (kg/m)
6 mm # 2 ( 1/4'') 28.27 0.222
10 mm # 3 ( 3/8'') 78.54 0.616
12 mm # 4 ( 1/2'') 113.10 0.888
16 mm # 5 ( 5/8'') 201.10 1.579
20 mm # 6 1 3/4'') 314.20 2.466
25 mm # 8 (1 ) 491.90 3.854
28 mm # 9 (1 1/8..) 615.75 4.833
32 mm # 10 (1 1/4'') 804.25 6.313
36 mm #11 (1 3/8'') 1017.90 7.991
Philippine Philippine Standard
Standard Compared with
Designation ASTM Standard
6 mm 10.7% smaller than ASTM No.2 or 1/4" diameter
10 mm 10.22% larger than ASTM No.3 or 3/8" diameter
12 mm 10.7% smaller than ASTM No.4 or 1/2" diameter
16 mm 1.6% larger than ASTM No.5 or 5/8" diameter
20 mm 10.22% larger than ASTM No. 6 or 3/4" diameter
25 mm 2.9% smaller than ASTM No. 8 or 1" diameter
28 mm 4.49% smaller than ASTM No.9 or 1 1/8" diameter
32 mm 1.6% smaller than ASTM No. 10 or 1 1/4" diameter
3. 36 mm 0.97% larger than ASTM No. 11 or 1 3/8" diameter
Bar Size
Nominal
Diameter
mm
Nominal
Area
Nominal Mass kg/m
# 10 11.3 100 0.785
# 15 16.0 200 1.570
# 20 19.5 300 2.355
# 25 25.2 500 3.925
# 30 29.9 700 6.495
# 35 35.7 1000 7.850
# 45 43.7 1500 11.775
# 55 66.4 2500 19.625
Minimum Cover for Cast-in Place Concrete Recommended by ACI Code
(ACI 318M-83)
A) If concrete is not exposed to weather:
For beams and columnsβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.. 40mm
For slabs,walls, joists, with No. 35 bars or smallerβ¦β¦β¦β¦β¦β¦β¦β¦. 20mm
For slabs,walls, joists, with No. 45 bars or smallerβ¦β¦β¦β¦β¦β¦β¦β¦. 40mm
For shells and folded plate members with No. 20 or larger barsβ¦β¦.. 20mm
For shells and folded plate members with No. 10 or No. 15 barsβ¦β¦. 15mm
4. B) If concrete is exposed to the weather:
If reinforcing bars are No. 20 or largerβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦ 50mm
If reinforcing bars are No. 10 or No. 15 β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦. 40mm
C) In all concrete cast directly on the earthβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.. 70mm
a) Slabs with beams or drop panelsβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦. 120mm
b) Slabs without beams or drop panelsβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦... 100mm
c) Slabs with beams on all four edgesβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦. 90mm
Temp. bars for slabs. (Area requirements)
As = 0.002 bt for Grade 300 bars
As = 0.0018 bt for Grade 400 bars
Clear distance between bars
a) Not less than the bar diameter nor 25mm for beams
b) Not less than 1.5 bar diameter nor 40mm for columns
c) When bars are placed in several layers, the clear distance between layers
must not be less than 25mm.
Maximum spacing of main reinforcing bars in slabs and walls is equal to 3 times the wall or
slabs thickness or 500mm whichever is smaller.
