Lecture 21 – Splices and Shear

Lecture 21 – Splices Lecture 21 – Splices and Shearand Shear

February 5, 2003CVEN 444

Lecture GoalsLecture Goals

Spice Shear Shear Design

Bar SplicesBar Splices

Why do we need bar splices? -- for long spans

Types of Splices

1. Butted &Welded

2. Mechanical Connectors

3. Lay Splices

Must develop 125% of yield strength

Tension Lap SplicesTension Lap SplicesWhy do we need bar splices? -- for long spans

Types of Splices

1. Contact Splice

2. Non Contact Spice (distance 6” and 1/5 splice length)

Splice length is the distance the two bars are overlapped.

Types of SplicesTypes of Splices

Class A Spice (ACI 12.15.2)

When over entire splice length.

and 1/2 or less of total reinforcement is spliced win the req’d lay length.

2

dreq's

provideds A

A

Types of SplicesTypes of Splices

Class B Spice (ACI 12.15.2)

All tension lay splices not meeting requirements of Class A Splices

Tension Lap Splice Tension Lap Splice (ACI 12.15)(ACI 12.15)

where As (req’d) = determined for bending

ld = development length for bars (not allowed to use excess reinforcement modification factor)

ld must be greater than or equal to 12 in.

Tension Lap Splice (ACI Tension Lap Splice (ACI 12.15)12.15)

Lap Spices shall not be used for bars larger than No. 11. (ACI 12.14.2)

Lap Spices should be placed in away from regions of high tensile stresses -locate near points of inflection (ACI 12.15.1)

Compression Lap Splice Compression Lap Splice (ACI 12.16.1)(ACI 12.16.1)

Lap, req’d = 0.0005fy db for fy < 60000 psi Lap, req’d = (0.0009fy -24) db for fy > 60000 psi Lap, req’d 12 in

For fc 3000 psi, required lap splice shall be multiply by (4/3) (ACI 12.16.1)

Compression Lap Splice Compression Lap Splice (ACI 12.17)(ACI 12.17)

In tied column splices with effective tie area throughout splice length 0.0015 hs factor = 0.83

In spiral column splices, factor = 0.75 But final splice length 12 in.

Example – Splice Example – Splice TensionTension

Calculate the lap-splice length for 6 #8 tension bottom bars in two rows with clear spacing 2.5 in. and a clear cover, 1.5 in., for the following cases

When 3 bars are spliced and As(provided) /As(required) >2

When 4 bars are spliced and As(provided) /As(required) < 2

When all bars are spliced at the same location. fc= 5 ksi and fy = 60 ksi

a.

b.

c.


For #8 bars, db =1.0 in. and = ? =? = ? =?

yd

trb c

b

3

40

flc Kd fd


The As(provided) /As(required) > 2, class ? splice applies;

The As(provided) /As(required) < 2, class ? splice applies;

Example – Splice Example – Splice CompressionCompressionCalculate the lap splice length for a # 10 compression bar in tied column when fc= 5 ksi and when a) fy = 60 ksi and b) fy = 80 ksi

Example – Splice Example – Splice CompressionCompression

For #10 bars, db =? in.

ydy

b c

0.020.003

flf

d f

Check ls > 0.005 db fy

Example – Splice Example – Splice CompressionCompression

For #10 bars, db =? in. The ld = 2? in.

Check ls > (0.0009 fy –24) db

So use ls = ? in.

Shear DesignShear Design

Uncracked Elastic Beam Uncracked Elastic Beam BehaviorBehavior

Look at the shear and bending moment diagrams. The acting shear stress distribution on the beam.


The acting stresses distributed across the cross-section.

The shear stress acting on the rectangular beam.

Ib

VQ


The equation of the shear stress for a rectangular beam is given as:

Note: The maximum 1st moment occurs at the neutral axis (NA).

