lecture 04 design of rc members for shea and torsion

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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan

Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures

Lecture-04Design of RC Members for

Shear and Torsion

1

By: Prof Dr. Qaisar Ali

Civil Engineering Department

UET [email protected]

www.drqaisarali.com


Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 2

Topics Addressed

� Design of RC Members for Shear� Shear Stresses in Rectangular Beams

� Diagonal Tension in RC Beams Subjected to Flexure and ShearLoading

� Types of Cracks in RC Beams

� Shear Strength of Concrete

� Web Reinforcement Requirement

� ACI Code Provisions for Shear Design

� Effect of Axial Force on Shear Strength of Concrete

2


Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 3

Topics Addressed

� Design of RC Members for Torsion

� Torsional Stresses in Solid Concrete Members

� Torsional Strength of Concrete

� Reinforcement Requirement

� ACI Requirements for Design of RC Members Subjected toTorsion

� Steps for Design of RC Member Subjected to Torsion

� Example



Design of RC Members for Shear

4

3



Shear Stresses in Rectangular

Beams

� Homogeneous Elastic

Rectangular Beams

Shear stress (ν) within linear

elastic range along the depth

of a beam can be calculated

by:

ν = VQ/Ib

5

As example, Q (statical moment) at level cd about the n.a will be equal to

Aabcd y′ , where Aabcd is area abcd.




Beams

6

a b

e fh

b

h/4

� Homogeneous Elastic

Rectangular Beams

For shear at neutral axis:

Q = Aabef × h/4 = bh2/8

I = bh3/12

Therefore,

ν = VQ/Ib =

V(bh2/8)/{(bh3/12)×b}

= (3/2)V/bh

4




Beams

� RC Beams in Non-Linear Inelastic Range

� When load on the beam is such that stresses are no longer

proportional to strain, then equation v = VQ/Ib for shear stress

calculation does not govern.

� The exact distribution of shear stresses over the depth of

reinforced concrete member in such a case is not fully known.

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Beams

� RC Beams in Non-Linear Inelastic Range

� Tests have shown that the average shear stress prior to crack

formation is:

� νav = V/ bd

8

5



Diagonal Tension in RC Beams

subjected to Flexure and Shear

Loading

9



Diagonal Tension in RC Beams

subjected to Flexure and Shear

Loading

10

� Conclusions from Previous Discussion

� The tensile stresses are not confined to horizontal bending

stresses that are caused by bending alone.

� Tensile stresses of various inclinations and magnitudes resulting

from shear alone (at the neutral axis) or from the combined

action of shear and bending, exist in all parts of a beam and can

impair its integrity if not adequately provided for.

� It is for this reason that the inclined tensile stresses, known as

diagonal tension stress must be carefully considered in

reinforced concrete design.

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Types of Cracks in Reinforced Concrete

Beam

11

• These are of two types

1. Flexure Cracks

2. Diagonal Tension Cracks

I. Web-shear cracks: These cracks are formed at locations where flexural

stresses are negligibly small.

II. Flexure shear cracks: These cracks are formed where shear force and

bending moment have large values.



Shear Strength of Concrete

12

� Tensile Strength of Concrete

� The direct tensile strength of concrete ranges from 3 to 5√fc′ as

given in table below:

Table: Approximate range of tensile strengths of concrete

Normal-Weight Concrete, psi Lightweight Concrete, psi

Direct tensile strength ft' 3 to 5√fc' 2 to 3√fc'

Split-cylinder strength fct 6 to 8√fc' 4 to 6√fc'

Modulus of rupture fr 8 to 12√fc' 6 to 8√fc'

7




13

� Shear Strength of Concrete in Presence of Cracks

� A large number of tests on beams have shown that in regions

where small moment and large shear exist (web shear crack

location) the nominal or average shear strength is taken as

3.5√fc′.

� However in the presence of large moments (for which adequate

longitudinal reinforcement has been provided), the nominal

shear strength corresponding to formation of diagonal tension

cracks can be taken as:

νcr = 2√fc′




14

� Shear Strength of Concrete in Presence of Cracks

� The same has been adopted by ACI code (refer to ACI 11.3.1).

