68402Slide # 1 Design of Beams for Flexure Monther Dwaikat Assistant Professor Department of...

68402 Slide # 1

Design of Beams for Flexure

Monther DwaikatAssistant Professor

Department of Building EngineeringAn-Najah National University

68402: Structural Design of Buildings II61420: Design of Steel Structures62323: Architectural Structures II

68402 Slide # 2

Introduction

Moment Curvature Response

Sectional Properties

Serviceability Requirements (Deflections)

Compact, Non-compact and Slender Sections

Lateral Torsional Buckling

Design of Beams

Design of Beams for Flexure

68402 Slide # 3

Beams under Flexure Members subjected principally

to transverse gravity loading

• Girders (important floor beams, wide spacing)

• Joists (less important beams, closely spaced)

• Purlins (roof beams, spanning between trusses)

• Stringers (longitudinal bridge beams)

• Lintels (short beams above window/door openings)

68402 Slide # 4

Design for Flexure

Limit states considered

• Yielding

• Lateral-Torsional Buckling

• Local Buckling

• Compact

• Non-compact

• Slender

68402 Slide # 5

Design for Flexure – LRFD Spec.

Commonly Used Sections:

• I – shaped members (singly- and doubly-symmetric)

• Square and Rectangular or round HSS

• Tees and Double Angles

• Rounds and Rectangular Bars

• Single Angles

• Unsymmetrical Shapes

Will not be covered in this course

68402 Slide # 6

Section Force-Deformation Response & Plastic Moment (MP)

A beam is a structural member that is subjected primarily to transverse loads and negligible axial loads.

The transverse loads cause internal SF and BM in the beams as shown in Fig. 1

w P

V(x)

M(x)

x

w P

V(x)

M(x)

x

Fig. 1- SF & BM in a SS Beam

68402 Slide # 7


These internal SF & BM cause longitudinal axial stresses and shear stresses in the cross-section as shown in the Fig. 2

Curvature = = 2/d (Planes remain plane)

Fig. 2 - Longitudinal axial stresses caused by internal BM

V(x)M(x)

yd

b

dF = b dy

V(x)M(x)

yd

b

dF = b dy

2/d

2/d

dybF ydybM2/d

2/d

68402 Slide # 8


Steel material follows a typical stress-strain behavior as shown in Fig 3 below. E = 200 GPa

y

y u

u

y

y u

u

Fig 3 - Typical steel stress-strain behavior.

68402 Slide # 9


If the steel stress-strain curve is approximated as a bilinear elasto-plastic curve with yield stress equal to y, then the section Moment - Curvature (M-) response for monotonically increasing moment is given by Fig. 4.

In Fig. 4, My is the moment corresponding to first yield and Mp is the plastic moment capacity of the cross-section.

• The ratio of Mp to My - the shape factor f for the section.

• For a rectangular section, f = 1.5. For a wide-flange section, f ≈ 1.1.

68402 Slide # 10

Moment-Curvature (NEW)

• Beam curvature is related to its strain and thus to the applied moment

EI

M

y

yy

(1)(1) (2)(2) (3)(3) (4)(4)

68402 Slide # 11

Moment-Curvature (NEW)

xxFSM

• When the section is within elastic range

yxy FSM

yxp FZM aA

Z x )2

(

xxx S

M

I

yMF

Where S is the elastic section modulus

• When the moment exceeds the yield moment My

• Then

Where Z is the plastic section modulus 1.1 S

a

68402 Slide # 12


My

Mp

A: Extreme fiber reaches y B: Extreme fiber reaches 2y C: Extreme fiber reaches 5yD: Extreme fiber reaches 10y E: Extreme fiber reaches infinite strain

A

B C ED

Curvature,

Sec

tion

Mo

men

t, M

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

A B C D E

My

Mp

A: Extreme fiber reaches y B: Extreme fiber reaches 2y C: Extreme fiber reaches 5yD: Extreme fiber reaches 10y E: Extreme fiber reaches infinite strain

A

B C ED

Curvature,

Sec

tion

Mo

men

t, M

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

A B C D E

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

A B C D E

Fig. 4 - M- response of a beam section

68402 Slide # 13


Calculation of MP: Cross-section subjected to either +y or - y at the plastic limit. See Figure 5 below.