Ib

VQ

avemax

2

max

3

5.1*2

3

84*

2Q

Inertia ofMoment 12

bh

V

bhhbh

bhI


The ideal shear stress distribution can be described as:

Ib

VQ


A realistic description of the shear distribution is shown as:


The shear stress acting along the beam can be described with a stress block:

Using Mohr’s circle, the stress block can be manipulated to find the maximum shear and the crack formation.

Inclined Cracking in Inclined Cracking in Reinforced Concrete Reinforced Concrete BeamsBeams

Typical Crack Patterns for a deep beam

Inclined Cracking in Inclined Cracking in Reinforced Concrete Reinforced Concrete BeamsBeamsFlexural-shear crack - Starts out as a flexural crack and propagates due to shear stress.

Flexural cracks in beams are vertical (perpendicular to the tension face).


For deep beam the cracks are given as:

The shear cracks Inclined (diagonal) intercept crack with longitudinal bars plus vertical or inclined reinforcement.


For deep beam the cracks are given as:

The shear cracks fail due two modes:

- shear-tension failure - shear-compression failure

Shear Strength of RC Beams Shear Strength of RC Beams without Web Reinforcementwithout Web Reinforcement

vcz - shear in compression zone

va - Aggregate Interlock forces

vd = Dowel action from longitudinal bars

Note: vcz increases from (V/bd) to (V/by) as crack forms.

Total Resistance = vcz + vay + vd (when no stirrups are used)

Strength of Concrete in Shear Strength of Concrete in Shear (No Shear Reinforcement)(No Shear Reinforcement)

(1) Tensile Strength of concrete affect inclined cracking load

Strength of Concrete in Strength of Concrete in Shear (No Shear Shear (No Shear Reinforcement)Reinforcement)

(2) Longitudinal Reinforcement Ratio, w

dbfV

db

A

wccw

w

sw

2:0025.00075.0for

cracks restrains


(3) Shear span to depth ratio, a/d (M/(Vd))

2d

a

2

d

a Deep shear spans more detail design required

Ratio has little effect


(4) Size of BeamIncrease Depth Reduced shear stress at

inclined cracking


(5) Axial Forces - Axial tension Decreases inclined cracking load - Axial Compression Increases inclined cracking

load (Delays flexural cracking)

Function and Strength of Function and Strength of Web ReinforcementWeb Reinforcement

Web Reinforcement is provided to ensure that the full flexural capacity can be developed. (desired a flexural failure mode - shear failure is brittle)

- Acts as “clamps” to keep shear cracks from widening

Function:


Uncracked Beam Shear is resisted uncracked concrete.

Flexural Cracking Shear is resisted by vcz, vay, vd

bars. allongitudin fromAction Dowl

forceInterlock Aggregate ofcomponent Vertical

zonen compressioin Shear

d

ay

cz

V

V

V


Flexural Cracking Shear is resisted by vcz, vay, vd and vs

Vs increases as cracks widen until yielding of stirrups then stirrups provide constant resistance.

Designing to Resist Designing to Resist ShearShear

Shear Strength (ACI 318 Sec 11.1)

n u

capacity demand

V V

u

n

factored shear force at section

Nominal Shear Strength

0.75 shear strength reduction factor

V

V

Designing to Resist Designing to Resist ShearShear

Shear Strength (ACI 318 Sec 11.1)

n c sV V V

c

s

V

V

Nominal shear provided by the shear reinforcement

Nominal shear resistance provided by concrete

Shear Strength Provided by Shear Strength Provided by ConcreteConcrete

Bending onlyBending only

Simple formula

More detailed

Note:

Eqn [11.5]

Eqn [11.3]

dbf

dbfV

wc

wcc

3.5

2

dbf

dbM

dVfV

wc

w

u

uwcc

3.5

25001.9

1

u

u

M

dV

Shear Strength Provided by Shear Strength Provided by ConcreteConcrete

Bending and Axial CompressionBending and Axial Compression

Nu is positive for compression and Nu/Ag are in psi.