� This reduction of shear strength of concrete is due to the pre-

existence of flexural cracks.

� It is important to mention here that this value of shear strength of

concrete exists at the ultimate i.e., just prior to the failure

condition.

8



Web Reinforcement Requirement in

Reinforced Concrete Beams

15

� Nominal Shear Capacity (Vn) of Reinforced Concrete

Beam

� Now general expression for shear capacity of reinforced

concrete beam is given as:

Vn = Vc + Vs

Where,

Vc = Nominal shear capacity of concrete,

Vs = Nominal shear capacity of shear reinforcement.

Note: in case of flexural capacity Mn = Mc + Ms (Mc = 0, at ultimate load)





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Beam

� Unlike moment, in expression for shear, the term Vc ≠ 0 because

test evidence have led to the conservative assumption that just

prior to failure of a web-reinforced beam, the sum of the three

internal shear components is equal to the cracking shear Vcr.

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17


Beam

� This sum is generally (somewhat loosely) referred to as the

contribution of the concrete to the total shear resistance, and is

denoted by Vc. Thus Vc = Vcr and

Vc = Vcz + Vd + Viy

Where,

Viy = Vertical component of sizable interlock forces. (Vi) have in fact been

measured, amounting to about one-third of the total shear force.

Vcz = Internal vertical forces in the uncracked portion of the concrete.

Vd = Internal vertical forces across the longitudinal steel, acting as a dowel.




Beam

� Vc = Vcz + Vd + Viy = 2√ (fc′)bwd [ACI 11.3.1.1]

� Vs = Avfyd/s

� Vu = ΦVc + ΦVs

� Vu = ΦVc + ΦAvfyd /s

� s = ΦAvfyd/ (Vu – ΦVc)



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10



ACI Code Provisions for Shear

Design

19

� Strength Reduction Factor

� The strength reduction factor Φ is to be taken equal to 0.75 for

shear.




Design

20

� Spacing Requirements for Stirrups

Note: (Vu- Φ Vc) ≤ Φ 8 √f’c bwd ; otherwise increase the depth

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Design

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� Location of Critical Section for Shear Design




Design

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� Location of Critical Section for Shear Design

Typical support conditions

Typical support conditions

Concentrated load within distance “d”

A beam loaded near its bottom edge, e.g. inverted T-beam

End of a monolithic vertical element

12



Effect of Axial Force on Shear

Strength of Concrete Members

23

� Axial Compression

� The ACI Code provides that for members carrying axial

compression as well as bending and shear, the contribution of

the concrete be taken as:

Vc = 2{1+ Nu/ (2000Ag)}√(fc′)bwd

Where,

Ag = Gross area of concrete

Nu = Axial load

Nu

Nu



Effect of Axial Force on Shear

Strength of Concrete Members

24

� Axial Tension

� The ACI Code provides that, for members carrying significant

axial tension as well as bending and shear, the contribution of

the concrete be taken as:

Vc = 2{1+ Nu/ (500Ag)}√(fc′)bwd

but not less than zero, where Nu is negative for tension.

Nu

Nu

13



Design of RC Members for Torsion

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Torsional Stresses

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� Introduction

� The shear stress induced due to applied torque on a member is

called as torsional shear stress or torsional stress.

Applied Torque, T

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� Circular Members Torsional stresses in solid circular members

can be computed as:

� τ = Tρ/J

Where,

T = applied torque,

ρ = radial distance,

J = polar moment of inertia.

Torsional Stresses

27



Torsional Stresses

28

� Rectangular Members

� Torsional stress variation in rectangular members is relatively

complicated.

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Torsional Stresses

29


� The largest stress occurs at the middle of the wide face “a”.

� The stress at the corners is zero.

� Stress distribution at any other location is less than that at the

middle and greater than zero.