Plastic centroid. A1

A2

y

y

yA1

yA2

y1

y2

22

11

21y

21

2y1y

Aofcentroidy

Aofcentroidy,Where

)yy(2

AM

2/AAA

0AAF


A2

y

y

yA1

yA2

y1

y2

22

11

21y

21

2y1y

Aofcentroidy

Aofcentroidy,Where

)yy(2

AM

2/AAA

0AAF

(a) General cross-section

(b) Stress distribution

(c) Force distribution


A2

y

y

yA1

yA2

y1

y2

22

11

21y

21

2y1y

Aofcentroidy

Aofcentroidy,Where

)yy(2

AM

2/AAA

0AAF


A2

y

y

yA1

yA2

y1

y2

22

11

21y

21

2y1y

Aofcentroidy

Aofcentroidy,Where

)yy(2

AM

2/AAA

0AAF

Figure 5. Plastic centroid and MP for general cross-section.

68402 Slide # 14

Moment-Curvature• When the whole section is yielding a plastic hinge

will be formed

plastic hinge

• Structural analysis by assuming collapsing mechanisms of a structure is known as “Plastic analysis”

• The plastic moment Mp is therefore the moment needed at the section to form a plastic hinge

68402 Slide # 15


The plastic centroid for a general cross-section corresponds to the axis about which the total area is equally divided, i.e., A1 = A2 = A/2

• The plastic centroid is not the same as the elastic centroid or center of gravity (c.g.) of the cross-section.

• As shown below, the c.g. is defined as the axis about which A1y1 = A2y2.

c.g. = elastic N.A.

A1, y1

A2, y2

About the c.g. A1y1 = A2y2

y1

y2c.g. = elastic N.A.

A1, y1

A2, y2

About the c.g. A1y1 = A2y2

y1

y2

68402 Slide # 16


• For a cross-section with at-least one axis of symmetry, the neutral axis corresponds to the centroidal axis in the elastic range. However, at Mp, the neutral axis will correspond to the plastic centroidal axis.

For a doubly symmetric cross-section, the elastic and the plastic centroid lie at the same point.

Mp = y x A/2 x (y1+y2)

As shown in Figure 5, y1 and y2 are the distance from the plastic centroid to the centroid of area A1 and A2, respectively.

68402 Slide # 17

Section Force-Deformation Response & Plastic Moment (MP) A/2 x (y1+y2) is called Z, the plastic section

modulus of the cross-section. Values for Z are tabulated for various cross-sections in the properties section of the LRFD manual.

bMp = 0.90 Z Fy

b - strength reduction factor

Mp - plastic moment, which must be 1.5 My for homogenous cross-sections

My - moment corresponding to onset of yielding at the extreme fiber from an elastic stress distribution = Fy S for homogenous cross-sections and = Fyf S for hybrid sections.

Z - plastic section modulus from the Properties section of the AISC manual.

S - elastic section modulus, also from the Properties section of the AISC manual.

68402 Slide # 18

Ex. 4.1 – Sectional Properties Determine the elastic section modulus, S, plastic section

modulus, Z, yield moment, My, and the plastic moment MP, of the cross-section shown below. What is the design moment for the beam cross-section. Assume A992 steel.

12 in.

16 in.

15 in.

0.75 in.

1.0 in.

F1

W

F2

tw = 0.5 in.

12 in.

16 in.

15 in.

0.75 in.

1.0 in.

F1

W

F2

12 in.

16 in.

15 in.

0.75 in.

1.0 in.

F1

W

F2

tw = 0.5 in.

300 mm

15 mm

10 mm

25 mm

400 mm

400 mm

68402 Slide # 19

Ag = 300 x 15 + (400 - 15 - 25) x 10 + 400 x 25 = 18100 mm2

Af1 = 300 x 15 = 4500 mm2

Af2 = 400 x 25 = 10000 mm2

Aw = 10 x (400 - 15 - 25) = 3600 mm2

distance of elastic centroid from bottom =

Ix = 400x253/12 +10000(12.5-145.3)2 + 10x3603/12 +3600(205-

145.3)2 + 300x153/12 +4500(392.5-145.3)2 = 503.7x106 mm4

Sx = 503.7x106 / (400-145.3) = 1977.5x103 mm3

My-x = Fy Sx = 680.2 kN-m.

