5.8 SHEAR STRENGTH5.8 SHEAR STRENGTH

Where:

–Vu = maximum shear based on the controlling combination of

factored loads

–Ø = resistance factor for shear = 0.90

–Vn = nominal shear strength

Consider a simple beam as shown in

Fig. a. At a distance x from the left

end and at the neutral axis of the

cross section, the state of stress is as

shown in Fig. d. Because this element

is located at the neutral axis, it is not

subjected to flexural stress.

From elementary mechanics of materials, the shearing stress isFrom elementary mechanics of materials, the shearing stress is

This equation based on the assumption that the stress is constant This equation based on the assumption that the stress is constant

across the width b, and it is therefore accurate only for small across the width b, and it is therefore accurate only for small

values of b, so the equation can not be applied to the flange of a values of b, so the equation can not be applied to the flange of a

W shape in the same manner as for the web.W shape in the same manner as for the web.

The average stress in the web w is V/ AThe average stress in the web w is V/ Aww

No big difference between the average stress and the maximum No big difference between the average stress and the maximum

stressstress

The web will completely yield long before the flanges begin to The web will completely yield long before the flanges begin to

yield.yield.

The following Figure shows the shearing stress distribution for a W-shape.The following Figure shows the shearing stress distribution for a W-shape.

We can write the equation for the stress in the web at failure as:We can write the equation for the stress in the web at failure as:

Where AWhere Aww is the area of the web. is the area of the web.

The nominal strength corresponding to this limit state is thereforeThe nominal strength corresponding to this limit state is therefore

The relationship between shear strength and the width-thickness The relationship between shear strength and the width-thickness

ratio is analogous to that between flexural strength and the ratio is analogous to that between flexural strength and the

width thickness ratio (for FLB or WLB) and between flexural width thickness ratio (for FLB or WLB) and between flexural

strength and unbraced length (for LTB) strength and unbraced length (for LTB)

wAy0.60FnV

y0.60FwAnV

vf

This relationship is illustrated in the Figure below :This relationship is illustrated in the Figure below :

This relationship given in the AISC as follows:This relationship given in the AISC as follows:

wyn

yw

AFV

andyinstabilitwebnoisThere

F

E

t

hFor

60.0

,

,45.2

w

y

wyn

ywy

t

h

F

E

AFV

andoccurcanbucklingwebInelastic

F

E

t

h

F

EFor

45.2

60.0

,

07.3,45.2

2

52.4

260,07.3

w

wn

wy

t

h

EAV

t

h

F

EFor

bucklingwebelasticisstatelimitThe

Where, AWhere, Aww is the area of the web = d*t is the area of the web = d*tww

And d is the overall depth of the web.And d is the overall depth of the web.

If h/tIf h/tw w is greater than 260, web stiffeners are required.is greater than 260, web stiffeners are required.

Shear is rarely a problem in rolled steel beams; the usual practice Shear is rarely a problem in rolled steel beams; the usual practice

is to design a beam for flexure and then to check it for shear.is to design a beam for flexure and then to check it for shear.

Example 5.7Example 5.7

Check the beam in Example 5.6 for shear.Check the beam in Example 5.6 for shear.

Solution:Solution:

From Example 56: A W14X90 with FFrom Example 56: A W14X90 with Fyy = 50 ksi is used. = 50 ksi is used.

WWuu = 2.080 kips/ft and L= 40 ft = 2.080 kips/ft and L= 40 ft

kips41.602

40*2.0802

LuwnV

From the ManualFrom the Manual

0.5950

2900045.245.2

9.25

y

w

F

E

t

h

The strength is governed by shear yielding of the webThe strength is governed by shear yielding of the web

yw F

E2.45

t

hsince

(OK)kips41.6kips166184.8*0.90 nφV

kips184.80.440*14.0*50*0.60wAy0.6FnV

The shear design strength is greater than the factored load shear, The shear design strength is greater than the factored load shear,

so the beam is satisfactory.so the beam is satisfactory.

BLOCK SHEARBLOCK SHEAR

To facilitate the connection of beams to other beams so that the

top flanges are at the same elevation, a short length of the top

flange of one of the beams may be cut away, or coped. If a

coped beam is connected with bolts as in Figure, segment ABC

will tend to tear out.