τmax = T/ (αb2a)

Table: Variation of α with ratio a/b.

a/b 1 1.5 2 3

α 0.2 0.23 0.24 0.267



Torsional Stresses

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� Torsional stress close to the faces of the rectangular member is

much greater than that of interior section.

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Torsional Stresses

31


� From the previous discussion it is concluded that torsional

stresses are concentrated in a thin outer skin of the solid cross

section.

� This leads to the concept of thin walled tube analogy.



Torsional Stresses

32

� Torsional Stresses From Thin Walled Tube Analogy

� According to ACI code, torsional stresses in a solid rectangular

section are computed using Thin Walled Tube Analogy.

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Torsional Stresses

33


� In the thin-walled tube analogy the resistance is assumed to be

provided by the outer skin of the cross section roughly centered on

the closed stirrups.

� According to thin walled tube analogy, shear stress and thus shear

flow remains constant within the thin walls of the tube i.e.,

� τ (shear stress)= shear force/ shear area = constant

� q (shear flow)= shear force per unit length = constant



Torsional Stresses

34


τ1 = V1/xot (shear stress = shear force/ shear area)

According to thin-walled tube theory,

V1/xot = V2/yot = V3/xot = V4/yot = τ = constant

Also as,

q1 = V1/xo {shear force/ unit length (shear flow)}

Therefore,

τ = q1/t OR q1 = τt

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Torsional Stresses

35


� As the terms V1 through V4 are induced shear and cannot be easily determined,

therefore, they can be expressed in terms of torque. Taking moments about

centerline of thin walled tube,

T = V4xo/2 + V2xo/2 + V1yo/2 + V3yo/2

T = (V4 + V2) (xo/2) + (V1 + V3) (yo/2)

� As, V4 = V2 and V3 = V1, therefore,

T = (2V2) (xo/2) + (2V1) (yo/2)

� As, q1 = V1/xo and q2 = V2/yo. Therefore,

T = 2qyoxo/2 + 2qxoyo/2 = 2qAo

� As, q = τt, therefore,

τ = T/ (2Aot)



Torsional Strength of Concrete

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� Torsional Strength of Concrete as per ACI Code

� We know that in case of shear, shear strength of concrete is given as

υc = 2√ (fc′), and

� As average shear stress (υav) is equal to V/bd, therefore V/bd = 2√ (fc′)

� V = Vc = 2√ (fc′)bd

� In case of torsion induced shear stresses (torsional stresses), ACI 11.5.1

says that “cracking is assumed when tensile stresses reach 4√ (fc′)”.

� τc = 4√ (fc′) ………………………….….. (a)

� From the previous discussion on torsional stresses in thin walled tube:

τ = T/ (2Aot) ………….………………… (b)

� Equating (a) and (b) gives: Tc = 4√ (fc′) (2Aot)

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37

� Torsional Strength of Concrete as per ACI Code

� According to ACI R11.5.1,

� Ao = (2/3)Acp, t = (3/4)Acp/ pcp

� Where, Acp = xy, pcp = 2(x + y) {full section of the member}

� Substituting values of “Ao ” and “t” in eqn. for Tc gives:

Tc = 4√ (fc′) Acp2/ pcp

� This equation represents the capacity of rectangular concrete member in

torsion.

x

yt




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� Definition of various terms

� Ao = gross area enclosed by shear flow path.

� Aoh = area enclosed by centerline of the

outermost closed transverse torsional

reinforcement;

� According to thin-walled analogy, as the

resistance is assumed to be provided by the

outer cross section roughly centered on the

closed stirrups, Ao should be equal to Aoh;

however to account for cracking ACI R11.5.3.6

requires Ao to be calculated as Ao = 0.85Aoh.

� Therefore Aoh = xoyo & Ao = 0.85Aoh.

Cross hatched area =Aoh

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� Definition of various terms

� Acp = area enclosed by outside perimeter of

concrete cross section

� pcp = outside perimeter of the concrete

cross section

� ph = perimeter of centerline of outermost

closed transverse torsional reinforcement

Cross hatched area =Aoh



Reinforcement Requirement

40

� Review of Reinforcement Requirement for Flexure

� To understand the reinforcement requirement for torsion, recall

the concept of flexural design of RC beam.

