reinforced concrete design-ii - dr.qaisar ali
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Reinforced Concrete
Design-II
By: Prof Dr. Qaisar Ali
Civil Engineering Department
UET Peshawarwww.drqaisarali.com
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Course Content
Mid Term
Introduction
One-Way Slab System Design
ACI Coefficient Method for Analysis of One-Way Slabs
Two Way Slab System Design
ACI Analysis Method for Slabs Supported on Stiff Beams or Walls
ACI Direct Design Method for Slabs with or without Beams
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Course Content
Final Term
Introduction to Earthquake Resistant Design of RC Structures
Introduction to Pre-stressed Concrete
Introduction to Bridge Engineering
Introduction to Retaining Walls
Introduction to Miscellaneous Topics
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Grading Policy
Midterm = 25 %
Final Term = 50 %
Session Performance = 25 %
Assignments = 10 % (6 Assignments )
Quizzes = 15 % (6 Quizzes)
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Lectures Availability
All lectures and related material will be available on
the website:
www.drqaisarali.com
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Lecture-01
Introduction
By: Prof Dr. Qaisar Ali
Civil Engineering Department
UET Peshawarwww.drqaisarali.com
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Concept of Demand and Capacity
Flexural Design of Beams using ACI Recommendations
Shear Design of Beams using ACI Recommendations
Example
7
Topics
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Concept of Capacity and Demand
Demand
Demand on a structure refers to all external actions.
Gravity, wind, earthquake, snow are external actions.
These actions when act on the structure will induce internal
disturbance(s) in the structure in the form of stresses (such
as compression, tension, bending, shear, and torsion).
The internal stresses are also called load effects.
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Concept of Capacity and Demand
Capacity
The overall ability of a structure to carry an imposed
demand.
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Beam will resist the
applied load up to its
capacity and will fail
when demand exceeds
capacity
Applied Load
(Demand)
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Concept of Capacity and Demand
Failure
Occurs when Capacity is less than Demand.
To avoid failure, capacity to demand ratio should be kept
greater than one, or at least equal to one.
It is, however, intuitive to have some margin of safety i.e., to
have capacity to demand ratio more than one. How much?
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Concept of Capacity and Demand
Failure
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Failure (Capacity < Demand)
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Concept of Capacity and Demand
Example 1.1
Calculate demand in the form of stresses or load effects on
the given concrete pad of size 12″ × 12″.
12
Concrete pad
50 Tons
12″
12″
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Concept of Capacity and Demand
Example 1.1
Solution: Based on convenience either the loads or the load
effects as demand are compared to the load carrying
capacity of the structure in the relevant units.
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Demand in the form of load:
Load = 50 Tons
Demand in the form of Load effects:
The effect of load on the pad will be
a compressive stress equal to load
divided by the area of the pad.
Load Effect=(50 × 2204)/ (12 × 12)
= 765.27 psi
Capacity of the pad in the form
of resistance should be able to
carry a stress of 765.27 psi.
In other words, the compressive
strength of concrete pad
(capacity) should be more than
765.27 psi (demand).
50 Tons
12″
12″
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Concept of Capacity and Demand
Example 1.2
Determine capacity to demand ratio for the pad of example
1.1 for the following capacities given in the form of
compressive strength of concrete (i) 500 psi (ii) 765.27 psi
(iii) 1000 psi (iv) 2000 psi. Comment on the results?
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50 Tons
12″
12″
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Concept of Capacity and Demand
Example 1.2
Solution: As calculated in example 1.1, demand = 765.27 psi.
Therefore capacity to demand ratios are as under:
i. Capacity/ Demand = 500 / 765.27 = 0.653 (Failure)
ii. 765.27/ 765.27 = 1.0 (Capacity just equal to Demand)
iii. 1000/ 765.27 = 1.3 (Capacity is 1.3 times greater than Demand)
iv. 2000/ 765.27 = 2.6 (Capacity is 2.6 times greater than Demand)
In (iii) and (iv), there is some margin of safety normally called as
factor of safety.
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Concept of Capacity and Demand
Safety Factor
It is always better to have a factor of safety in our designs.
It can be achieved easily if we fix the ratio of capacity to
demand greater than 1.0, say 1.5, 2.0 or so, as shown in
example 1.2.
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Concept of Capacity and Demand
Safety Factor
For certain reasons, however, let say we insist on a factor of
safety such that capacity to demand ratio still remains 1.0.
