The document outlines the course content and structure for a Reinforced Concrete Design-II class taught by Prof. Dr. Qaisar Ali, including topics that will be covered in the midterm and final terms such as one-way slab design, two-way slab design, earthquake design, and retaining walls. The grading policy and availability of course materials online are also mentioned. The document serves as an introduction and overview for students taking the Reinforced Concrete Design-II course.

1. Department of Civil Engineering, University of Engineering and Technology Peshawar
Reinforced Concrete
Design-II
By: Prof Dr. Qaisar Ali
Civil Engineering Department
UET Peshawar
www.drqaisarali.com
Prof. Dr. Qaisar Ali
Reinforced Concrete Design – II
1
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


Prof. Dr. Qaisar Ali
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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2. 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 various Types of Retaining Walls and Design
of Cantilever RW

Introduction to Bridge Engineering

Design of Stairs
Prof. Dr. Qaisar Ali
Reinforced Concrete Design – II
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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)
Prof. Dr. Qaisar Ali
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3. 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
Prof. Dr. Qaisar Ali
Reinforced Concrete Design – II
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Department of Civil Engineering, University of Engineering and Technology Peshawar
Lecture-01
Introduction
By: Prof Dr. Qaisar Ali
Civil Engineering Department
UET Peshawar
drqaisarali@nwfpuet.edu.pk
Prof. Dr. Qaisar Ali
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4. Department of Civil Engineering, University of Engineering and Technology Peshawar
Topics
 Concept of Capacity and Demand
 Flexure Design of Beams using ACI Recommendations
 Shear Design of Beams using ACI Recommendations
 Example
Prof. Dr. Qaisar Ali
Reinforced Concrete Design – II
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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.
Prof. Dr. Qaisar Ali
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5. 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.
Beam
will
resist
the
Applied Load
(Demand)
applied load up to its
capacity and will fail
when demand exceeds
capacity
Prof. Dr. Qaisar Ali
Reinforced Concrete Design – II
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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?
Prof. Dr. Qaisar Ali
Reinforced Concrete Design – II
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6. Department of Civil Engineering, University of Engineering and Technology Peshawar
Concept of Capacity and Demand
 Failure
Failure (Capacity < Demand)

Reinforced Beam Test Video
Prof. Dr. Qaisar Ali
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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″.
50 Tons
Concrete pad
12″
12″
Prof. Dr. Qaisar Ali
Reinforced Concrete Design – II
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7. 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.
50 Tons
Demand in the form of load:
Capacity of the pad in the form
Load = 50 Tons
of resistance should be able to
carry a stress of 765.27 psi.
Demand in the form of Load effects:
In other words, the compressive
The effect of load on the pad will be
12″
a compressive stress equal to load
strength
of
concrete
pad
(capacity) should be more than
divided by the area of the pad.
12″
Load Effect=(50 × 2204)/ (12 × 12)
765.27 psi (demand).
= 765.27 psi
Prof. Dr. Qaisar Ali
Reinforced Concrete Design – II
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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?
50 Tons
12″
12″
Prof. Dr. Qaisar Ali
Reinforced Concrete Design – II
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8. 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.
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.

Capacity/ Demand = 500 / 765.27 = 0.653 (Failure)
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.
Prof. Dr. Qaisar Ali
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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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9. 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?
Prof. Dr. Qaisar Ali
Reinforced Concrete Design – II
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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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10. 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 318-02 (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.

Prof. Dr. Qaisar Ali
Note: We shall be following ACI 318-02 throughout this course
Reinforced Concrete Design – II
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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.
Prof. Dr. Qaisar Ali
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11. 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.
Prof. Dr. Qaisar Ali
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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)
200 Tons
Concrete pad
12″
12″
Prof. Dr. Qaisar Ali
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12. Department of Civil Engineering, University of Engineering and Technology Peshawar
Concept of Capacity and Demand
200 Tons
Concrete pad
 Example 1.3

12″
Solution:
12″

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.
Prof. Dr. Qaisar Ali
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Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Concept of Capacity and Demand
200 Tons
Concrete pad
 Example 1.3

12″
Solution:

12″
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).
Prof. Dr. Qaisar Ali
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13. Department of Civil Engineering, University of Engineering and Technology Peshawar
Concept of Capacity and Demand
 Example 1.4

Check the capacity of the concrete beam given in figure below
against flexural stresses within the linear elastic range.
Concrete compressive strength (fc′) = 3 ksi
2.0 kip/ft
20″
20′-0″
12″
Beam section
Prof. Dr. Qaisar Ali
Reinforced Concrete Design – II
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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:


