Lecture 11 – Analysis Lecture 11 – Analysis and Designand DesignFebruary 7, 2003CVEN 444

Lecture GoalsLecture Goals

One-way Slab designResistance Factors and LoadsDesign of Singly Reinforced Rectangular Beam Unknown section dimensions Known section dimensions

Pattern LoadsPattern LoadsUsing influence lines to determine pattern loads

Largest moments in a continuous beam or frame occur when some spans are loaded and others are not.

Influence lines are used to determine which spans to load and which spans not to load.

Pattern LoadsPattern Loads

Influence Line: graph of variation of shear, moment, or other effect at one particular point in a structure due to a unit load moving across the structure.

Pattern Pattern LoadsLoads

Quantitative Influence Lines Ordinate

are calculated (“exact”)

MacGregor (1997)

Pattern LoadsPattern LoadsQualitative Influence Lines Mueller-Breslau Principle

Used to provide a qualitative guide to the shape of the influence line

Pattern LoadsPattern LoadsQualitative Influence Lines (cont.) For moments

Insert pin at location of interestTwist beam on either side of pinOther supports are unyielding, so

distorted shape may be easily drawn.

For frames, joints are assumed free to rotate, assume members are rigidly connected (angle between members does not change)

Qualitative Influence Lines The Mueller-Breslau principle can be stated as follows:

If a function at a point on a structure, such as reaction, or shear, or moment is allowed to act without restraint, the deflected shape of the structure, to some scale, represents the influence line of the function.

Pattern LoadsPattern LoadsQualitative Influence Lines

Fig. 10-7 (b,f) from MacGregor (1997)

Pattern Pattern LoadsLoads

Frame Example:• Maximize +M at point B.• Draw qualitative

influence lines.

• Resulting pattern load:“checkerboard pattern”

Pattern LoadsPattern LoadsArrangement of Live Loads – (ACI 318-02, Sec. 8.9.1) It shall be permitted to assume that:

The live load is applied only to the floor or roof under consideration, and

The far ends of columns built integrally with the structure are considered to be fixed.

Pattern LoadsPattern LoadsArrangement of Live Loads – ACI 318-99, Sec. 8.9.2: It shall be permitted to assume that the

arrangement of live load is limited to combinations of:Factored dead load on all spans with

full factored live load on two adjacent spans.

Factored dead load on all spans with full factored live load on alternate spans.

MomentMomentEnvelopesEnvelopes

Fig. 10-10; MacGregor (1997)

The moment envelope curve defines the extreme boundary values of bending moment along the beam due to critical placements of design live loading.

MomentMomentEnvelopes ExampleEnvelopes ExampleGiven following beam with a dead load of 1 k/ft and live load 2 k/ft obtain the shear and bending moment envelopes

MomentMomentEnvelopes ExampleEnvelopes ExampleUse a series of shear and bending moment diagrams

Wu = 1.2wD + 1.6wL

0

1

2

3

4

5

0 5 10 15 20 25 30 35 40

(ft)

kips

-80

-60

-40

-20

0

20

40

60

80

0 5 10 15 20 25 30 35 40

ft

kips

-250

-200

-150

-100

-50

0

50

100

150

0 5 10 15 20 25 30 35 40

ft

k-ft

Shear Diagram Moment Diagram

MomentMomentEnvelopes ExampleEnvelopes ExampleUse a series of shear and bending moment diagrams

Wu = 1.2wD + 1.6wL

Shear Diagram Moment Diagram

0

0.2

0.4

0.6

0.8

1

1.2

1.4

0 5 10 15 20 25 30 35 40

ft

k/ft

-20

-15

-10

-5

0

5

10

15

20

0 5 10 15 20 25 30 35 40

ft

kips

-80

-60

-40

-20

0

20

40

0 5 10 15 20 25 30 35 40

ft

k-ft

MomentMomentEnvelopes ExampleEnvelopes ExampleUse a series of shear and bending moment diagrams

Wu = 1.2wD + 1.6wL

Shear Diagram Moment Diagram

00.5

11.5

22.5

33.5

44.5

5

0 5 10 15 20 25 30 35 40

ft

k/ft

-60-50-40-30-20-10

01020304050

0 5 10 15 20 25 30 35 40

ft

kips

-200

-150

-100

-50

0

50

100

150

200

0 5 10 15 20 25 30 35 40

ft

k-ft

MomentMomentEnvelopes ExampleEnvelopes ExampleThe shear envelope

MomentMomentEnvelopes ExampleEnvelopes ExampleThe moment envelope

Moment Envelope

-300

-200

-100

0

100

200

0 5 10 15 20 25 30 35 40

ft

k-ft

Minimum Moment Maximum Moment

Approximate Analysis of Continuous Approximate Analysis of Continuous Beam and One-Way Slab Beam and One-Way Slab

SystemsSystemsACI Moment and Shear Coefficients

Approximate moments and shears permitted for design of continuous beams and one-way slabsSection 8.3.3 of ACI Code