Spacing of tie wires for columns:
a) Not more than 16 bar diameter
b) Not more than 48 tie diameter
c) Not more than the least dimension of the cross section
Modulus of Elasticity of Concrete:
5. Ec = wc
1.5
0.043βfcβ Mpa for values of wc between 1500-2500 kg/m3
Ec = 4700βfcβ for normal density concrete
fc = 0.45fcβ
fc = 165 MPa for Grade 400
fc = 140 MPa for Grade 300
The ratio 0.45 gives a factor of safety of
1
0.45
= 2.22
Es = 2,040,000 kg/cm2
Es = 200,000 MPa for all types of steel
ALLOWABLE STRESSES IN CONCRETE (S.I. VERSION)
(ACI 318-63)
Description Symbol fcβ=17.78 MPa fcβ=20.73 MPa fcβ=27.65MPa
6. Modulus of n =
πΈπ
πΈπ
10
Elasticity
Flexure:
Compression fcβ = 0.45fcβ 7.78
Tension in
Plain Conc. fc= 0.13βfcβ 0.54
Footing &
Walls
Shear:
Diagonal
tension at
distance βdβ v = 0.09βfcβ 0.37
from face of
support
Joist with
no web v = 0.10βfcβ 0.42
reinforcement
Punching shear
at distance βdβ/2 vp = 0.17βfcβ 0.71
7. Bearing:
On full area fc = 0.25 fcβ 4.32
On one third
area or less fc = 0.375 fcβ 6.48
ALLOWABLE STRESSES IN CONCRETE (S.I. VERSION)
(ACI 318-63)
Description Symbol fcβ=176 kg/cm2
fcβ=211 kg/cm2
fcβ=281 kg/cm2
Modulus of n =
πΈπ
πΈπ
10 9 8
Elasticity
Flexure:
Compression fcβ = 0.45fcβ 79 94.8 128.5
Tension in
Plain Conc. fc= 0.424βfcβ 5.63 6.16 7.11
Footing &
Walls
Shear:
8. Diagonal
tension at
distance βdβ v = 0.292βfcβ 3.87 4.24 4.89
from face of
support
Punching shear
for slabs and vp = 0.53βfcβ 7.03 7.73 8.84
footings
Bearing:
On full area fc = 0.25 fcβ 44 52.7 70.3
On one third
area or less fc = 0.375 fcβ 66 79.1 105.5
Allowable Stresses (Working Stress Design)
A) Modulus of Elasticity of Concrete:
Ec = 4730βfcβ MPa (S.I. System)
Ec = 15100βfcβ kg/cm2 (MKS System)
9. B) Modulus of Elasticity of Steel:
Es = 200,000 MPa (S.I. System)
Es = 2,040,000 kg/cm2
(MKS System)
C) Allowable Shearing Stress:
Va = 0.09βfcβ MPa (S.I. System)
Va = 0.29βfcβ kg/cm2
(MKS System)
D) Allowable Punching Stress:
Vp = 0.17βfcβ MPa (S.I. System)
Vp = 0.53βfcβ kg/cm2
(MKS System)
E) Allowable Bond Stress:
1) For top bars in tension:
Ua =
7.18βππβ
π·
MPa (S.I. System)
Ua =
2.29βππβ
π·
kg/cm2 (MKS System)
2) Other than top bars in tension:
Ua =
10.14βππβ
π·
MPa (S.I. System)
Ua =
3.23βππβ
π·
kg/cm2 (MKS System)
10. 3) For all compression bars:
Ua = 0.54βππβ MPa (S.I. System)
Ua = 1.70βππβ kg/cm2
(MKS System)
ALLOWABLE BOND STRESS
(ACI 318 - 63)
1. For tension bars with sizes and deformations conforming to ASTM A305 top bars
(Horizontal bars placed that more than 300 mm of concrete is cast in the
member below the bar).
Top barsβ¦.u =
7.18βππβ
π·
nor 2.42 MPa β¦u =
2.29βππβ
π·
nor 24.6 kg/cm2
Bars other than top bars β¦.. u =
10.14βππβ
π·
nor 3.46 MPa β¦β¦β¦..
...u =
3.23βππβ
π·
nor 35.2 kg/cm2
For plain bars, use one half of these values but not more than 1.11 MPa or 11.25
kg/cm2
2. For tension bars with sizes and deformations conforming to ASTM A408 top bars
β¦.. u = 0.18βππβMPa β¦.u = 0.556βππβkg/cm2
Bars other than top bars ... . u = 0.25βππβMPa β¦β¦. u = 0.795βππβkg/cm2
11. 3. For all deformed compression bars:
u = 0.54βππβnor 2.76 MPa β¦. u = 1.72 βππβnor 28.1 kg/cm2
WORKING STRESS DESIGN
THE STRAIGHT-LINE OR ELASTIC-LOAD THEORY
This is based from the assumption that plate sections remain plane
and normal to the longitudinal fiber stress before and after bending and
that both materials are elastic. The stress of concrete varies from fc in
the extreme fiber to zero at the neutral axis.