Simple formula

Eqn [11.4]

Eqn [11.7]

g

uwc

wc

g

uc

5001 3.5

2000

1 2

A

Ndbf

dbfA

NV

Typical Shear Typical Shear ReinforcementReinforcement

Stirrup - perpendicular to axis of members (minimum labor - more material)

ACI Eqn 11-15

s

dfAV

cossinyvs

s

dfAV yv

so90

Typical Shear Typical Shear ReinforcementReinforcement

Bent Bars (more labor - minimum material) see req’d in 11.5.6

ACI 11-5.6

s

dfAV

cossinyvs

s

dfAV yv

so 41.1

45

Stirrup Anchorage Stirrup Anchorage RequirementsRequirements

Vs based on assumption stirrups yield

Stirrups must be well anchored.


each bend must enclose a long bar

# 5 and smaller can use standard hooks 90o,135o, 180o

#6, #7,#8( fy = 40 ksi )

#6, #7,#8 ( fy > 40 ksi ) standard hook plus a minimum embedment

Refer to Sec. 12.13 of ACI 318 for development of web reinforcement. Requirements:


Also sec. 7.10 requirement for minimum stirrups in beams with compression reinforcement, beams subject to stress reversals, or beams subject to torsion

Design Procedure for Design Procedure for ShearShear

(1) Calculate Vu

(2) Calculate Vc Eqn 11-3 or 11-5 (no axial force)

(3) Check

cu VV 2

1

If yes, add web reinforcement (go to 4)

If no, done.


(4) cuc VVV 2

1 If

v

w

ysv

y

wv A

b

fAs

f

sbA min for

50or 50 maxmin

Also:

(Done) 11.5.4 "24

2max d

s

Provide minimum shear reinforcement


(5)

cu

scus

scnu

scu

VV

VVVV

VVVV

VVV

d)(req' calulate , If

Check:

11.5.4 illegal otherwise, 8 dbfV wcs


Solve for required stirrup spacing(strength) Assume # 3, #4, or #5 stirrups

s

ysv

V

dfAs

(6)

from 11-15


(7) Check minimum steel requirement (eqn 11-13)

50max

w

ysv

b

fAs


(8) Check maximum spacing requirement (ACI 11.5.4)

illegal 8 If :Note

"124

4 If

"242

4 If

c

maxc

maxc

dbfV

dsdbfV

dsdbfV

ws

ws

ws


(9) Use smallest spacing from steps 6,7,8

Note: A practical limit to minimum stirrup spacing is 4 inches.

Location of Maximum Shear Location of Maximum Shear for Beam Designfor Beam Design

Non-pre-stressed members:Sections located less than a distance d from face of support may be designed for same shear, Vu, as the computed at a distance d.

Compression fan carries load directly into support.

Location of Maximum Shear for Beam Design

The support reaction introduces compression into the end regions of the member.

No concentrated load occurs with in d from face of support .

1.

2.

When:

Location of Maximum Shear for Beam Design

Compression from support at bottom of beam tends to close crack at support

HomeworkHomeworkDetermine the development length required for the bars shown . fc =4-ksi and fy = 60-ksi. Check the anchorage in the column. If it is not satisfactory, design an anchorage using a 180o hook and check adequacy.

HomeworkHomeworkConsidering the anchorage of the beam bars into a column, determine the largest bar that can be used with out a hook. fc = 3-ksi and fy= 40ksi

HomeworkHomeworkA simple supported uniformly loaded beam carries a total factored design load of 4.8 k/ft (including self-weight) on a clear span of 34 ft. fc =3 ksi and fy=40 ksi. Assume that the supports are 12 in wide and assume that the bars are available in 30 ft lengths.

Design a rectangular beam

Determine bar cutoffs.

Locate splices and determine the lap length.

Lecture 21 – Splices and Shear

Documents

shear cracks

sheartension failure

shear distribution

maximum shear

web reinforcementvcz

compression lap splice

lap splice length

splicestension lap splice