The applied load in this case will be

the vertical beam reaction.

Shear will occur along line AB and

there will be tension along BC. Thus

the block shear strength will be a

limiting value of the reaction.

We covered the computation of block shear strength in Chapter3

Example 5-8

5.9 Deflection5.9 Deflection

Steel beams are designed for the factored design loads. The moment Steel beams are designed for the factored design loads. The moment

capacity, i.e., the factored moment strength (φcapacity, i.e., the factored moment strength (φbbMMnn) should be greater than ) should be greater than

the moment (Mthe moment (Muu) caused by the factored loads. ) caused by the factored loads.

A serviceable structure is one that performs satisfactorily, not causing A serviceable structure is one that performs satisfactorily, not causing

discomfort or perceptions of unsafety for the occupants or users of the discomfort or perceptions of unsafety for the occupants or users of the

structure. structure.

For a beam, being serviceable usually means that the deformations, primarily For a beam, being serviceable usually means that the deformations, primarily

the vertical sag, or deflection, must be limited.the vertical sag, or deflection, must be limited.

The maximum deflection of the designed beam is checked at the service-level The maximum deflection of the designed beam is checked at the service-level

loads. The deflection due to service-level loads must be less than the loads. The deflection due to service-level loads must be less than the

specified values.specified values.

Appropriate limits for deflection can be found from the governing building Appropriate limits for deflection can be found from the governing building

code. code.

For the common case of a simply supported, uniformly loaded For the common case of a simply supported, uniformly loaded beam such as that in the following Figure, the maximum vertical beam such as that in the following Figure, the maximum vertical

deflection is:deflection is:

Deflection limit:Deflection limit:

Example 5-9 :Example 5-9 :

solutionsolution

Ponding is one deflection problem that does affect the safety of a Ponding is one deflection problem that does affect the safety of a

structure.structure.

The AISC specification requires that the roof system have The AISC specification requires that the roof system have

sufficient stiffness to prevent ponding.sufficient stiffness to prevent ponding.

5.10 DESIGN:5.10 DESIGN:

Beam design entails the selection of a cross-sectional shape

that will have enough strength and that will meet

serviceability requirements.

The design process can be outlined as follows:

1. Compute the factored load moment.

2. Select a shape that satisfies this strength requirement.

This can be done in one of two ways:

Assume a shape, compute the design strength, and

compare it with the factored load moment.

Use beam design chart in part 5 at Manual (preferred).

3. Check the shear strength.

4. Check the defection.

Beam Design Charts::

– Many graphs, charts, and tables are available for the

practicing engineer, and these aids can greatly simplify the

design process.

– To determine the efficiency, they are used in design offices.

– You should approach their use with caution and not allow

basic principles to become obscured.

– The curves of design moment versus un braced length given

in Part 5 of the Manual.

– All curves were generated with Fy =50 ksi and Cb = 1.0.

– For other values of Cb simply multiply the design moment

from the chart by Cb.

The following curve described the design moment ФbMn as a function of unbraced length Lb for a particular compact shape.

Remember that design strength can never exceed plastic momentRemember that design strength can never exceed plastic moment

Noncompact shapes may fail due to local buckling

Two sets of curves are available, one for W-shapes and one for C-shapes and MC-

shapes.

Example 5.12

Use A992 steel and select a rolled shape for the beam shown

below. The concentrated load is a service live load, and the

uniform load is 30% dead load and 70% live load. Lateral

bracing is provided at the ends and at mid span. There is no

restriction on deflection.

5.11 FLOOR AND ROOF FRAMING SYSTEMS5.11 FLOOR AND ROOF FRAMING SYSTEMS

5.12 HOLES IN BEAMS5.12 HOLES IN BEAMS

If beam connections are made with bolts, holes will be punched or If beam connections are made with bolts, holes will be punched or

drilled in the beam web or flange.drilled in the beam web or flange.

Sometimes electrical conduits and ventilation ducts need large Sometimes electrical conduits and ventilation ducts need large

holes.holes.

Ideally, holes should be placed in the web only at section of low Ideally, holes should be placed in the web only at section of low

shear, and holes should be made in the flanges at points of low shear, and holes should be made in the flanges at points of low

bending moment.bending moment.

