l - 01 introduction

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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

1



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

2





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

3



Grading Policy

Midterm = 2 5 %

Final Term = 5 0 %

Session Performance = 2 5 %

Assignments = 10 % (6 Assignments )

Quizzes = 15 % (6 Quizzes)

4





Lectures Availability

All lectures and related material will be available on

the website:

www.drqaisarali.com

5



Lecture-01

Introduction

By: Prof Dr. Qaisar Ali

Civil Engineering DepartmentUET Peshawar [email protected]

6





Topics

Concept of Capacity and Demand

Flexure Design of Beams using ACI Recommendations

Shear Design of Beams using ACI Recommendations

Example

7




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.

8






Capacity

The overall ability of a structure to carry an imposed

demand.

9

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?

10






Failure

Reinforced Beam Test Video

11

Failure (Capacity < 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 pad50 Tons

12″

12″






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.

13

50 TonsDemand 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 loaddivided by the area of the pad.

Load Effect=(50 × 2204)/ (12 × 12)

= 765.27 psi

12″12″

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).




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?

14

50 Tons

12″

12″






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.

15




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.

16






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?

17




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.

18






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.

Note: We shall be following ACI 318-02 throughout this course

19




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.

20






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.

21




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 (f c′) = 3 ksi, therefore

Allowable strength = f c′/2 = 1.5 ksi (for Working Stress method)

22

Concrete pad200 Tons

12″

12″






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.

23


12″

12″




Example 1.3

Solution:

Let us assume that we want to use steel bar reinforcement of yield strength f y =

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 = Acf c′/2 + Asf y/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).

24


12″

12″






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 (f c′ ) =3ks i

25

2.0 kip/ft

20″

12″

Beam section

20′-0″




Example 1.4

Solution:

M = wl 2/8 = {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

26






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).

Therefore, ƒtension = 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

27




Example 1.5

Check the shear capacity of the same beam.

28

2.0 kip/ft

20′-0″Solution:

Shear Demand:

Vu = (20/10) × {10 – (17.5/12)} = 17.1 kips

Shear Capacity:

Vc = 2 f′ bh (ACI 11.3.1.1)

= 2 3000 ×12×20/1000

= 26.29 kips > 17.1 kips

20 kips17.1 kips

17.5″ = 1.46′





Flexural Design of Beams Using ACI

Recommendations Design:

ΦMn ≥ Mu (ΦMn is Mdesign or Mcapacity)

For ΦMn = Mu

As = Mu/ {Φf y (d – a/2)}

31



Flexural Design of Beams Using ACIRecommendations

Design:

ρmin = 3 f c′ /f y ≥ 200/f y (ACI 10.5.1)

ρmax = 0.85β1(f c′/f y){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 f c′ ≤ 4000 psi, ACI 10.2.7.3)

32





Flexural Design of Beams Using ACI

Recommendations Design:

ρmax and ρmin for various values of f c′ and f y

33

Table 01: Maximum & Minimum Reinforcement Ratios

f c′ (psi) 3000 4000 5000

f y (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



Shear Design of Beams using ACIRecommendations

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.

34





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 = Avf y/(50bw),

d/2

24 inches

Avf y/ {0.75 f′ bw}

35




When ΦVc < Vu, web reinforcement is required as:

Vu = ΦVc + ΦVs

ΦVs = Vu – ΦVc

ΦAvf yd / s=Vu – ΦVc

s = Φ Avf yd/(Vu – ΦVc)

36





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 satisf ied, reduce maximum spacing requirement by onehalf.

37




38





Flexural and Shear Design of Beam as per ACI:

Design the beam shown below as per ACI 318-02.

39

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

20′-0″

Example 1.6

Take f ′ c = 3 ksi & f y = 40 ksi



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)

40

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= 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

41

Example 1.6




Solution:

Step No. 03: Analysis.

Flexural Analysis:

Mu = Wu l 2/8= 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

42

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:

ΦMn ≥ Mu (ΦMn is Mdesign or Mcapacity)

For ΦMn = Mu

ΦAsf y(d – a/2) = Mu

As= M

u/ {Φf

y(d – a/2)}

Calculate “As” by trial and success method.

43



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 = Asf y/ (0.85f c′bw)

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

44





Example 1.6


Solution:


Design for flexure:

Second Trial:

Third Trial:

“Close enough to the previous value of “a” so that As = 4.37in2 O.K

45

• 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

• A

s= 2370.24 / [0.9 × 40 × {17.5 – (4.88/2)}] = 4.37 in2

• a = 4.37 × 40/ (0.85 × 3 × 14) = 4.90 inches



Example 1.6


Solution:


Design for flexure:

Check for maximum and minimum reinforcement allowed by ACI:

ρmin = 3 f′ /f y ≥ 200/f y (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





Example 1.6


Solution:


Design for flexure:

ρmax = 0.85β1(f c′/f y){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 f c′ ≤ 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



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.

48





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.

49



Example 1.6


Solution:


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)

As ΦVc < Vu, Shear reinforcement is required.

50





Example 1.6


Solution:


Design for Shear:

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

Spacingrequired (s) = ΦAvf yd/ (Vu – ΦVc)

= 0.75×0.22×40×17.5/ (33.74–20.13) ≈ 8.5″

51



Example 1.6


Solution:


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

= Av

f y

/(50bw

) =0.22 × 40000/(50 × 14) = 12.57″

smax = d/2 = 17.5/2 = 8.75″

smax = 24″

Avf y/ 0.75√(f c′)bw = 0.22×40000/ {(0.75×√(3000)×14} =15.30″

Therefore smax = 8.75″

52





Example 1.6


Solution:


Design for Shear:

Other checks:

Check for depth of beam:

ΦVs ≤ Φ8 f′ bwd (ACI 11.5.6.9)

Φ8 f′ bwd = 0 . 7 5 × 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



Example 1.6


Solution:


Design for Shear:

Other checks:

Check if “ΦVs ≤ Φ4 f′ bwd” (ACI 11.5.4.3):

If “ΦVs ≤ Φ4 f′ bwd”, the maximum spacing (smax) is O.K. Otherwise

reduce spacing by one half.

13.61 kips < 40.26 kips O.K.

54





Example 1.6


Solution:


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 = (ΦAvf yd)/ 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)

55



Example 1.6

56

#3 @ 8.5″ c/c#3 @ 8.75″ c/c

Not Required

theoretically

ΦVn





Example 1.6


Solution:

Step No. 05: Drafting.

57

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.



References

ACI 318-02

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

Darwin and Dolan

58





Appendix

59



Appendix

60





Appendix

61



The End

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l - 01 introduction

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