Sx - elastic section modulus

Ex. 4.1 – Sectional Properties

y

4500(400 15 / 2) 3600 205 10000 12.5145.3

18100y mm

68402 Slide # 20

Ex. 4.1 – Sectional Properties distance of plastic centroid from bottom =

y1 = centroid of top half-area about plastic centroid

= mm

y2 = centroid of bottom half-area about plas. cent.

= mm

Zx = A/2 x (y1 + y2) = 9050 x (256.7 + 11.3) = 2425400 mm3

Zx - plastic section modulus

18100400 9050

222.6

p

p

y

y mm

pypypy

py

18100 145.3 9050 22.6 / 222.6 256.7

9050

22.6 / 2 11.3

68402 Slide # 21

Ex. 4.1 – Sectional Properties Mp-x = Zx Fy = 2425400 x 344/106 = 834.3 kN.m

Design strength according to AISC Spec. F1.1= bMp= 0.9 x 834.3 = 750.9 kN.m

Check = Mp 1.5 My

Therefore, 834.3 kN.m < 1.5 x 680.2 = 1020.3 kN.m - OK!

68402 Slide # 22

Flexural Deflection of Beams - Serviceability Steel beams are designed for the factored design loads.

The moment capacity, i.e., the factored moment strength (bMn) should be greater than the moment (Mu) caused by the factored loads.

A serviceable structure is one that performs satisfactorily, not causing discomfort or perceptions of unsafety for the occupants or users of the structure. • For a steel beam, being serviceable usually means that the

deformations, primarily the vertical slag, or deflection, must be limited.

• 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 specified values.

68402 Slide # 23

Serviceability Requirements

• Steel beams need to satisfy SLS in addition to ULS

• Serviceability limit states are usually checked using non-factored loads.

• Deflection under live loads shall be limited to L/360

• Dead load deflections can be compensated by cambering beams.

• SLS might also include limiting stresses in bottom or top flanges if fatigue is a concern in design (Will be further discussed with plate girders).

• Standard equations to calculate deflection for different load cases:

68402 Slide # 24

Flexural Deflection of Beams - Serviceability The AISC Specification gives little guidance other than a

statement, “Serviceability Design Considerations,” that deflections should be checked. Appropriate limits for deflection can be found from the governing building code for the region.

The following values of deflection are typical max. allowable deflections.

LLDL+LL

• Plastered floor construction L/360 L/240

• Unplastered floor construction L/240 L/180

• Unplastered roof construction L/180 L/120

• DL deflection – normally not considered for steel beams

68402 Slide # 25

Flexural Deflection of Beams - Serviceability

In the following examples, we will assume that local buckling and lateral-torsional buckling are not controlling limit states, i.e, the beam section is compact and laterally supported along the length.

68402 Slide # 26

Ex. 4.2 - Deflections Design a 9 m long simply supported beam subjected to

UDL of 6 kN/m dead load and a UDL of 8 kN/m live load. The dead load does not include the self-weight of the beam.

• Step I. Calculate the factored design loads (without self-weight).

wu = 1.2 wD + 1.6 wL = 20 kN/m

Mu = wu L2 / 8 = 20 x 92 / 8 = 202.5 kN.m (SS beam)

• Step II. Select the lightest section from the AISC Manual design tables.

Zx = Mu/(bFy) = 202.5x106/(0.9x344) = 654x103

select W16 x 26 made from A992 steel with bMp = 224 kN.m

68402 Slide # 27

Ex. 4.2 - Deflections• Step III. Add self-weight of designed section and check design

wsw = 0.38 kN/mTherefore, wD = 6.38 kN/m wu = 1.2 x 6.38 + 1.6 x 6 = 20.46 kN/mTherefore, Mu = 20.46 x 92 / 8 = 207.2 kN.m < bMp of W16 x 26.OK!