12. fc = compressive unit stress on the concrete at the surface most
remote from the neutral axis.
fs = tensile unit stress in the longitudinal reinforcement.
b = width of beam.
d = depth from the center of the steel bars to the extreme fiber of
concrete.
kd = distance from the neutral axis to the extreme fiber of concrete.
jd = distance between the compressive force C and the tensile force
T.
As = cross-sectional area of steel bars.
p = ratio of the area of steel to that of the effective area of concrete.
Mc = resisting moment of concrete.
Ms = resisting moment of steel.
Es = modulus of elasticity of concrete.
Ξc = deformation per unit length of concrete.
Ξs = deformation per unit length of steel.
n = ratio of modulus of elasticity of steel to that of concrete.
17. 1. Compute live load moment, depending on type of load and
support
π =
π€πΏ2
8
(simply supported)
π =
π€πΏ2
12
(continuous beam)
2. Compute the design constant
πΈπ
πΈπ
= π
ππ = 0.45ππβ²
π =
π
π+
ππ
ππ
π = 1 β
π
3
π =
1
2
ππ ππ
3. Solve for the depth βdβ by assuming a value of βbβ ranging from
250 mm to 300 mm.
π = β
π
π π
4. Add 50 mm to 100 mm to this computed βdβ to take care of the
dead load. Use the relation
π =
π
2
ππ π =
π
1.75
5. Add the dead load to the live load to obtain the total load.
6. Recompute the total moment
7.Recheck the computed π = β
π
π π
(The result would be the same as
that values which you have added 50 mm or 100 mm)
8. Check for shear π£ =
π
ππ
18. 9. If v > va use 10 mm stirrups
10. Spacing of stirrups
π =
π΄π£ππ£
π£β²π
π£β² = π£ β π£π
π£π = .09βππβ²πππ΄
π ππ πππ‘ πππππ‘ππ π‘βππ
π
2
(max. spacing of stirrups)
11. Check for Bond Stress
ππ =
π
β ππ
π
π = max π βπππ
β = πππππππ‘ππ ππ π‘πππ πππ ππππ
π
12. If π’ > ππ, π‘βππ πππ ππππ π π‘πππ ππππ
19.
20. MOMENT COEFFICIENT FOR CONTINUOUS BEAMS AND SLABS
Limitations to this coefficients:
1. Adjacent clear spans should not differ by more than 20% of the shorter pan.
2. Ratio of live load to deadload should not exceed 3.
3. The loads should be uniformly distributed only.
For Positive Moment.
End spans:
If discontinuous end is unrestrained- - - - - - - - - - - - - -
π€πΏ2
11
If discontinuous end is integral with the support- - - - -
π€πΏ2
14
Interior spans- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
π€πΏ2
16
For Negative Moment:
Negative moment at exterior face of first interior support:
Two spans- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
π€πΏβ²2
9
More than two spans- - - - - - - - - - - - - - - - - - - - - - - - -
π€πΏβ²2
10
Negative moment at other faces interior supports- - - -
π€πΏβ²2
11
Negative moment at face of all supports for:
a) Slabs with spans less than 3 m.- - - - - - - - - - - - - - - - -
π€πΏβ²2
12
b) Beams and girders where the ratio of the sum of column stiffness to beam stiffness exceeds 8 at each
end of span- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
π€πΏβ²2
12
Negative moment at interior faces of exterior supports for members built integrally with their supports.
a) When the support is a spandrel beam of girder- - - - - -
π€πΏ2
24
b) When the support is a column- - - - - - - - - - - - - - - - - - - -
π€πΏ2
16
Shear Forces:
a) Shear in end members at first interior support- - - - - - - - 1.15
π€πΏ
2
21. b) Shea0r at all other supports- - - - - - - - - - - - - - - - - - - - - - -
π€πΏ
11
The figure below shows a free body diagram of a segment of a concrete beam dx in length. The vertical
shear V.
βππ΄=0
T(jd) + V(dx) = (T + π₯π)jd (1) π₯π =
π(ππ₯)
ππ
T(jd) + V(dx) β T(jd) = π₯π(jd) βπΉβ= 0
V(dx) = π₯π(jd) H + T = T + π₯π
H = vb(dx) (shearing force along the horizontal plane)
bv(dx) + T = T + π₯π
(2) π₯π = vb(dx)
(1) & (2)
π(ππ₯)
ππ
= vb(dx)
v =
π
πππ
(1956 ACI CODE)
v =
π
ππ
(1963 ACI CODE)
V = maximum shear at a distance βdβ from support
v = shearing stress
d = effective depth
b = width of beam
Allowable shearing stress is
va = 0.09βππβ²MPa
va = 0.09βππβ²kg/cmΒ²
22.