� Elastic Range Flexural Capacity

� For an uncracked concrete beam, the flexural stresses are given by:

� f = My/I

� According to ACI code, at f = fr =7.5 √fc′, concrete is at the verge of failure in

tension. Therefore capacity of section at that stage is:

� M = Mcr = fry/I

21




41


� Ultimate Flexural Capacity

� For a cracked RC beam at ultimate stage, the flexural capacity is given as:

� Mn = Mc + Ms

� Where,

� Mc = Nominal flexural capacity of concrete in tension,

� Ms = Nominal flexural capacity of tension steel.




42


� Ultimate Flexural Capacity

� As concrete is weak in tension (Refer ACI 9.5.2.1 for concrete tensile

strength), Mc ≈ 0, therefore,

� Mn ≈ Ms = Asfy (d – a/2)

� So the tension reinforcement along with the concrete in compression acts

as a couple to resists the flexural demand on the member.

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43

� Review of Reinforcement Requirement for Shear

� Similarly, recall the concept of shear design of RC beam.

� Elastic Range Shear Capacity

� For an uncracked concrete beam, the shear stresses are given by:

� υ = VQ/Ib

� According to ACI code, υ = υcr = 2√fc′ is the nominal shear strength

corresponding to formation of diagonal tension cracks. Therefore shear

capacity of section at that stage is:

� Vcr= υcr Ib/Q = 2√fc′ Ib/Q




44


� Ultimate Shear Capacity

� For a cracked RC beam at ultimate stage, the shear capacity is given as:

� Vn = Vc + Vs

� Where,

� Vc = Nominal shear capacity of concrete,

� Vs = Nominal shear capacity of shear reinforcement.

23




45


� Ultimate Shear Capacity

� Unlike flexure, the term Vc ≠ 0 because from test evidence:

� Vc = Vcz + Vd + Viy = 2√ (fc′)bwd [ACI 11.3.1.1]

� Therefore shear steel (stirrups) along with the contribution of concrete (Vc)

acts together to resists the shear demand due to applied load on the

member.




46

� Reinforcement requirement for Torsion

� In the case of Torsion

� ΦTn= ΦTc+ ΦTs

� ACI Code requires that ΦTc shall be taken equal to zero.

� Therfore ΦTn= ΦTs

� We will see in the next slides that how ΦTs is calculated.

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47

� Reinforcement requirement for Torsion

� Reinforcement requirement for reinforced concrete members

subjected to torsion is determined using thin walled tube and

space truss analogies.

As discussed earlier, from

thin walled tube analogy,

internal effects in the form of

induced shear forces (V1 to

V4) will generate due to

applied torque (T) as

shown.




48

� Space Truss Analogy

Such internally induced

shear forces will crack the

member as shown.

Due to cracks the member

splits up into diagonal

compressive portions or

struts.

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49


C

CHCV

If the compressive force in the strut is

C, then it can be resolved into two

components:

CH = Horizontal component

CV = Vertical component

Longitudinal reinforcement shall be

provided to resist CH and vertical

stirrups shall be provided to resist CV.

This leads to space truss analogy.




50


� In space truss analogy, the concrete compression diagonals (struts),

vertical stirrups in tension (ties), and longitudinal reinforcement

(tension chords) act together as shown in figure.

� The analogy derives that torsional stress will be resisted by the vertical

stirrups as well as by the longitudinal steel.

Vertical Reinforcement acts as ties

Long. Reinforcement Acts as tension chordConcrete Compression strut

C

CH

CV

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51

� Vertical Stirrup Reinforcement

� Refer to figure (a), from space truss analogy, Cv = V4

� Refer figure (b),

� V4 = nAtfyv = (yocotθ/s) Atfyv [where n = no of stirrups]

� As V4 = V2, therefore V2 = (yocotθ /s) Atfyv

� And V1 = V3 = (xocotθ /s) Atfyv (b)

(a) Face 4 of the member.