Approximate Analysis of Approximate Analysis of Continuous Beam and One-Way Continuous Beam and One-Way Slab SystemsSlab SystemsACI Moment and Shear Coefficients -

Requirements:Two or more spansApproximately Equal Spans Larger of 2 adjacent spans not greater

than shorter by > 20%Uniform LoadsLL/DL 3 (unfactored)

Approximate Analysis of Approximate Analysis of Continuous Beam and One-Way Continuous Beam and One-Way Slab SystemsSlab SystemsACI Moment and Shear Coefficients -

Requirements: ( cont.)Prismatic members Same A, I, E throughout member length

Beams must be in braced frame without significant moments due to lateral forces Not state in Code, but necessary for

coefficients to apply.

** All these requirements must be met to use the coefficients!**

Approximate Analysis of Approximate Analysis of Continuous Beam and One-Way Continuous Beam and One-Way Slab SystemsSlab Systems

2

)( 2

nuvu

numu

lwCV

lwCMwu = Total factored dead and live

load per unit lengthCm = Moment coefficientCv = Shear coefficientln = Clear span length for span in

question for –Mu at interior face of exterior support, +Mu

and Vu

ln = Average of clear span length for adjacent spans for –Mu at interior supports

ACI Moment and Shear Coefficients – Methodology:

Approximate Analysis of Continuous Approximate Analysis of Continuous Beam and One-Way Slab SystemsBeam and One-Way Slab Systems

ACI Moment and Shear Coefficients

See Section 8.3.3 of ACI Code

ExampleExampleDesign the eight-span east west in figure. A typical 1-ft wide design strip is shaded. A partial section through this strip is shown. The beams are assumed to be 14 in. wide. The concrete strength is 3750 psi and the reinforcement strength is 60 ksi. The live load is 100 psf and dead load of 50 psf.

Example – One-way Example – One-way SlabSlab

Use table 9.5(a) to determine the minimum thickness of the slab. 12 in15 ft 180 in

ftl

180 in.min. h = 7.5 in.

24 24l

End bay:

180 in.min h = 6.43 in.28 28l

Interior bays:

Use 7.5 in.

Example – One-way Example – One-way SlabSlab

Compute the trial factored loads based on thickness.

D 3 2

1 ft lb lb7.5 in 150 93.75 12 in ft ft

w

u D L1.2 1.6 1.2 50 psf + 93.75 psf 1.6 100 psf332.5 psf

w w w

Factored load

L D3w wCheck ratio for 8.3.3

OK!

Example – One-way Example – One-way SlabSlab

Compute factored external moment.

22U

U

332.5 psf 15 ft6801. lb-ft/ft

C 11 81.61 k-in/ft

w LM

UN

81.61 k-in/ft 90.68k-in/ft0.9

MM

Nominal moment

Example – One-way Example – One-way SlabSlab

The thickness is 7.5 in. so we will assume that the bar is located d = 7.5in – 1.0 in. = 6.5 in. (From 3.3.2 ACI 318 0.75 in + ~0.25 in( 0.5*diameter of bar) = 1.0 in

N s y

2Ns

y

0.92

90.68 k-in/ft 0.258 in /ft0.9 60 ksi 0.9 6.5 in

aM T d A f d

MAf d

Assume that the moment arm is 0.9d

Example – One-way Example – One-way SlabSlab

Recalculate using As = 0.2 in2

s y

c

NN s y s

y

s

2

0.258 in. 60 ksi0.405 in.

0.85 0.85 3.75 ksi 12 in

22

90.68 k-in/ft0.405 in.60 ksi 6.5 in.

20.240 in /ft

A fa

f b

MaM A f d Aaf d

A

Example – One-way Example – One-way SlabSlab

Check the yield of the steel

1

t cu

0.405 in. 0.476 in.0.85

6.5 in. 0.476 in. 0.0030.476 in.

0.038 0.005

ac

d cc

Steel has yielded so we can use = 0.9

Example – One-way Example – One-way SlabSlab

Check to minimum requirement for every foot

s

y

min minc

y

0.24 in. 0.0030112 in. 6.5 in.

200 200 0.0033360000

0.003333 3 3750 0.0031

60000

Abd

f

ff

Problem!

Example – One-way Example – One-way SlabSlab

What we can do is rework the spacing between the bars by change b Use a #4 bar As = 0.2 in2

2

s s 0.2 in 9.23 in.0.00333 6.5 in.

Use b = 9 in.

A Abbd d

Example – One-way Example – One-way SlabSlab

Check for shrinkage and temperature reinforcement for min = 0.0018 As = minbh from 7.12.2.1 ACI

2s min

2

2

0.0018 12 in. 7.5 in. 0.162 in /ft

0.2 inspacing = 12 in. =14.8 in.0.162 in

A bd

Use 1 # 4 bar every 9 in.

HomeworkHomework

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