Then there are three ways of doing this:
Take an increased demand instead of actual demand (load),
e.g. 70 ton instead of 50 ton in the previous example,
Take a reduced capacity instead of actual capacity such as
1500 psi for concrete whose actual strength is 3000 psi
Doing both.
How are these three situations achieved?
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Concept of Capacity and Demand
Working Stress Method
In the Working Stress or Allowable Stress Design method,
the material strength is knowingly taken less than the actual
e.g. half of the actual to provide a factor of safety equal to
2.0.
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Concept of Capacity and Demand
Strength Design Method
In the Strength Design method, the increased loads and the
reduced strength of the material are considered, but both based on
scientific rationale. For example, it is quite possible that during the
life span of a structure, dead and live loads increase.
The factors of 1.2 and 1.6 used by ACI 318-11 (Building code
requirements for structural concrete, American Concrete Institute
committee 318) as load amplification factors for dead load and live
load respectively are based on probability based research studies.
Note: We shall be following ACI 318-11 throughout this course
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Concept of Capacity and Demand
Strength Design Method
Similarly, the strength is not reduced arbitrarily but
considering the fact that variation in strength is possible due
to imperfections, age factor etc. Strength reduction factors
are used for this purpose.
Factor of safety in Strength Design method is thus the
combined effect of increased load and reduced strength,
both modified based on a valid rationale.
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Concept of Capacity and Demand
About Ton
1 metric ton = 1000 kg or 2204 pound
1 long ton: In the U.S., a long ton = 2240 pound
1 short ton: In the U.S., a short ton = 2000 pound
In Pakistan, the use of metric ton is very common; therefore
we will refer to Metric Ton in our discussion.
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Concept of Capacity and Demand
Example 1.3
Design the 12″ × 12″ pad to carry a load of 200 tons. The
area of the pad cannot be increased for some reasons.
Concrete strength (fc′) = 3 ksi, therefore
Allowable strength = fc′/2 = 1.5 ksi (for Working Stress method)
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Concrete pad
12″
12″
200 Tons
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Concept of Capacity and Demand
Example 1.3
Solution:
Demand in the form of load (P) = 200 Tons = 200 × 2204/1000 = 440.8 kips
Demand in the form of load effects (Stress) = (200 × 2204)/ (12 × 12)
= 3061.11 psi = 3.0611 ksi
Capacity in the form of strength = 1.5 ksi (less than the demand of 3.0611 ksi).
There are two possibilities to solve this problem:
Increase area of the pad (geometry); it cannot be done as required in the example.
Increase the strength by using some other material; using high strength concrete,
steel or other material; economical is to use concrete and steel combine.
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Concept of Capacity and Demand
Example 1.3
Solution:
Let us assume that we want to use steel bar reinforcement of yield strength fy =
40 ksi. Then capacity to be provided combinely by both materials should be at
least equal to the demand. And let us follow the Working Stress approach, then:
{P = Rc + Rs (Demand=Capacity)} (Force units)
Capacity of pad = Acfc′/2 + Asfy/2 (Force units)
Therefore,
440.8 = (144 × 3/2) + (As × 40/2)
As = 11.24 in2 (Think on how to provide this much area of steel? This is how
compression members are designed against axial loading).
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Concept of Capacity and Demand
Example 1.4
Check the capacity of the plain concrete beam given in figure
below against flexural stresses within the linear elastic range.
Concrete compressive strength (fc′) = 3 ksi
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2.0 kip/ft
20″
12″
Beam section
20′-0″
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Concept of Capacity and Demand
Example 1.4
Solution:
M = wl2/8 = {2.0 × (20)2/8} × 12 = 1200 in-kips
Self-weight of beam (w/ft) = (12 × 20 × 0.145/144) = 0.24167 k/ft
Msw (moment due to self-weight of beam) = (0.24167×202×12/8) = 145
in-kips
M (total) = 1200 + 145 = 1345 in-kips
In the linear elastic range, flexural stress in concrete beam can be
calculated as:
ƒ = My/I (linear elastic range)
Therefore, M = ƒI/y
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Concept of Capacity and Demand
Example 1.4
Solution:
y = (20/2) = 10″ ; I = 12 × 203/12 = 8000 in4
ƒ =?
The lower fibers of the given beam will be subjected to tensile
stresses. The tensile strength of concrete (Modulus of rupture) is
given by ACI code as 7.5 f′c, (ACI 18.3.3).