C CV

CH

∆N4

V4




� If V1 to V4 are known, At can be determined from previous eqns. However as

discussed earlier, it is convenient to express V1 to V4 in terms of T by taking

moments about centerline of thin walled tube,

Tn = V4xo/2 + V2xo/2 + V1yo/2 + V3yo/2

� With V4 = V2 and V1 = V3, we have,

Tn = 2V2xo/2 + 2V1yo/2 OR Tn = V2xo + V1yo

Putting values of V1 and V2 derived earlier,

Tn = {(yocotθ/s)Atfyv}xo + {(xocotθ/s)Atfyv}yo

Tn = 2(At/s)fyvxoyocotθ

� With xoyo = Ao, we have,

Tn = 2(At/s)fyvAocotθ ………….. (i)


52

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53


� For no failure, torsional capacity of the member shall be greater than or equal to

torsional demand i.e., ΦTn ≥ Tu [ with Φ = 0.75 ]

� For ΦTn = Tu, equation (i) becomes,

Tu = Φ2(At/s)fyvAocotθ

� Where, At = Steel area in one leg of stirrup, therefore from above equation:

At (one legged) = Tus/ (Φ2fyvAocotθ)

� For two legs, we have,

At (2 legged) = 2Tus/ (Φ2fyvAocotθ)

At (2 legged) = Tus/ (ΦfyvAocotθ)




54


� If θ = 45o (for non-prestressed members, ACI 11.5.3.6), then:

� At (2 legged) = Tus/ (ΦfyvAo) [where Ao = 0.85Aoh (ACI R11.5.3.6)]

� This expression can be used to find shear reinforcement requirement due to

torsion.

� The total shear reinforcement requirement is therefore the sum of the shear

reinforcement requirements due to direct shear and torsion both i.e.,

� Atotal = At (2 legged) + Av (2 legged)

� Where, Av (2 legged) = (Vu – ΦVc)s/ (Φfyvd)

28




55

� Longitudinal Steel Reinforcement

� Refer figure (b) and (c), for face 4 we have,

� ΔN4 = V4 cot θ = V4 [ for θ = 45o ]

� Similarly, ΔN1 = V1; ΔN2 = V2; ΔN3 = V3;

� Total longitudinal reinforcement for torsion is:

� ∆N = ∆N1 + ∆N2 + ∆N3 + ∆N4

� ∆N = V1 + V2 + V3 + V4

� As V1 = V3 and V2 = V4, therefore,

� ∆N = 2V1 + 2V4

(c)


C CV

CH ∆N4

V4




56


� ∆N = 2V1 + 2V4

� With,

� V4 = (yo/s)Atfyv, and,

� V1 = (xo/s)Atfyv

∆N = 2 (xo/s)Atfyv + 2 (yo/s)Atfyv

∆N = 2 (xo+ yo)Atfyv/s

� As ΔN = Alfyl,and therefore,

Alfyl = 2 (xo+ yo)Atfyv/s

� As (2xo + 2yo) = ph , therefore,

� Al = (At /s) ph (fyv /fyl )

(c)


C CV

CH

∆N4

V4

29




57


� Now as derived earlier,

At (one legged) = Tus/ (Φ2fyvAo)

� Therefore expression for Al becomes,

Al = {(Tus/ (Φ2fyvAo) /s} ph (fyv /fyl )

Al = Tu ph / (Φ2fylAo)

� This expression can be used to find the longitudinal reinforcement requirement

due to torsion.



ACI Requirements

58

� Concrete Part in Torsional Capacity of RC Member

� According to ACI R11.5.3.5, in the calculation of Tn, all the

torque is assumed to be resisted by stirrups and longitudinal

steel with Tc = 0. At the same time, the shear resisted by

concrete Vc is assumed to be unchanged by the presence of

torsion.