Prof. Dr. Qaisar Ali
ƒ = My/I (linear elastic range)
Therefore, M = ƒI/y
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14. 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′ , (ACI 9.5.2.3).
f′ = 7.5 × 3000 = 411 psi

Therefore, ƒtension = 7.5

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
Prof. Dr. Qaisar Ali
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Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Concept of Capacity and Demand
 Example 1.5

Check the shear capacity of the same beam.
2.0 kip/ft
Solution:
20′-0″
Shear Demand:
Vu = (20/10) × {10 – (17.5/12)} = 17.1 kips
20 kips
17.1 kips
Shear Capacity:
Vc = 2
f′ bh (ACI 11.3.1.1)
= 2 3000 ×12×20/1000
17.5″ = 1.46′
= 26.29 kips > 17.1 kips
Prof. Dr. Qaisar Ali
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15. Department of Civil Engineering, University of Engineering and Technology Peshawar
Flexural Design of Beams Using ACI
Recommendations
 Load combinations: ACI 318-02, Section 9.2.
Load Combinations: ACI 318-02, Section 9.2.
U = 1.4(D + F)
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
Prof. Dr. Qaisar Ali
(9-1)
(9-7)
Reinforced Concrete Design – II
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Department of Civil Engineering, University of Engineering and Technology Peshawar
Flexural Design of Beams Using ACI
Recommendations
 Strength Reduction Factors: ACI 318-02, Section 9.3.
Prof. Dr. Qaisar Ali
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16. 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)}
Prof. Dr. Qaisar Ali
Reinforced Concrete Design – II
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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 10.3.5). When εt = 0.005, Φ = 0.9 for
flexural design.
β1= 0.85 (for fc′ ≤ 4000 psi, ACI 10.2.7.3)
Prof. Dr. Qaisar Ali
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17. 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
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
Prof. Dr. Qaisar Ali
Reinforced Concrete Design – II
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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 11.5.5.1 recommends minimum web
reinforcement.
Prof. Dr. Qaisar Ali
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18. 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 11.5.4 and
11.5.5.3 shall be minimum of:

smax = Avfy/(50bw),

d/2

24 inches

Avfy/ {0.75 f′ bw}
Prof. Dr. Qaisar Ali
Reinforced Concrete Design – II
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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)
Prof. Dr. Qaisar Ali
Reinforced Concrete Design – II
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19. 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′ bwd (ACI 11.5.6.9)
If not satisfied, increase depth of beam.
 Check for Spacing:
ΦVs ≤ Φ4
f′ bwd (ACI 11.5.4.3)
If not satisfied, reduce maximum spacing requirement by one
half.
Prof. Dr. Qaisar Ali
Reinforced Concrete Design – II
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Department of Civil Engineering, University of Engineering and Technology Peshawar
Shear Design of Beams using ACI
Recommendations
Prof. Dr. Qaisar Ali
Reinforced Concrete Design – II
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20. Department of Civil Engineering, University of Engineering and Technology Peshawar
Example 1.6
 Flexural and Shear Design of Beam as per ACI:

Design the beam shown below as per ACI 318-02.
W D.L = 1.0 kip/ft
W L.L = 1.5 kip/ft
20′-0″
Take f ′c = 3 ksi & fy = 40 ksi
Prof. Dr. Qaisar Ali
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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)
W D.L = 1.0 kip/ft
W L.L = 1.5 kip/ft
20′-0″
20″
14″
Beam section
Prof. Dr. Qaisar Ali
Reinforced Concrete Design – II
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21. 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. 02: Loads.

Self weight of beam = γcbwh = 0.15 × (14 × 20/144) = 0.292 kips/ft

W u = 1.2D.L + 1.6L.L (ACI 9.2)
= 1.2 × (1.0 + 0.292) + 1.6 × 1.5 = 3.9504 kips/ft
Prof. Dr. Qaisar Ali
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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. 03: Analysis.

Flexural Analysis:
Mu = W u l2/8 = 3.9504 × (20)2 × 12/8 = 2370.24 in-kips
3.9504 kip/ft

Analysis for Shear in beam:
Vu = 39.5 × {10 – (17.5/12)}/10 = 33.74 k
33.74 kips
39.50
SFD
2370.24
BMD
Prof. Dr. Qaisar Ali
Reinforced Concrete Design – II
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22. 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.
Prof. Dr. Qaisar Ali
Reinforced Concrete Design – II
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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)

Prof. Dr. Qaisar Ali
= 4.25 × 40/ (0.85 × 3 × 14) = 4.76 inches
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23. 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:
• 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
Third Trial:
• a = 4.37 × 40/ (0.85 × 3 × 14) = 4.90 inches