23.
24. DIAGONAL TENSION
When a simple beam is subjected to bending the fibers above the neutral axis are either in
compression or tension while tensile stresses occur in the fibers below or above the neutral axis. In
addition to these stresses, there are also inclined stresses of diagonal stresses which tend to produce
cracks. These cracks appear to be vertical at the center of span and become more inclined as they
approach the supports in which it makes an angle of about 45o
with the horizontal. The stresses that
cause these cracks are called diagonal tension. To resist these diagonal tension, reinforcing bars called
stirrups are used. The vertical stirrup is assumed to carry the vertical component of the diagonal tension
while the longitudinal bars carries the horizontal component of the diagonal tension.
Consider a strip having a length of S(mm).
Assume the diagonal tension as Td acting at an angle of 45o
with the horizontal. T is the component of Td,
both in horizontal and vertical. Since we could not place reinforcement in inclined position, it is then
placed at a vertical position to carry a load of T. But the horizontal force T is the only force that could
resist shearing stress as shown.
T = vβ bs
π΄π£π
π£ = vβ bs
S =
π΄π£ππ£
vβ b
Spacing of stirrups
S = Spacing of stirrups
Av = cross- sectional area of stirrups (10mm)
25. (min. 3/8 dia. with an area of 0.11 in2
but
we would be using U-stirrups therefore
Av = 0.11(2)
Av = 0.22(157mm2
)
π
π¦= allowable tensile stress of stirrups if not given, it is equal to fs.
vβ = v - π£π
v =
π
ππ
(actual shearing stress)
π
π = allowable shearing stress
π
π = 0.29 βππβ² kg/cm2
= 0.09 βππβ² MPa
Max. spacing of stirrups =
π
2
HOW TO DETERMINE THE PORTION OF THE
BEAM REQUIRING WEB REINFORCEMENT
(STIRRUPS)
vβ = v β π£π (excess shearing stress)
π£π = allowable shearing stress
L = span of beam
X = length of beam from support in which
stirrups are required.
26. For Simply Supported Beams
π£β²
π₯
=
π£
πΏ/2
x =
πΏ π£β²
2π£
For Cantilever Beam
π£β²
π₯
=
π£
πΏ
x =
πΏ π£β²
π£
BOND STRESS
β MA = 0
V(dx) + Tjd = (T + βT)jd
V(dx) + Tjd - Tjd = βTjd
βT =
π(ππ₯)
ππ
πd = β0 = perimeter of steel bar
27. F = uπd(dx)
F = uβ0(dx)
βT = F
βT = uβ0(dx)
π(ππ₯)
ππ
= uβ0(dx)
u =
π
β0ππ
(bond stress)
Allowable Bond Stresses:
Tension Bars: (Deformed Bars)
A) ASTM A 305 U =
7.18βππβ²
π·
MPa
Top Bars
Bar other than top bars:
B) ASTM A 408 U =
10.14βππβ²
π·
MPa
Top Bars U = 0.18 βππβ² Mpa
Bars other than top bars: U = 0.25 βππβ² Mpa
Compression bars U = 0.54 βππβ² Mpa
Note: For plain bars, use one half of ASTM A 305 but not more than 1.14 MPa.
28. ANCHORAGE BOND
T = π΄π π
π
F = β0 UL
T = F
π΄π π
π = β0 UL
ππ2
4
= πdUL
L =
πππ
4π’
Where L = length of embedment π
π = tensile stress of steel bars
d = diameter of steel bars u = bond stress of concrete
ELASTIC THEORY IS BASED ON THE FOLLOWING ASSUMPTIONS
1. Plane sections remain plane before and after bending occurs.
2. Concrete is elastic; that is the stress of concrete varies from zero at the neutral axis to a maximum at
the extreme fibers.
3. Concrete is not the food in carrying tension and only the steel bars are carrying all the stresses due to
bending which is tension.