This is not always be possible, so the effect of the holes must be This is not always be possible, so the effect of the holes must be

accounted for.accounted for.

The effect of small holes will be small, particularly for flexure.The effect of small holes will be small, particularly for flexure.

AISC B10 permits bolt holes in flanges to be ignored when: AISC B10 permits bolt holes in flanges to be ignored when:

0.75 F0.75 Fuu A Afnfn ≥ 0.90 F ≥ 0.90 Fyy A Afgfg

where, Awhere, Afg fg is the gross tension flange areais the gross tension flange area

and, Aand, Afn fn is the net tension flange area.is the net tension flange area.

If this condition is not met, the previous equation is solved for an If this condition is not met, the previous equation is solved for an

effective tension flange area that satisfies that criterion. effective tension flange area that satisfies that criterion.

fnAyF

uF

6

5feA

or

y0.9F

fnAu0.75F

feAfgA

See examples 5.14 and 5.15See examples 5.14 and 5.15

5.13 OPEN-WEB STEEL JOISTS5.13 OPEN-WEB STEEL JOISTS

Open-web steel joists are prefabricated trusses of the type shown Open-web steel joists are prefabricated trusses of the type shown

in the Figure.in the Figure.

They are used in floor and roof systems.They are used in floor and roof systems.

For a given span, it is lighter in weight and its more easier for For a given span, it is lighter in weight and its more easier for

electrical conduits and ventilation ducts than a rolled shape .electrical conduits and ventilation ducts than a rolled shape .

More economical than a rolled shape More economical than a rolled shape

5.14 Bearing plates and column base plates :5.14 Bearing plates and column base plates :

– The function of the plate is to distribute a concentrated load to The function of the plate is to distribute a concentrated load to

the supporting materialthe supporting material

– Two types of beam bearing plates are considered Two types of beam bearing plates are considered

One that transmits the beam reaction to a supportOne that transmits the beam reaction to a support

One that transmits a load to the top flange of a beam.One that transmits a load to the top flange of a beam.

–The design of the bearing plate consists of three steps.The design of the bearing plate consists of three steps.

Determine dimension N so that web yielding and web Determine dimension N so that web yielding and web

crippling are prevented crippling are prevented

Determine Dimension B so that the area N Determine Dimension B so that the area N ××B is sufficient to B is sufficient to

prevent the supporting material from being crushed in prevent the supporting material from being crushed in

bearing.bearing.

Determine the thickness t so that the plate has sufficient Determine the thickness t so that the plate has sufficient

bending strength.bending strength.

Web YieldingWeb Yielding

Web yielding is the compressive crushing of a beam web caused by a force Web yielding is the compressive crushing of a beam web caused by a force

acting on the flange directly above or below the webacting on the flange directly above or below the web

When the load is transmitted through a plate, web yielding is assumed to take When the load is transmitted through a plate, web yielding is assumed to take

place on the nearest section of width tplace on the nearest section of width tww..

In rolled shape, this section will be at the toe of the fillet, a distance k from the In rolled shape, this section will be at the toe of the fillet, a distance k from the

outside face of the flangeoutside face of the flange

If the load is assumed to distribute itself at a slop of 1:2.5 as shownIf the load is assumed to distribute itself at a slop of 1:2.5 as shown

The area at the support subject to yielding is (2.5k + N) tThe area at the support subject to yielding is (2.5k + N) tw w

The nominal strength for web yielding at the support is:The nominal strength for web yielding at the support is:

The bearing length N at the support should not be less than k.The bearing length N at the support should not be less than k.

At the interior load, the length of the section subject to yielding isAt the interior load, the length of the section subject to yielding is

The design strength is The design strength is ØRwØRw where where Ø = 1.0Ø = 1.0

Concrete Bearing StrengthConcrete Bearing Strength

Usually, concrete used as the material beam support.Usually, concrete used as the material beam support.