“Close enough to the previous value of “a” so that As = 4.37 in2 O.K
Prof. Dr. Qaisar Ali
Reinforced Concrete Design – II
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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:


ρmin = 3

3 × 3000 /40000 = 0.004

200/40000 = 0.005

Therefore, ρmin = 0.005

Prof. Dr. Qaisar Ali
Check for maximum and minimum reinforcement allowed by ACI:
Asmin = ρminbwd = 0.005 × 14 × 17.5 = 1.225 in2
f′ /fy ≥ 200/fy (ACI 10.5.1)
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24. 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 10.3.5). 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
Prof. Dr. Qaisar Ali
Reinforced Concrete Design – II
47
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,


6 #8 bars (4.74 in2),

Prof. Dr. Qaisar Ali
8 #7 bars (4.80 in2),
or combination of two different size bars.
Reinforced Concrete Design – II
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24

25. 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.
Prof. Dr. Qaisar Ali
49
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′ bwd
= 0.75×2× 3000 ×14×17.5/1000 = 20.13 k (Φ=0.75, ACI 9.3.2.3)

Prof. Dr. Qaisar Ali
As ΦVc < Vu, Shear reinforcement is required.
Reinforced Concrete Design – II
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25

26. 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 (s) = ΦAvfyd/ (Vu – ΦVc)
= 0.75×0.22×40×17.5/ (33.74–20.13) ≈ 8.5″
Prof. Dr. Qaisar Ali
Reinforced Concrete Design – II
51
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 11.5.4 and 11.5.5.3 is minimum of:

smax = 24″

Prof. Dr. Qaisar Ali
smax = d/2 = 17.5/2 = 8.75″


smax = Avfy/(50bw) =0.22 × 40000/(50 × 14) = 12.57″

Avfy/ 0.75√(fc′)bw = 0.22×40000/ {(0.75×√(3000)×14} =15.30″
Therefore smax = 8.75″
Reinforced Concrete Design – II
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26

27. 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
Φ8
f′ bwd (ACI 11.5.6.9)
f′ 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.
Prof. Dr. Qaisar Ali
Reinforced Concrete Design – II
53
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

If “ΦVs ≤ Φ4
f′ bwd” (ACI 11.5.4.3):
f′ bwd”, the maximum spacing (smax) is O.K. Otherwise
reduce spacing by one half.

Prof. Dr. Qaisar Ali
13.61 kips < 40.26 kips O.K.
Reinforced Concrete Design – II
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27

28. 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:

Arrangement of stirrups in the beam: Shear capacity of RC beam is
given as: ΦVn = ΦVc + ΦVs

ΦVc = 20.13 kips

With #3, 2 legged vertical stirrups @ 8.75″ c/c (maximum spacing and
minimum reinf. requirement as permitted by ACI),

ΦVs = (ΦAvfyd)/ smax

ΦVs = (0.75 × 0.22 × 40 × 17.5/8.75) = 13.2 kips

Therefore ΦVn = 20.13 + 13.2 = 33.33 k < (Vu = 33.74 k)
Prof. Dr. Qaisar Ali
55
Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Example 1.6
ΦVn
#3 @ 8.5″ c/c
Prof. Dr. Qaisar Ali
#3 @ 8.75″ c/c
Reinforced Concrete Design – II
Not Required
theoretically
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28

29. 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. 05: Drafting.
Note that some nominal negative
reinforcement has been provided at
the beam ends to care for any
incidental negative moment that may
develop due to partial restrain as a
result of friction etc. between beam
ends and walls. In other words,
though the beam has been analyzed
assuming hinge or roller supports at
the ends, however in reality there will
always be some partial fixity or
restrain at the end.
Prof. Dr. Qaisar Ali
Reinforced Concrete Design – II
57
Department of Civil Engineering, University of Engineering and Technology Peshawar
References
 ACI 318-02
 Design of Concrete Structures (13th Ed.) by Nilson,
Darwin and Dolan
Prof. Dr. Qaisar Ali
Reinforced Concrete Design – II
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30. Department of Civil Engineering, University of Engineering and Technology Peshawar
Appendix
Prof. Dr. Qaisar Ali
Reinforced Concrete Design – II
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Department of Civil Engineering, University of Engineering and Technology Peshawar
Appendix
Prof. Dr. Qaisar Ali
Reinforced Concrete Design – II
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30

31. Department of Civil Engineering, University of Engineering and Technology Peshawar
Appendix
Prof. Dr. Qaisar Ali
Reinforced Concrete Design – II
61
Department of Civil Engineering, University of Engineering and Technology Peshawar
The End
Prof. Dr. Qaisar Ali
Reinforced Concrete Design – II
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