30



ACI Requirements

59

� Size of Cross-Section: ACI 11.5.3.1

� The cross-section dimensions for solid section shall be such that:

√ [{Vu/ (bwd)} 2 + {Tuph/ (1.7Aoh2)} 2] ≤ Φ {2√ (fc′) + 8√ (fc′)}

Shear stress Torsional shear stress Shear capacity ACI restriction

� ACI R11.5.3.1 —The size of a cross section is limited for two reasons, first to

reduce unsightly cracking and second to prevent crushing of the surface concrete

due to inclined compressive stresses due to shear and torsion.

� If the above condition does not meet, increase cross-section.

� For Hollow section:

Vu/ (bwd) + Tuph/ (1.7Aoh2) ≤ Φ {2√ (fc′) + 8√ (fc′)}

Shear stress Torsional shear stress Shear capacity ACI restriction



ACI Requirements

60

� Reinforcement Requirement for Torsion: ACI 11.5.3.5,

11.5.1 & 11.5.2

� Statically Indeterminate Case

� For statically indeterminate case, no torsional reinforcement is

required if ΦTc ≥ Tu

31



ACI Requirements

61

� Reinforcement Requirement for Torsion: ACI 11.5.3.5,

11.5.1 & 11.5.2

� Statically Determinate Case:

� For statically determinate case, no torsional reinforcement is

required if ΦTc/4 ≥ Tu

Tc = 4√(fc′)Acp2/pcp.



ACI Requirements

62

� Maximum Spacing of the Stirrups (Torsional Stirrups)

� Least of smax = ph/8, 12 inches or Atotalfy/(50bw) [Atotal = Av + At (two legged)]

� Minimum Longitudinal Reinforcement: ACI 11.5.5

� Almin = 5√ (fc′)Acp/fyl – (At/s)phfyv/fyl

where At is single legged reinforcement. If At has been calculated for 2-legged,

then use At/2 in the above equation, and, At/s ≥ 25bw/fyv, with fyv in psi.

� Spacing of longitudinal bars along depth of beam shall be ≤ 12 in and should be

distributed around perimeter.

� Dia. of longitudinal bars ≥ 3/8″,

� Dia. of longitudinal bar > 0.042s {s = stirrups spacing}

32



Steps for Design of Reinforced Concrete

Members Subjected to Torsion

63

� Determine ΦTcr , if ΦTcr/4 < Tu for statically determinate case or ΦTcr < Tu for

indeterminate case, torsional reinforcement is required.

� Check dimensions according to 11.5.3.1.

� Calculate At (2 legged) = Tus/ (ΦfyvAo) {where Ao = 0.85Aoh) and Av (2 legged) = (Vu – ΦVc)s/

(Φfyd)

� Atotal = At + Av. Then determine spacing.

� Check spacing with maximum spacing requirements of ACI.

� Al = Tuph/ (Φ2Aofyl)

� Check maximum spacing, minimum reinforcement and diameter of longitudinal bars.



Example

64

� Design of Cantilever Beam:

� A cantilever beam supports its own dead load plus a

concentrated load. The beam is 54 inches long, and the

concentrated load acts at a point 6 inches from the end of the

beam and 6 inches away from the centroidal axis of the member.

� The un-factored concentrated load consists of a 20-kip dead and

20 kip live load. The beam also supports an un-factored axial

compressive dead load of 40 kip.

� Use normal weight concrete with fc′ = 3000 psi and both fy and fyt

= 60000 psi.

33



Example

65

20 kips DL & 20 kips LL

� Design of Cantilever Beam



Example

66

� Solution:

� Step No. 01: Sizes.

� According to ACI 9.5.2.1, table 9.5 (a):

Minimum thickness of beam (cantilever) = hmin = l/8

� As l = 54″, therefore

hmin = (54/8) = 6.75″ (Minimum requirement by ACI 9.5.2.1).

� Though any depth of beam greater than 6.75″ can be taken as per ACI

minimum requirement, we will use a depth equal to 24″.

� Let the width of beam section (bw) be equal to 14″.