4. There must be no slippage between concrete and steel bars.
29. Balanced design = a design so proportional such that the maximum stresses in the concrete and
steel are reach simultaneously so they fail together.
Under-reinforced design = the steel reinforcement is lesser than what is needed for a balanced
design. In this particular type of design, the steel fails first while the concrete has not yet reached its
allowable values but the failure is gradual with the steel yielding.
Over-reinforced design = the steel reinforcement is more than what is needed for a balanced
design. This type of design is not advisable cause concrete fails suddenly in compression.
CE BOARD β AUGUST 1965
Design a section of a concrete beam reinforced for tension only. The beam is simply supported
on a span of 6 m and carries a load of 18 KN/m. What stirrups will be required? Use ACI Specs, with fcβ =
17.2 MPa, fs = 124 MPa, n = 12. Assume b = d/2
Solution:
fc = 0.45 fcβ
fc = 0.45 (17.2)
fc = 7.74 MPa
k =
π
π+
ππ
ππ
k =
12
12+
124
7.74
k = 0.428 M =
π€πΏ2
8
j = 1 -
π
3
M =
21473(6)2(1000)
8
j = 1 -
0.428
3
M = 96.63 x 106
N.mm
j = 0.857 d = βπ
π π
30. R =
1
2
fc kj d = β
96.63 x 106
1.42(250)
= 522 < 525
R =
7.74(.428)(.857)
2
Steel requirements:
R = 1.42 As =
π
ππ ππ
M =
π€πΏ2
8
As =
96.63 x 106
124(.857)(525)
M =
18000(6)2(1000)
8
As = 1732 ππ2
M = 81 x 106
N.mm Try 28 mmΟ΄;
d = βπ
π π
A =
π
4
(282
) = 615.75 ππ2
Try b = 250 mm 615.75 N = 1732
d = β
81000000
1.42(250)
N = 2.81 say 3 bars
d = 477.7 mm
Try d = 525 mm
Total depth = 525 + 65 = 590mm
Dead Load = 0.25(.59)(2400)(9.81)
Dead Load = 3473 N/m
Total Load = 18000 + 3473 = 21473 N/m
Check for Shear;
V =
1
2
(21473) (6) β 21473 (.525)
V = 53146
v =
π
ππ
v =
53146
250(525)
v = 0.40 MPa
va = 0.09 βfcβ²
va = 0.09 β17.2 = 0.37 MPa
32. CE BOARD β FEBRUARY 1975
A concrete beam reinforced for tension only is required to carry a uniformly distributed load of
27000 N/m (including its own weight) on a simple span of 9 m. The overall depth ββDββ is twice the width
ββbββ and the center of the steel reinforcement is to be located at a distance 1/10 D from the underside of
the beam. Compute the dimension of the beam and the area of reinforcement required. fs = 124 MPa, fc
= 5.53 MPa, n = 15.
Solution:
k =
π
π+
ππ
ππ
k =
15
15+
124
5.53
k = 0.401
j = 1 -
π
3
j = 1 -
.401
3
j = .866
R =
1
2
fc kj
R =
5.53(.401)(.866)
2
R =0.96 Recheck;
M = R b π2
d = βπ
π π
M =
π€πΏ2
8
d = β
273375000
0.96(450)
M =
27000(9)2(1000)
8
d = 795mm < 810 mm (safe)
M = 273375 x 103
N.mm As =
π
ππ ππ
33. 273375 x 103
= 0.96 bπ2
As =
273375000
124(.866)(810)
bπ2
= 284765625 As = 3143 ππ2
(
π·
2
)(0.90D)2
= 284765625 Use 450 mm x 900 mm
D = 889 say 900 mm with As = 3143 πππ
b = 450 mm
d = 0.90(900) = 810 mm
Problem 3.
A reinforced concrete beam rectangular in cross-section, is loaded to produce a
maximum moment of 7100 N-m. if π = 200 ππ, π = 250 ππ, ππβ²
= 4.8 πππ and ππ =
124 πππ, determine As for balanced design.
