course content design-ii c

8/9/2019 Course Content Design-II C

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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 Peshawar

www.drqaisarali.com

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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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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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Grading Policy

Midterm = 2 5 %

Final Term = 5 0 %

Session Performance = 2 5 %

Assignments = 10 %(6 Assignments )

Quizzes = 15 %(6 Quizzes)

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Lectures Availability

All lectures and related material will be available on

the website:

www.drqaisarali.com

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Lecture-01

Introduction

By: Prof Dr. Qaisar Ali

Civil Engineering DepartmentUET Peshawarwww.drqaisarali.com

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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 exceedscapacity

Applied Load

(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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Example 1.1

Calculate demand in the form of stresses or load effects on

the given concrete pad of size 12 12.

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Concrete pad50 Tons

12

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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 wil l 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


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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

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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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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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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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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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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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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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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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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




Example 1.3

Solution:

Demandin 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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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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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 ) =3ks i

24

2.0 kip/ft

20

12

Beam section

20-0


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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) = (0.2416720212/8) = 145

in-kips

M (total) = 1200 + 145 = 1345 in-kips

In the linear elastic range, flexural stress in concrete beam can be

calculated as: = My/I (linear elastic range)

Therefore, M = I/y

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Example 1.4

Solution:

y = (20/2) = 10 ; I = 12 203/12 = 8000 in4

=?

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 18.3.3).

Therefore, tension = 7.5 f = 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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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(Lror S or R) (9-2)

U= 1.2D + 1.6(Lror S or R) + (1.0L or 0.8W) (9-3)

U= 1.2D + 1.6W+ 1.0L + 0.5(Lror 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)



Flexural Design of Beams Using ACIRecommendations

Strength Reduction Factors:ACI 318-11, Section 9.3.

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Recommendations Design:

Mn Mu (Mn is Mdesign or Mcapacity)

For Mn = Mu

As = Mu/ {fy (d a/2)}

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Flexural Design of Beams Using ACIRecommendations

Design:

min = 3 fc /fy 200/fy (ACI 10.5.1)

max = 0.851(fc/fy){u/(u + t)}

Where,

u = 0.003

t = Net tensile strain(ACI 10.3.4). When t = 0.005, = 0.9 forflexural design.

1= 0.85 (for fc 4000 psi,ACI 10.2.7.3)

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Recommendations Design:

max and min for various values of fc and f y

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Table 01: Maximum & Minimum Reinforcement Ratios

fc (psi) 3000 4000 5000

fy (psi) 40000 60000 40000 60000 40000 60000

min 0.005 0.0033 0.005 0.0033 0.0053 0.0035

max 0.0203 0.0135 0.027 0.018 0.0319 0.0213



Shear Design of Beams using ACIRecommendations

When Vc/ 2 Vu, no web reinforcement is required.

When Vc Vu, theoretically no web reinforcement is

required. However as long as Vc/2 is not greater

than Vu, ACI 11.1.1 recommends minimum web

reinforcement.

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Shear Design of Beams using ACI

Recommendations

Maximum spacing and minimum reinforcement

requirement as permitted byACI 11.4.5 and 11.4.6

shall be minimum of:

smax = Avfy/(50bw),

d/2

24 inches

Avfy/ {0.75 f bw}

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When Vc < Vu, web reinforcement is required as:

Vu = Vc + Vs

Vs = Vu Vc

Avfyd / s=Vu Vc

s = Avfyd/(Vu Vc)

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Recommendations

Check for Depth of Beam:

Vs 8 fbwd (ACI 11.4.7.9)

If not satisfied, increase depth of beam.

Check for Spacing:

Vs 4 fbwd (ACI 11.4.5.3)

If not satisf ied, reduce maximum spacing requirement by onehalf.

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Recommendations

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Placement of Shear Reinforcement

Sd = Design Spacing (Vc < Vu )Smax = Maximum Spacing (Vc > Vu)

(Vc/2> Vu)

Vu is the shear force at distance d from the face of the support.Vc and Vc/2 are plotted on shear force diagram.



Flexural and Shear Design of Beam as per ACI:

Design the beam shown below as perACI 318-11.