Therefore, ƒtension = 7.5 f′c = 7.5 × 3000 = 411 psi
Hence M = Capacity of concrete in bending = 411 × 8000/ (10 × 1000)
= 328.8 in-kips
Therefore, Demand = 1345 in-kips and Capacity = 328.8 in-kips
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Flexural Design of Beams Using ACI
Recommendations
Load combinations: ACI 318-11, Section 9.2.
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Load Combinations: ACI 318-11, Section 9.2.
U = 1.4(D + F) (9-1)
U = 1.2(D + F + T) + 1.6(L + H) + 0.5(Lr or S or R) (9-2)
U = 1.2D + 1.6(Lr or S or R) + (1.0L or 0.8W) (9-3)
U = 1.2D + 1.6W + 1.0L + 0.5(Lr or S or R) (9-4)
U = 1.2D + 1.0E + 1.0L + 0.2S (9-5)
U = 0.9D + 1.6W + 1.6H (9-6)
U = 0.9D + 1.0E + 1.6H (9-7)
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Flexural Design of Beams Using ACI
Recommendations
Strength Reduction Factors: ACI 318-11, Section 9.3.
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Flexural Design of Beams Using ACI
Recommendations
Design:
ΦMn ≥ Mu (ΦMn is Mdesign or Mcapacity)
For ΦMn = Mu
As = Mu/ {Φfy (d – a/2)}
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Flexural Design of Beams Using ACI
Recommendations
Design:
ρmin = 3 fc′ /fy ≥ 200/fy (ACI 10.5.1)
ρmax = 0.85β1(fc′/fy){εu/(εu + εt)}
Where,
εu = 0.003
εt = Net tensile strain (ACI 10.3.4). When εt = 0.005, Φ = 0.9 for
flexural design.
β1= 0.85 (for fc′ ≤ 4000 psi, ACI 10.2.7.3)
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Flexural Design of Beams Using ACI
Recommendations
Design:
ρmax and ρmin for various values of fc′ and fy
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Table 01: Maximum & Minimum Reinforcement Ratios
fc′ (psi) 3000 4000 5000
fy (psi) 40000 60000 40000 60000 40000 60000
ρmin 0.005 0.0033 0.005 0.0033 0.0053 0.0035
ρmax 0.0203 0.0135 0.027 0.018 0.0319 0.0213
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Shear Design of Beams using ACI
Recommendations
When ΦVc/2 ≥ Vu, no web reinforcement is required.
When ΦVc ≥ Vu, theoretically no web reinforcement is
required. However as long as ΦVc/2 is not greater
than Vu, ACI 11.1.1 recommends minimum web
reinforcement.
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Shear Design of Beams using ACI
Recommendations
Maximum spacing and minimum reinforcement
requirement as permitted by ACI 11.4.5 and 11.4.6
shall be minimum of:
smax = Avfy/(50bw),
d/2
24 inches
Avfy/ {0.75 f′c bw}
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Shear Design of Beams using ACI
Recommendations
When ΦVc < Vu, web reinforcement is required as:
Vu = ΦVc + ΦVs
ΦVs = Vu – ΦVc
ΦAvfyd/s = Vu – ΦVc
s = ΦAvfyd/(Vu – ΦVc)
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Shear Design of Beams using ACI
Recommendations
Check for Depth of Beam:
ΦVs ≤ Φ8 f′cbwd (ACI 11.4.7.9)
If not satisfied, increase depth of beam.
Check for Spacing:
ΦVs ≤ Φ4 f′c bwd (ACI 11.4.5.3)
If not satisfied, reduce maximum spacing requirement by one
half.
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Shear Design of Beams using ACI
Recommendations
37
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Shear Design of Beams using ACI
Recommendations
38
Placement of Shear Reinforcement
Sd = Design Spacing (ΦVc < Vu )
Smax = Maximum Spacing (ΦVc > Vu)
“Vu” is the shear force at distance “d” from the face of the support.
“ΦVc” and “ΦVc/2” are plotted on shear force diagram.
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Flexural and Shear Design of Beam as per ACI:
Design the beam shown below as per ACI 318-11.
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WD.L = 1.0 kip/ft
WL.L = 1.5 kip/ft
20′-0″
Example 1.6
Take f ′c = 3 ksi & fy = 40 ksi
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Example 1.6
Flexural and Shear Design of Beam as per ACI:
Solution:
Step No. 01: Sizes.
For 20′ length, a 20″ deep beam would be appropriate
(assumption).
Width of beam cross section (bw) = 14″ (assumption)
40
20″
14″
Beam section
20′-0″
WD.L = 1.0 kip/ft
WL.L = 1.5 kip/ft
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Flexural and Shear Design of Beam as per ACI:
Solution:
Step No. 02: Loads.