• Step IV. Check deflection at service loads.w = 8 kN/m D = 5 w L4 / (384 E Ix) = 5 x (8) x103 x (9)4 / (384 x 200x125) D = 27.3 mm > L/360 - for plastered floor construction

• Step V. Redesign with service-load deflection as design criteriaL /360 = 25 mm > 5 w L4/(384 E Ix)Therefore, Ix > 136.7x106 mm4

68402 Slide # 28

Ex. 4.2 - Deflections

Select the section from the moment of inertia selection from Section Property Tables – select W16 x 31 with Ix = 156x106 mm4

Note that the serviceability design criteria controlled the design and the section

68402 Slide # 29

Ex. 4.3 – Beam Design Design the beam shown below. The unfactored dead and live loads

are shown in Fig. 6 below.

Step I. Calculate the factored design loads (without self-weight).

wu = 1.2 wD + 1.6 wL = 1.2 x 10 + 1.6 x 11 = 29.6 kN/m

Pu = 1.2 PD + 1.6 PL = 1.2 x 0 + 1.6 x 40 = 64 kN

Mu = wU L2 / 8 + PU L / 4 = 299.7 + 144 = 443.7 kN.m

0.67 k/ft. (dead load)10 kips (live load)

30 ft.

15 ft.

0.75 k/ft. (live load)

40 kN10 kN/m

11 kN/m

4.5 m

9 m

68402 Slide # 30

Ex. 4.3 – Beam Design Step II. Select W21 x 44 Zx = 1563x103 mm3

bMp = 0.9x1563x103x344/1000000 = 483.9 kN.m

Self-weight = wsw = 0.64 kN/m.

Step III. Add self-weight of designed section and check designwD = 10 + 0.64 = 10.64 kN/m

wu = 1.2 x 10.64 + 1.6 x 11 = 30.4 kN/m

Therefore, Mu = 30.4 x 92/8 + 144 = 451.8 < bMp of W21 x 44.OK!

Step IV. Check deflection at service live loads.Service loads

• Distributed load = w = 11 kN/m

• Concentrated load = P = L = 40 kN

68402 Slide # 31

Ex. 4.3 – Beam DesignDeflection due to uniform distributed load = d = 5 w L4 / (384 EI)

Deflection due to concentrated load = c = P L3 / (48 EI)

Therefore, service-load deflection = = d + c

= 5x11x94x109/(384x351x106x200) +40x93x109/(48x351x106x200) = 13.4 + 8.7 = 22.1 mmL = 9 m.

Assuming unplastered floor construction, max = L/240 = 9000/240 = 37.5 mm

Therefore, < max - OK!

68402 Slide # 32

Local Buckling of Beam Section – Compact and Non-compact Mp, the plastic moment capacity for the steel shape, is

calculated by assuming a plastic stress distribution (+ or - y) over the cross-section.

The development of a plastic stress distribution over the cross-section can be hindered by two different length effects: • Local buckling of the individual plates (flanges and webs) of the

cross-section before they develop the compressive yield stress y.

• Lateral-torsional buckling of the unsupported length of the beam / member before the cross-section develops the plastic moment Mp.

The analytical equations for local buckling of steel plates with various edge conditions and the results from experimental investigations have been used to develop limiting slenderness ratios for the individual plate elements of the cross-sections.

68402 Slide # 33

M

M

M

M

Figure 7. Local buckling of flange due to compressive stress (s)

Local Buckling of Beam Section – Compact and Non-compact

68402 Slide # 34

Steel sections are classified as compact, non-compact, or slender depending upon the slenderness (l) ratio of the individual plates of the cross-section. • Compact section if all elements of cross-section have p

• Non-compact sections if any one element of the cross-section has p r

• Slender section if any element of the cross-section has r

It is important to note that:

• If p, then the individual plate element can develop and sustain y for large values of e before local buckling occurs.

• If p r, then the individual plate element can develop y at some locations but not in the entire cross section before local buckling occurs.

• If r , then elastic local buckling of the individual plate element occurs.


68402 Slide # 35

Classification of Sections

• Classifications of bending elements are based on limits of local buckling

• The dimensional ratio l represents

• Two limits exist p and r

p represents the upper limit for compact sections

r represents the upper limit for non-compact sections

f

f

t

b

2

wt

h

fb

hwt

ft

compactP

compactnonrP

slenderr

68402 Slide # 36

Compact

Non-Compact

Slender

y

Co

mpr

essi

ve a

xial

str

ess,

Effective axial strain,

Compact

Non-Compact

Slender

y

Co

mpr

essi

ve a

xial

str

ess,

Effective axial strain,

Figure 8. Stress-strain response of plates subjected to axial compression and local

buckling.