This material must resist the bearing load applied by steel plateThis material must resist the bearing load applied by steel plate

If the plate covers the full area of the support, If the plate covers the full area of the support,

The nominal strength is:The nominal strength is:

If the plate does not cover the full area of the supportIf the plate does not cover the full area of the support

WhereWhere

ffcc is the compressive strength of concrete after 28 days is the compressive strength of concrete after 28 days

AA11 is the bearing area is the bearing area

AA22 is the full area of the support is the full area of the support

1'cp A0.85fP

1

21

'cp A

AA0.85fP

If area AIf area A22 is not concentric with A is not concentric with A11, then A, then A22 should be taken as the should be taken as the

largest concentric area that is geometrically similar to Alargest concentric area that is geometrically similar to A11, as , as

illustrated in the following Figureillustrated in the following Figure

AISC also requiresAISC also requires

The design bearing strength is The design bearing strength is ФФccPPpp where where ФФcc=0.60 =0.60

21

2

A

A

Plate ThicknessPlate Thickness

WhereWhere

RRuu is the support reaction is the support reaction

B is width of the bearing plateB is width of the bearing plate

N is length of the bearing plateN is length of the bearing plate

y

2u

BNF

n2.222Rt

Example 5.17Example 5.17

Design a bearing plate to distribute the reaction of a W 21 x 68 with a Design a bearing plate to distribute the reaction of a W 21 x 68 with a

span length of 15 ft 10 inches center to center of supports. The total span length of 15 ft 10 inches center to center of supports. The total

service load, including the beam weight , is 19 kips with equal parte service load, including the beam weight , is 19 kips with equal parte

dead and live load. The beam is to be supported on reinforced concrete dead and live load. The beam is to be supported on reinforced concrete

wall with fwall with fcc = 3500 psi. the beam made of A992 steel and the plate is = 3500 psi. the beam made of A992 steel and the plate is

A36.A36.

SolutionSolution

The factored load is = 1.2*4.5 + 1.6*4.5 = 12.600 kips/ftThe factored load is = 1.2*4.5 + 1.6*4.5 = 12.600 kips/ft

The reaction is = 12.6*15.83/2 = 99.3 kipsThe reaction is = 12.6*15.83/2 = 99.3 kips

Determine the length of bearing N required to prevent web yielding.Determine the length of bearing N required to prevent web yielding.

RRuu = 2.5k + N) F = 2.5k + N) Fyy t tw w = (2.5(1.438) + N) 50*0.430 ≥ 99.73= (2.5(1.438) + N) 50*0.430 ≥ 99.73

N ≥ 1.044N ≥ 1.044

Determine the length of bearing N required to prevent web Determine the length of bearing N required to prevent web crippling. Assume N/d > 0.2 crippling. Assume N/d > 0.2

N ≥ 3.0 inN ≥ 3.0 inCheck the assumptionCheck the assumptionN/d = 3.0/21.1 = 0.14 < 0.2 (N.G.) so for N/d < 0.2N/d = 3.0/21.1 = 0.14 < 0.2 (N.G.) so for N/d < 0.2

99.730.43

0.685*50*290001.5

0.685

0.430.2

21.1

4N1

20.43*0.40*075

uRwt

ftyEF1.5

ft

wt0.2d

4N12

wt0.40φ

99.730.43

0.685*50*290001.5

0.685

0.43

21.1

N1

20.43*0.40*075

uRwt

ftyEF1.5

ft

wt

d

N12

wt0.40φ

3

3

N ≥ 2.59 in, and N/d = 2.59/21.1 = 0.12< 0.2 (OK)N ≥ 2.59 in, and N/d = 2.59/21.1 = 0.12< 0.2 (OK)

Try N = 6.0Try N = 6.0

Assume full area of support is used, the required plate area isAssume full area of support is used, the required plate area is

A = A = ФФ * 0.85 * f * 0.85 * fcc * A ≥ R * A ≥ Ru u

A = 0.60 * 0.85 * 3.5 * A ≥ 99.73A = 0.60 * 0.85 * 3.5 * A ≥ 99.73

A ≥ 55.87 inA ≥ 55.87 in22

B = 55.87/6 = 9.31 inB = 55.87/6 = 9.31 in

The flange width of a W21x68 is 8.27 < 9.31, use B as 10 inThe flange width of a W21x68 is 8.27 < 9.31, use B as 10 in

Compute the required plate thickness, n = (B – 2k)/2 Compute the required plate thickness, n = (B – 2k)/2

= (10 – 2*1.19)/2 = 3.81 in = (10 – 2*1.19)/2 = 3.81 in

Use a PL 1.25 * 6 * 10.Use a PL 1.25 * 6 * 10.

in1.22

36*6*10

3.81*99.73*2.222

BNF

n2.222R t

2

y

2u

Column Base PlatesColumn Base Plates

Major differences between bearing and base plates are:Major differences between bearing and base plates are:

Bending in bearing plates in one direction, whereas column base Bending in bearing plates in one direction, whereas column base

plates are subjected to two-way bending.plates are subjected to two-way bending.