� Therefore, bw = 14″ ; h = 24″ ; d = h – 2.5 = 21.5″

34



Example

67

� Solution:

� Step No. 02: Loads.

� Factored self weight of beam per running foot = 1.2×(14×24/144)×0.15

= 0.42 kip/ft

� Factored concentrated load = 1.2DL + 1.6LL

= 1.2 × 20 + 1.6 × 20 = 56 kips

� Factored axial load (Nu) = 1.2 × 40 = 48 kips



Example

68

� Solution:

� Step No. 03: Analysis.

35



Example

69

� Solution:

� Step No. 03: Analysis.

� Analysis for Shear:

� Vmax = 57.9 kips

� Vu = 57.1 kips

� Analysis for Flexure:

� Mu = - 226 kip-ft

� Analysis for Torsion:

� Tu = 28.0 kip-ft



Example

70

� Solution:

� Step No. 04: Design.

� Check if Pu ≥ 0.1 fc′Ag

Pu = 48 kip

0.1fc′Ag = 0.1 × 3 × 14 × 24 = 100.8 kip

� Pu < 0.1 fc′Ag

� Therefore, axial force effect can be neglected in flexural design. Incase

Pu > 0.1 fc′Ag, member shall be designed for bending and axial load

both.

36



Example

71

� Solution:


� Design for Flexure:

� Mu = 228 ft-kip = 2736 in-kip

� By trial and success method, As = 2.62 in2 (6 #6 bars)

� Asmax = ρmaxbwd

ρmax = 0.85 × 0.85 × (3/60) × {0.003/ (0.003 + 0.005)} = 0.01355

Asmax = 0.01355 × 14 × 21.5 = 4.07 in2

� Asmin = ρminbwd = (greater of 3 √fc′/fy OR 200/fy)bwd

= 0.0033 × 14 × 21.5 = 0.993 in2

� Asmin<As<Asmax, O.K.



Example

72

� Solution:


� Design for Torsion and Shear both: The design parameters required

for torsion design are as follows,

� bw = 14″; d = 21.5″; h = 24″; hf = 6″

� Acp, pcp, Aoh, ph, Ao:

Acp = bwh = 14 × 24 = 336 in2

pcp = 2h +2bw = 2 × 24 + 2 × 14 = 76 in

� With 1 ¾ in cover to the centre of the stirrup bars from all faces

xo = 14 – 2 × 1 ¾ = 10.5 in

yo = 24 – 2 × 1 ¾ = 20.5 in

37



Example

73

� Solution:


� Design for Torsion and Shear both: The design parameters required

for torsion design are as follows,

� Thus,

Aoh = xoyo = 10.5 × 20.5 = 215.25 in2

Ao = 0.85Aoh = 182.96 in2

ph = 2 (xoh + yoh) = 2 (10.5 + 20.5) = 62 inches

� Finally,

Acp = 336 in2; pcp = 76 inches; Aoh = 215.25 in2; ph = 62 inches; Ao =

182.96 in2



Example

74

� Solution:


� Design for Torsion and Shear both: The design parameters required for

torsion design are as follows,

� (i) Check for size of beam: ACI 11.5.3,

√ [{Vu/ (bwd)} 2 + {Tuph/ (1.7Aoh2)} 2] ≤ Φ {2√ (fc′) + 8√ (fc′)}

� Where, Vu and Tu at critical section. Therefore,

√[{57.1/(21.5×12)}2+{28×12×62/(1.7×215.252)}2]≤0.75×10×√ (3000)/1000

0.345 ksi ≤ 0.411 ksi

� Therefore size of the beam is O.K.

38



Example

75

� Solution:


� Design for Torsion and Shear both:

� (ii) Check if ΦTc/4 ≥ Tu (For statically determinate case):

� ΦTc = 0.75 × 4√ (fc′)Acp2/pcp

= 0.75 × 4 × {√ (3000)/ (12 × 1000)} × 3362/ 76 = 20.34 ft-kip

� No reinforcement is required if ΦTc/4 ≥ Tu

ΦTc/4 = 20.34/4 = 5.085 ft-kip

� As ΦTc/4 < Tu, therefore torsional reinforcement is required.