Thus, slender sections cannot develop Mp due to elastic local buckling. Non-compact sections can develop My but not Mp before local buckling occurs. Only compact sections can develop the plastic moment Mp.


68402 Slide # 37

All rolled wide-flange shapes are compact with the following exceptions, which are non-compact.• W21x48, W40x174, W14x99, W14x90, W12x65, W10x12, W8x10,

W6x15 (made from A992)


68402 Slide # 38

Classification of Sections• The limits are

fb

hwt

ft

f

f

t

b

2

wt

h

yp F

E76.3

yp F

E38.0

9.6883.0

yr F

Ey

r F

E70.5

WebFlange

68402 Slide # 39

Lateral-Torsional Buckling (LTB)

M

M

M

M

(a)

(b)

M

M

M

M

M

M

M

M

(a)

(b)

The laterally unsupported length of a beam-member can undergo LTB due to the applied flexural loading (BM).

Figure 9. Lateral-torsional buckling of a wide-flange beam subjected to constant moment.

68402 Slide # 40

Lateral-Torsional Buckling (LTB) LTB is fundamentally similar to the flexural buckling or

flexural-torsional buckling of a column subjected to axial loading. • The similarity is that it is also a bifurcation-buckling type

phenomenon.

• The differences are that lateral-torsional buckling is caused by flexural loading (M), and the buckling deformations are coupled in the lateral and torsional directions.

There is one very important difference. For a column, the axial load causing buckling remains constant along the length. But, for a beam, usually the LTB causing bending moment M(x) varies along the unbraced length. • The worst situation is for beams subjected to uniform BM along the

unbraced length. Why?

68402 Slide # 41

Lateral-Torsional Buckling (LTB) –Uniform BM Consider a beam that is simply-supported at the ends and

subjected to four-point loading as shown below. The beam center-span is subjected to uniform BM (M). Assume that lateral supports are provided at the load points.

Laterally unsupported length = Lb. If the laterally unbraced length Lb is less than or equal to a

plastic length LP then lateral torsional buckling is not a problem and the beam will develop its plastic strength MP.

Lb

PP

68402 Slide # 42

Lateral-Torsional Buckling (LTB) – Uniform BM

Lp = 1.76 ry x - for I members & channels

If Lb is greater than LP then lateral torsional buckling will occur and the moment capacity of the beam will be reduced below the plastic strength MP as shown in Fig. 10.

As shown in Fig. 10, the lateral-torsional buckling moment (Mn = Mcr) is a function of the laterally unbraced length Lb and can be calculated using the equation:

yFE /

68402 Slide # 43


Mn = Mcr =

Mn - moment capacity

Lb - laterally unsupported length.

Mcr - critical lateral-torsional buckling moment.

E – 200 GPa;

G – 77 GPa

Iy - moment of inertia about minor or y-axis (mm4)

J - torsional constant (mm4) from the Section Property Tables.

Cw - warping constant (mm6) from the Section Property Tables.

This Eq. is valid for ELASTIC LTB only (like the Euler equation). This means it will work only as long as the cross-section is elastic and no portion of the cross-section has yielded.

2 2

2 2

y w

b b

EI ECGJ

L L

68402 Slide # 44


Fig. 10 Lateral Torsional Buckling (Uniform Bending)

PlasticNo

Instability

InelasticLTB Elastic

LTB

Lr

Mom

ent C

apac

ity,

Mn

Unbraced length, Lb

Mn =

2

2

2

2

b

w

b

y

L

ECGJ

L

EI

pr

pbrppn LL

LLMMMM )(

Mn = Mp

Zx Fy= Mp

Sx (Fy – 10) = Mr

Lp Lr

Mom

ent C

apac

ity,

Mn

Mom

ent C

apac

ity,

Mn

Mn

Unbraced length, LbUnbraced length, Lb

Mn =

2

2

2

2

b

w

b

y

L

ECGJ

L

EI Mn =

2

2

2

2

b

w

b

y

L

ECGJ

L

EI

2

2

2

2

b

w

b

y

L

ECGJ

L

EI

pr

pbrppn LL

LLMMMM )(

Mn = Mp

Zx Fy= Mp

Sx (Fy – 10) = Mr

Lp

(0.7Fy)

68402 Slide # 45


As soon as any portion of the cross-section reaches the yield stress Fy, the elastic LTB equation cannot be used.