Web crippling and web yielding are not factors in column base Web crippling and web yielding are not factors in column base

plate design.plate design.

Design steps:Design steps:

Determine the allowable strength of the foundationDetermine the allowable strength of the foundation

Determine the required column base plate area Determine the required column base plate area

Select column base plate dimension (not less than column Select column base plate dimension (not less than column

dimension)dimension)

Determine the thicknessDetermine the thickness

Where, Where, y

u

0.9BNF

2PL t

pc

u

f

f

f

P

P

bd

dbX

X

X

bBn

2

4

111

2

2

8.02

0.95d-Nm

)nn(m,maxL

60.04

1

c

fdbn

Pp = nominal bearing strength from Pp = nominal bearing strength from AISC equation AISC equation

1'cp A0.85fP

1

21

'cp A

AA0.85fP

Example 5.18Example 5.18

A W 10 x 49 is used as a column and is supported by a concrete pier as A W 10 x 49 is used as a column and is supported by a concrete pier as

shown in the Figure. The top surface of pier is 18 in by 18 in. Design an shown in the Figure. The top surface of pier is 18 in by 18 in. Design an

A36 base plate for a column dead load of 98 kips and a live load of 145 A36 base plate for a column dead load of 98 kips and a live load of 145

kips the concrete strength is fkips the concrete strength is fcc = 3000 psi. = 3000 psi.

The factored load = 1.2 * 98 + 1.6 * 145 = 349.6The factored load = 1.2 * 98 + 1.6 * 145 = 349.6

Compute the required bearing area (assume that plat area < pier Compute the required bearing area (assume that plat area < pier

area)area)

(OK)21.41161.1

18*18

A

A

check

in161.1A

349.6A

18*18A*3*085*0.6

PA

AA(0.85)fφ

1

2

21

11

u1

21

'cc

Also the plate must be at least as large as the column, soAlso the plate must be at least as large as the column, so

BBffd = 10 * 9.98 = 99.8 < 161.1 ind = 10 * 9.98 = 99.8 < 161.1 in22 (OK) (OK)

For B = N = 13 in, AFor B = N = 13 in, A11 provided = 3 * 13 = 169 > 161.1 in provided = 3 * 13 = 169 > 161.1 in22

Check the assumptionCheck the assumption

Plate area = 169 < pier area = 18 * 18 (OK)Plate area = 169 < pier area = 18 * 18 (OK)

Determine the required thicknessDetermine the required thickness

in2.49710*9.984

1db

4

1n

in2.52

813

2

0.8bBn

in1.762

9.4813

2

0.95d-Nm

f

f

in0.8936*13*13*0.9

349.6*22.5

0.9BNF

2PL t

y

u

As a conservative simplification, let As a conservative simplification, let λλ =1, giving =1, giving

L = max(m, n, L = max(m, n, λλn’) = max (1.76, 2.5, 2.497) = 2.5 inn’) = max (1.76, 2.5, 2.497) = 2.5 in

The required plate thickness isThe required plate thickness is

Use a PL 1 x 13 x 13Use a PL 1 x 13 x 13

5.15 BIAXIAL BENDING5.15 BIAXIAL BENDING

Please try to understand it alonePlease try to understand it alone

May be necessary in some of your projects.May be necessary in some of your projects.

If there is a time, we will discuss it laterIf there is a time, we will discuss it later

Any problem, you can see me in my office within the office hours.Any problem, you can see me in my office within the office hours.

The last section 5.16 canceled.The last section 5.16 canceled.

لله“ ومماتي ومحياي نسكي و صالتي ان لله“ قل ومماتي ومحياي نسكي و صالتي ان قل

وأنا * أمرت وبذلك له شريك ال العالمين وأنا * رب أمرت وبذلك له شريك ال العالمين رب

" العظيم الله صدق المسلمين " أول العظيم الله صدق المسلمين أول