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WD.L = 1.0 kip/ftWL.L = 1.5 kip/ft

20-0

Example 1.6

Take f c = 3 ksi & fy = 40 ksi


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Example 1.6


Solution:

Step No. 01: Sizes.

For 20 length, a 20 deep beam would be appropriate

(assumption).

Width of beam cross section (bw) = 14 (assumption)

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20

14

Beam section

20-0

WD.L = 1.0 kip/ftWL.L = 1.5 kip/ft




Solution:

Step No. 02: Loads.

Selfweight of beam= cbwh = 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


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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



Example 1.6


Solution:

Step No. 04: Design.

Design for flexure:

Mn Mu (Mn is Mdesign or Mcapacity)

For Mn = Mu

Asfy(d a/2) = Mu

As = Mu/ {fy (d a/2)}

Calculate As by trial and success method.

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Example 1.6


Solution:


Design for flexure:

First Trial:

Assume a = 4

As = 2370.24 / [0.9 40 {17.5 (4/2)}] = 4.25 in2

a = Asfy/ (0.85f

cb

w)

= 4.25 40/ (0.85 3 14) = 4.76 inches

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Example 1.6


Solution:


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


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Example 1.6


Solution:


Design for flexure:

Check for maximum and minimum reinforcement allowed by ACI:

min = 3 f /fy 200/fy (ACI 10.5.1)

3 3000 /40000 = 0.004

200/40000 = 0.005

Therefore, min = 0.005

Asmin = minbwd = 0.005 14 17.5 = 1.225 in2

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Example 1.6


Solution:


Design for flexure:

max = 0.851(fc/fy){u/(u + t)}

t = Net tensile strain (ACI 10.3.4). When t = 0.005, = 0.9 for flexural design.

1= 0.85(for fc 4000 psi,ACI 10.2.7.3)

max = 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

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Example 1.6


Solution:


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 #7bars (4.80 in2),

6 #8bars (4.74 in2),

or combination of two different size bars.

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Example 1.6


Solution:


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.

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Example 1.6


Solution:


Design for Shear:

Vu = 33.74 kips

Vc = (Capacity of concrete in shear) = 2 f bwd

= 0.752 3000 1417.5/1000 = 20.13 k (=0.75,ACI 9.3.2.3)

As Vc < Vu, Shear reinforcement is required.

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Example 1.6


Solution:


Design for Shear:

Assuming #3, 2 legged (0.22 in2), vertical stirrups.

Spacing required (Sd) = Avfyd/ (Vu Vc)

= 0.750.224017.5/ (33.7420.13) 8.5

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Example 1.6


Solution:


Design for Shear:

Maximum spacing and minimum reinforcement requirement as

permitted byACI 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 = 0.2240000/ {(0.75(3000)14} =15.30

Therefore smax = 8.75

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Example 1.6


Solution:


Design for Shear:

Other checks:

Check for depth of beam:

Vs 8 f bwd (ACI 11.4.7.9)

8 f bwd = 0 . 7 5 8 3000 14 17.5/1000 = 80.52 k

Vs = Vu Vc = 33.74 20.13 =13.61 k < 80.52 k, O.K.

Therefore depth is O.K. If not, increase depth of beam.

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Example 1.6


Solution:


Design for Shear:

Other checks:

Check if Vs 4 fbwd (ACI 11.4.5.3):

If Vs 4 fbwd, the maximum spacing (smax) is O.K. Otherwise

reduce spacing by one half. 13.61 kips < 40.26 kips O.K.

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Example 1.6

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As Sd Smax we will provide Sdthough out


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Example 1.6


Solution:

Step No. 05: Drafting.

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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 fr iction etc. between beam

ends and walls. In other words,

though the beam has been analyzed

assuming hinge or roller supports atthe ends, however in reality there will

always be some partial fixity or

restrain at the end.



3D Model

SketchUp Model

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References

ACI 318-11

Design of Concrete Structures (13th Ed.) by Nilson,

Darwin and Dolan

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Appendix

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Appendix

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Appendix

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The End

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course content design-ii c

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