• Lr is the unbraced length that corresponds to a LTB moment

Mr = Sx (0.7Fy).

• Mr will cause yielding of the cross-section due to residual stresses.

When the unbraced length is less than Lr, then the elastic LTB Eq. cannot be used.

When the unbraced length (Lb) is less than Lr but more than the plastic length Lp, then the LTB Mn is given by the Eq. below:

68402 Slide # 46

Lateral-Torsional Buckling – Uniform BM

2

0

0

7.076.611

7.095.1

Jc

hS

E

F

hS

Jc

F

ErL xy

xytsr

pr

pbrppn LL

LLMMMM )(• If Lp Lb Lr, then

• This is linear interpolation between (Lp, Mp) and (Lr, Mr)

• See Fig. 10 again.

•

•

• For a doubly symmetric I-shape: c = 1

• h0 = distance between the flange centroids (mm)

x

wy

ts S

CIr 2

68402 Slide # 47

Moment Capacity of Beams Subjected to Non-uniform BM As mentioned previously, the case with uniform bending

moment is worst for lateral torsional buckling.

For cases with non-uniform bending moment, the LTB moment is greater than that for the case with uniform moment.

The AISC specification says that:• The LTB moment for non-uniform bending moment case

• Cb x lateral torsional buckling moment for uniform moment case.

68402 Slide # 48

Moment Capacity of Beams Subjected to Non-uniform BM

Cb is always greater than 1.0 for non-uniform bending moment.

• Cb is equal to 1.0 for uniform bending moment.

• Sometimes, if you cannot calculate or figure out Cb, then it can be conservatively assumed as 1.0. for doubly and singly symmetric sections

Mmax - magnitude of maximum bending moment in Lb

MA - magnitude of bending moment at quarter point of Lb

MB - magnitude of bending moment at half point of Lb

MC - magnitude of bending moment at three-quarter point of Lb

• Use absolute values of M

0.33435.2

5.12

max

max

cBA

b MMMM

MC

68402 Slide # 49

Flexural Strength of Compact Sections

0.33435.2

5.12

max

max

CBA

b MMMM

MC

Mmax

MA

MB

MC@ quarter

@ mid@ three-quarter

Moments determined between bracing points

68402 Slide # 50

Values of Cb3-1

68402 Slide # 51

Moment Capacity of Beams Subjected to Non-uniform Bending Moments The moment capacity Mn for the case of non-uniform

bending moment• Mn = Cb x {Mn for the case of uniform bending moment} Mp

• Important to note that the increased moment capacity for the non-uniform moment case cannot possibly be more than Mp.

• Therefore, if the calculated values is greater than Mp, then you have to reduce it to Mp

68402 Slide # 52

Moment Capacity of Beams Subjected to Non-uniform BM

LrLp

Mr

Mp

Cb = 1.0

Cb = 1.2

Cb = 1.5

Mom

ent C

apac

ity,

Mn

Unbraced length, Lb

Figure 11. Moment capacity versus Lb for non-uniform moment case

Uniform BMUniform BM

Non-uniform BMNon-uniform BM

Cb = 1.0 means uniform BM

68402 Slide # 53

Structural Design of Beams• Steps for adequate design of beams:

1) Compute the factored loads, factored moment and shear

2) Determine unsupported length Lb and Cb

3) Select a WF shape and choose Zx assuming it is a compact section with full lateral support

4) Check the section dimension for compactness and determine bMn

5) Use service loads to check deflection requirements

yb

ux F

MZ

nbu MM

ynbu

ypn

ZFMM

ZFMM

9.0

68402 Slide # 54

Ex. 4.4 – Beam Design Use Grade 50 steel to design the beam shown below. The

unfactored uniformly distributed live load is equal to 40 kN/m. There is no dead load. Lateral support is provided at the end reactions. Select W16 section.