Solution
πΆ =
1
2
ππβ²
(ππ)(200)
πΆ =
4.8π(250)(200)
2
πΆ = 120000 π
π = πΆππ
π = 120000(1 β
π
3
)(250π)
7100000 = 120000(250π)(1 β
π
3
)
3πβπ2
3
= 0.24
34. 3π β π2
= 0.72
π2
β 3π + 0.72 = 0
π =
3Β±2.40
2
π = 0.26
π = 1 β
π
3
π = 1 β
0.26
3
π = 0.912
π = π΄π ππ ππ
7100000 = π΄π (124)(.912)(250)
π΄π = 251.13 ππ2
PROBLEM:
A rectangular reinforced concrete beam shown, carries a total moment of 119.2 KN.m. If fc = 5.2
Mpa, fs = 138 MPa and n = 12, determine the steel area required for a balanced design.
35. SOLUTION:
fc
x
=
fs
n
450βx
5.2
x
=
138
12
450βx
5.2(450 β x) = 11.5x
x = 140.12 mm
I =
250x3
3
+ nAs (450 - x)2
I =
250(140.123)
12
+ 12As (450 β 140.12)2
I = 229255170.8 + 1152307.4 As
fc =
Mcβ
I
5.2 =
119.2( 140.12 )(10)6
229255170.8 + 1152307.4 As
1192126888 + 5991998.5 As = 119.2 (140.12) (10)6
As = 2588 mm2
36. PROBLEM:
The beam in the figure shown has a 20 Mpa concrete and is reinforced with 4-25 mm bars. If
it sustains a bending moment of 100 kN.m, calculate the actual bending stress in steel concrete n
= 10.
Solution:
2
2
19635
)
4
(
)
25
(
4
10 mm
nAs ο½
ο·
οΈ
οΆ
ο§
ο¨
ο¦
ο½
ο°
38. PROBLEM:
A rectangular simply supported reinforced concrete beam has a cross-section of 250 mm x
500 mm and a simple span of 6 m. It is reinforced with 4 - 20 mm Υ bars in one row, the center
of bars being 60 mm above the bottom of the beam. Using fs = 124 Mpa, fc = 7 Mpa, n = 12,
determine the additional load P that could be carried by this section at midspan besides a
uniform load of 12kN/m including the weight of beam.
Solution:
42. The cross-section shown is reinforced with 2-32mm ΓΈ and 2-28 mm ΓΈ bars
placed in one layer. This beam has single span of 8m and carries the moving loads
shown. Neglecting its own weight, determine whether this beam could carry such
load. fβc= 20 MPa, n= 12, v= 0.4 MPa, u= 1.40 MPa
Solution:
nAs = 12 [
Ο
4
(32)2
(2) +
Ο
4
(28)2
(2) ] = 34079.997 mm2
400x (
x
2
) = 34079.997 ( 600 β x )
x = 245.706 mm
600 β x = 354.294 mm
I =
400 ( 245.706 )3
3
+ 34079.997 ( 354.294 )2
= 6255.682 x 106
mm4
Moment capacity of concrete:
fβc =
Mcx
I
0.45( 20 ) =
Mc ( 246 )
6255.682 x 106
43. Mc = 228.866 kN.m
Moment capacity of steel:
fs
n
=
Ms ( 600βx )
I
124
12
=
Ms ( 354.294 )
6255.682 x 106
Ms = 182.453 kN.m
Safe resisting moment M = 182.453 kN.m
For maximum moment, place the moving loads so that the centerline of the
beam is midway between the biggest load and the resultant of the loads.
44. R = 50 + 40 + 10 = 100 kN
100x = 40(3) + 10(4)
x = 1.6 m
Ξ£MR2 = 0
R1 (8) = 100 (3.2)
R1 = 40 kN
Max. M occurs under the biggest load:
Max. m = 40(3.2) = 128 kN.m Λ 182.453 kN.m (ok)
Check shear:
For maximum shear, place the biggest load almost above the support R1,
Ξ£MR2 = 0
8R1 = 100(6.4)
R1 = 80 kN
Max. shear = R1 = 80 kN
v =
V
bd
=
80 ( 1000 )
400 ( 600 )
= 0.333 MPa
allow. v = 0.4 MPa Λ 0.333 MPa (ok)
Check bond:
u =
V
β o jd
Ξ£o = Ο (32) (2) + Ο (28) (2) = 377 mm
jd = 600 -
x
3
jd = 600 -
245.706
3
= 518.098 mm
u =
80 (1000)
377 (518.098)
= 0.419 MPa Λ 1.4 MPa (ok)
Therefore, the beam could carry the moving loads.