Example

76

� Solution:



� (a) Torsional Reinforcement (at critical section, two legged):

� At (2 legged) = Tus/(ΦfyvAo)

� Therefore,

� At (2 legged) = (28 × 12) s/ (0.75 × 60 × 182.96) = 0.041s

39



Example

77

� Solution:



� (b) Shear Reinforcement:

� As an axial load of 48 kips is acting on the beam, therefore ΦVc

shall be calculated using Vc = 2{1+ Nu/ (2000Ag)}√(fc′)bwd, where

Ag is the gross area and Nu is the axial force.

� ΦVc = Φ2{1+ Nu/ (2000Ag)}√(fc′)bwd

= 0.75×2×{1+48/(2000×336)}×√(3000)×14×21.5/1000

= 24.73 kips



Example

78

� Solution:



� (b) Shear Reinforcement:

� Av (2 legged) = (Vu – ΦVc)s/ (Φfyvd)

� Av (2 legged)= {57.1 – 24.73}s/ (0.75 × 60 × 21.5)

� Av (2 legged) = 0.03346s

40



Example

79

� Solution:



� Atotal = At (2 legged) + Av (2 legged) = 0.041s + 0.03346s

Atotal = 0.07446s

� Assuming 3/8″ Φ, 2 legged with total bar area Ab = 0.22 in2

sdesign = 0.22/0.07446 = 2.95 inches

� Trying 1/2″ Φ, 2 legged with total bar area Ab = 0.40 in2

sd = 0.40/0.07446 = 5.37 inches

We will use 1/2” Φ bars at 5” c/c



Example

80

� Solution:



� (c) Maximum Spacing Requirement (ACI 11.5.5)

� For torsion s = ph/8 or 12″= 62/8 = 7.75″ or 12″

� For shear s = d/2 or 24″= 21.5/2 = 10.75″ or 24″

� Therefore maximum spacing allowed = 7.75″

� First stirrup is to be placed at a distance (sd/2 = 5/2 ≈ 2.5″) from the

face of the support.

� Therefore 2.5 inches from face of the support provide s = 5 inches

throughout.

41



Example

81

� Solution:



� Longitudinal reinforcement.

Al = Tuph/ (Φ2Aofyl) = 28 × 12 × 62/ (0.75 × 2 × 182.96 × 60) = 1.265 in2

� Almin = 5{√(fc′)/fyl}Acp – (At/ 2s)phfyv/ fyl

= 5{√(3000) × 336/60000} – (0.041s/2s) × 62 × 60/60 = 0.262 in2



Example

82

� Solution:




� According to the ACI Code, the spacing must not exceed 12 in.,

and the bars may not be less than No.3 (No. 10) in size nor have a

diameter less than 0.042s.

� dmin = 0.042s = 0.042 × 5 = 0.21 inch

42



Example

83

� Solution:




� Reinforcement will be placed at the top, mid depth, and bottom of the

member, each level to provide not less than 1.265/3 = 0.422 in2. Two

No. 6 bars will be used at mid depth, and reinforcement to be placed for

flexure will be increased by 0.422 in2 at the top and bottom of member.

� Therefore final flexural reinforcement is:

� As, main = 2.62 + 0.422 = 3.042 in2 (7 #6 bars)

� Whereas 2 #6 bars will be provided for bottom and mid depth

reinforcement each.



Example

84

� Solution:

� Step No. 05: Drafting.

54″

(5+2) #6 bars

#4, 2 legged @ 5″ c/c

2 #6 bars24″

14″

2 #6 bars

(5+2) #6 bars

#4, 2 legged @ 5″ c/c

43



References

85

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

Dolan.

� Reinforced Concrete – Mechanics and Design [5th Ed.] by James

MacGregor.

� ACI 318.



The End

86

lecture 04 design of rc members for shea and torsion

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