Self weight of beam = γcbwh = 0.15 × (14 × 20/144) = 0.292 kips/ft
Wu = 1.2D.L + 1.6L.L (ACI 9.2)
= 1.2 × (1.0 + 0.292) + 1.6 × 1.5 = 3.9504 kips/ft
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Example 1.6
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Flexural and Shear Design of Beam as per ACI:
Solution:
Step No. 03: Analysis.
Flexural Analysis:
Mu = Wu l2/8 = 3.9504 × (20)2 × 12/8 = 2370.24 in-kips
Analysis for Shear in beam:
Vu = 39.5 × {10 – (17.5/12)}/10 = 33.74 k
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SFD
BMD
3.9504 kip/ft
33.74 kips
2370.24
39.50
Example 1.6
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Example 1.6
Flexural and Shear Design of Beam as per ACI:
Solution:
Step No. 04: Design.
Design for flexure:
ΦMn ≥ Mu (ΦMn is Mdesign or Mcapacity)
For ΦMn = Mu
ΦAsfy(d – a/2) = Mu
As = Mu/ {Φfy (d – a/2)}
Calculate “As” by trial and success method.
43
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Example 1.6
Flexural and Shear Design of Beam as per ACI:
Solution:
Step No. 04: Design.
Design for flexure:
First Trial:
Assume a = 4″
As = 2370.24 / [0.9 × 40 × {17.5 – (4/2)}] = 4.25 in2
a = Asfy/ (0.85fc′bw)
= 4.25 × 40/ (0.85 × 3 × 14) = 4.76 inches
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Example 1.6
Flexural and Shear Design of Beam as per ACI:
Solution:
Step No. 04: Design.
Design for flexure:
Second Trial:
Third Trial:
“Close enough to the previous value of “a” so that As = 4.37 in2 O.K
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• As = 2370.24 / [0.9 × 40 × {17.5 – (4.76/2)}] = 4.35 in2
• a = 4.35 × 40/ (0.85 × 3 × 14) = 4.88 inches
• As = 2370.24 / [0.9 × 40 × {17.5 – (4.88/2)}] = 4.37 in2
• a = 4.37 × 40/ (0.85 × 3 × 14) = 4.90 inches
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Example 1.6
Flexural and Shear Design of Beam as per ACI:
Solution:
Step No. 04: Design.
Design for flexure:
Check for maximum and minimum reinforcement allowed by ACI:
ρmin = 3 f′c /fy ≥ 200/fy (ACI 10.5.1)
3 × 3000 /40000 = 0.004
200/40000 = 0.005
Therefore, ρmin = 0.005
Asmin = ρminbwd = 0.005 × 14 × 17.5 = 1.225 in2
46
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Example 1.6
Flexural and Shear Design of Beam as per ACI:
Solution:
Step No. 04: Design.
Design for flexure:
ρmax = 0.85β1(fc′/fy){εu/(εu + εt)}
εt = Net tensile strain (ACI 10.3.4). When εt = 0.005, Φ = 0.9 for flexural design.
β1= 0.85 (for fc′ ≤ 4000 psi, ACI 10.2.7.3)
ρmax = 0.85 × 0.85 × (3/40) × (0.003/(0.003+0.005) = 0.0204 = 2 % of area of
concrete.
Asmax = 0.0204 × 14 × 17.5 = 4.998 in2
Asmin (1.225) < As (4.37) < Asmax (4.998) O.K
47
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Example 1.6
Flexural and Shear Design of Beam as per ACI:
Solution:
Step No. 04: Design.
Design for flexure:
Bar Placement: 10 #6 bars will provide 4.40 in2 of steel area which is
slightly greater than required.
Other options can be explored. For example,
8 #7 bars (4.80 in2),
6 #8 bars (4.74 in2),
or combination of two different size bars.
48
25
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Example 1.6
Flexural and Shear Design of Beam as per ACI:
Solution:
Step No. 04: Design.
Design for flexure:
Curtailment of flexural reinforcement:
Positive steel can be curtailed 50 % at a distance (l/8) from face of
the support.
For Curtailment and bent up bar details refer to the following figures
provided at the end of this lecture:
Graph A2 and A3 in “Appendix A” of Nilson 13th Ed.
Figure 5.15 of chapter 5 in Nilson 13th Ed.
49
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Example 1.6
Flexural and Shear Design of Beam as per ACI:
Solution:
Step No. 04: Design.