45. Problem: Page 72
What must be the maximum concentrated load at the third points of the
beam shown whose span is 6 m. Assume fβc = 9 MPa, fs = 1224 MPa, n = 12.
46. Solution:
Assume the location of the N.A. as shown.
nAs = 12 (
Ο
2
) (20)2
(4) = 30159.289 mm2
75x (
x
2
) (2) + 150 (
xβ150
2
) (x β 150) = 30159.289 (400-x)
x = 238.716 mm
x β 150 = 88.716 mm
400 β x = 161.284 mm
I =
75 ( 238.716 )3
3
(2) +
150 ( 88.716 )3
3
+ 30159.289 (161.284)3
= 127244.287
x106
mm4
For concrete:
fβc =
Mcx
I
9 =
Mc (238.716)
127244.287 x106
Mc = 4797.326 kN.m
For steel:
fs
n
=
Ms (400βx)
I
124
12
=
Ms ( 161.284 )
127244.287 x106
47. Ms = 8152.437 kN.m
Safe M = 4797.326 kN.m
max. M =
PL
3
+
wL2
8
w = [ 0.3(0.46) β 0.15(.15) ] (9.81) (2.4) = 2.719 kN/m
4797.326 =
P( 6 )
3
+
2.719 ( 6 )2
8
P = 2393.545 kN
Problem:
Find the actual stresses in concrete and steel if the reinforced beam shown
carries a bending moment (due to dead load and live load) equal to 150 kN.m.
Neglect the lack of symmetry n = 10.
48. Solution:
nAs = 10 (
Ο
4
) (28)2
(4) = 24630.086 mm2
24630.086 (750 β x) = 300x (
x
2
) + 200 (x β 150) (
x β 150
2
)
x = 265.702 mm
750 β x = 484.298 mm
X β 150 = 115.702 mm
I =
200 ( 115.702 )3
3
+
300 ( 265.702 )3
3
+ 24630.086 (484.298)3
= 2799697.166 x106
mm4
Stress in concrete:
fβc =
ππ±
π
=
πππ ( ππππ )π ( πππ.πππ )
πππππππ.πππ π±πππ
= 0.014 MPa
49. Stress in steel:
ππ¬
π§
=
π ( πππβπ± )
π
ππ¬
ππ
=
πππ ( ππππ)π ( πππ.πππ )
πππππππ.πππ π±πππ
= 0.026 MPa
5. Teaching and Learning Activities
A. Solve the following problems for mastery of the procedures and
specifications:
1. Design a section of a concrete beam reinforced for tension only. The beam is simply
supported on a span of 10m and carries a load of 120 KN/m. What stirrups will be
required? Use ACI Specs, with fcβ = 17.2 MPa, fs = 124 MPa, n = 12. Assume b = d/2
2. Design a section of a concrete beam reinforced for tension only. The beam is a
continuous beam with same spans of 5m and carries a load of 30 KN/m. What stirrups
will be required? Use ACI Specs, with fcβ = 20MPa, fs = 138MPa, n = 10. Assume 1.75b
= d
Note: Additional problems will also be given on investigation of rectangular and beams
of irregular sections.
B. Project: Prepare the floor plans, elevations and perspective drawings of a
building with a specified number of storeys to be assigned by the professor. The
architectural design of the building must be βmillennialβ or modern. To be submitted to
on Monday of the following week.
6. Recommended learning materials and resources for supplementary reading.
50. Reinforced Concrete Design by Gillesania, Chapter1
Other instructional materials may also be used by students like related textbooks and
videos to enrich their knowledge about the subject.
7. Flexible Teaching Learning Modality (FTLM) adopted
Remote
Asynchronous (modules, exercises, problem sets, etcβ¦)
8. Assessment Task
In this part, students are given Self-assessment Questions (SAQs) and asked to
consider broader aspects of the different topics taken up. Quizzes have been prepared
for this part and can be found in the original module by the author.
9. References
Besavilla, V., Reinforced Concrete Design, 2016
Esplana, Dindo Civil Engineering Review Manual, 2015
Gillesania, DIT Reinforced Concrete Design, 3rd
Edition, 2015
National Structural Code of the Philippines 2015
Nilson, W. Reinforced Concrete Design, 2010