Design for Shear:
Vu = 33.74 kips
ΦVc = (Capacity of concrete in shear) = Φ2 f′c bwd
= 0.75×2× 3000 ×14×17.5/1000 = 20.13 k (Φ=0.75, ACI 9.3.2.3)
As ΦVc < Vu, Shear reinforcement is required.
50
26
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Example 1.6
Flexural and Shear Design of Beam as per ACI:
Solution:
Step No. 04: Design.
Design for Shear:
Assuming #3, 2 legged (0.22 in2), vertical stirrups.
Spacing required (Sd) = ΦAvfyd/ (Vu – ΦVc)
= 0.75×0.22×40×17.5/ (33.74–20.13) ≈ 8.5″
51
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Example 1.6
Flexural and Shear Design of Beam as per ACI:
Solution:
Step No. 04: Design.
Design for Shear:
Maximum spacing and minimum reinforcement requirement as
permitted by ACI 11.4.5 and 11.4.6 is minimum of:
smax = Avfy/(50bw) =0.22 × 40000/(50 × 14) = 12.57″
smax = d/2 = 17.5/2 = 8.75″
smax = 24″
Avfy/ 0.75√(fc′)bw = 0.22×40000/ {(0.75×√(3000)×14} =15.30″
Therefore smax = 8.75″
52
27
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Example 1.6
Flexural and Shear Design of Beam as per ACI:
Solution:
Step No. 04: Design.
Design for Shear:
Other checks:
Check for depth of beam:
ΦVs ≤ Φ8 f′c bwd (ACI 11.4.7.9)
Φ8 f′c bwd = 0.75 × 8 × 3000 × 14 × 17.5/1000 = 80.52 k
ΦVs = Vu – ΦVc = 33.74 – 20.13 =13.61 k < 80.52 k, O.K.
Therefore depth is O.K. If not, increase depth of beam.
53
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Example 1.6
Flexural and Shear Design of Beam as per ACI:
Solution:
Step No. 04: Design.
Design for Shear:
Other checks:
Check if “ΦVs ≤ Φ4 f′c bwd” (ACI 11.4.5.3):
If “ΦVs ≤ Φ4 f′c bwd”, the maximum spacing (smax) is O.K. Otherwise
reduce spacing by one half.
13.61 kips < 40.26 kips O.K.
54
28
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Example 1.6
Flexural and Shear Design of Beam as per ACI:
Step 05: Drafting (Shear Reinforcement)
55
Note:
As Sd ≈ Smax we will provide sd up to 7.5 ft from
the face of support. Beyond this point, theoretically
no reinforcement is required, however, we will
provide #3 2-legged stirrups @ 12 in c/c.
x1
3.9504 kip/ft
33.74 kips39.50
kips 20.13 kips
10.06 kips
x2
x1 = (10.06)(10)/(39.50) ≈ 2.5 ft
x2 = (20.13)(10)/(39.50) ≈ 5.0 ft
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Example 1.6
Flexural and Shear Design of Beam as per ACI:
Step 05: Drafting (Flexural Reinforcement)
56
SECTION B-B
#3, 2 legged
stirrups @ 8.5" c/c20"
14"
2 #4 Bars
(5 + 5) #6 Bars
SECTION C-C
#3, 2 legged
stirrups @ 12" c/c20"
14"
2 #4 Bars
(5 + 5) #6 Bars
SECTION A-A
20"
5 #6 Bars
#3, 2 legged
stirrups @ 8.5" c/c
2 #4 Bars
14"
A
A
B
B
L = 20.0'
L/8 = 2.5'
s/2 = 4.25"
L/8 = 2.5'
#3, 2 legged vertical stirrups @ 8.5" c/c
1" Spacer bars @ 3' c/c
2 #4 bars
5 #6 Bars(5 + 5) #6 Bars
#3, 2 legged vertical stirrups @ 8.5" c/c#3, 2 legged
vertical stirrups @ 12" c/c
7.5' 5.0' 7.5'
C
C
29
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
3D Model
SketchUp Model
57
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
References
ACI 318-11
Design of Concrete Structures (13th Ed.) by Nilson,
Darwin and Dolan
58
30
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Appendix
59
Exact curtailments lengths for simply supported positive moments (to be measured from face of the support)
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Appendix
60
Figure: Cutoff for bars in approximately equal spans with uniformly distributed loads
L1 L2
L1
4
L1
3
L2
3
L2
3
L1
8
L2
8
L2
8
31
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
The End
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