24 ft.

wL = 3 kips/ft.

Lateral support / bracing

wwLL = 40 kN/m = 40 kN/m

7.5 m7.5 m

68402 Slide # 55

Ex. 4.4 – Beam Design• Step I. Calculate the factored loads assuming a reasonable self-

weight.

Assume self-weight = wsw = 1.46 kN/m.

Dead load = wD = 0 + 1.46 = 1.46 kN/m.

Live load = wL = 40 kN/m.

Ultimate load = wu = 1.2 wD + 1.6 wL = 65.8 kN/m.

Factored ultimate moment = Mu = wu L2/8 = 462.3 kN-m.

Is BM uniform?? Yes Cb =1.0

No Go to Step II

• Step II. Determine unsupported length Lb and Cb

There is only one unsupported span with Lb = 7.5 m

Cb = 1.14 for the parabolic bending moment diagram, See values of Cb shown in Table 3-1.

68402 Slide # 56

Ex. 4.4 – Beam Design

mFErL yyp 71.11000

3442000004.4076.1/76.1

mmS

CIr

x

wy

ts 1.48101327

10610105.153

96

• Step III. Select a wide-flange shape

• Compute Zx = 462.3*106/(0.9*344) = 1493x106 mm3.

• Select W16 x 50 steel section

• Zx = 1508x103 mm3 Sx = 1327x103 mm3 ry = 40.4 mm

• Cw = 610x109 mm6 Iy = 15.5x106 mm4 J = 0.63x106 mm4

•

•

•

2

0

0

7.076.611

7.095.1

Jc

hS

E

F

hS

Jc

F

ErL xy

xytsr

68402 Slide # 57


mLr 26.581.21398101327

11063.0

3447.0

200000

1000

1.4895.1

3

6

81.211063.0

398101327

200000

3447.076.61

7.076.61

2

6

32

0

Jc

hS

E

Fxy

• h0 = D - TF = 414 – 16 = 398 mm

•

•

• Lb > Lr

mkNMmkNmmkN

L

ECGJ

L

EICM

p

b

w

b

ybn

.8.51810

344101508.222.10222

7500

106102001063.077

7500

105.1520014.1

6

33

2

926

2

62

2

2

2

2

68402 Slide # 58

• Step IV. Check if section is adequate

• Mu > Mn Not OK

• Step V. Try a larger section.

• After few trials select W16 x 67 Mn = 497.7 > Mu OK

• Step VI. Check for local buckling. = Bf / 2Tf = 7.7; Corresponding p = 0.38 (E/Fy)0.5 = 9.19

Therefore, < p - compact flange

= H/Tw = 35.9; Corresponding p = 3.76 (E/Fy)0.5 = 90.5

Therefore, < p - compact web

Compact section. - OK! This example demonstrates the method for designing beams and

accounting for Cb > 1.0)

Values for Lr and Lp can be obtained from Tables too


68402 Slide # 59

Ex. 4.5 – Beam Design Design the beam shown below. The concentrated live

loads acting on the beam are shown in the Figure. The beam is laterally supported at the load and reaction points. Use Grade 50 steel.

30 ft.

wsw = 0.1 kips/ft.


12 ft. 8 ft. 10 ft.

30 kips 30 kips135 KN135 KN 135 KN135 KN

1.5 KN/m1.5 KN/m

3.6 m3.6 m 2.4 m2.4 m 3.0 m3.0 m

9.0 m9.0 m

68402 Slide # 60

Ex. 4.5 – Beam Design• Step I. Assume a self-weight and determine the factored design

loads

Let, wsw = 1.5 kN/m

PL = 135 kN

Pu = 1.6 PL = 216 kN

wu = 1.2 x wsw = 1.8 kN/m

The reactions and bending moment diagram for the beam are shown below.

68402 Slide # 61


wsw = 0.12 kips/ft.

12 ft. 8 ft. 10 ft.

48 kips 48 kips

46.6 kips 53 kips

550.6 kip-ft. 524 kip-ft.

A

B CD

A B C D

30 ft.

wsw = 0.1 kips/ft.


12 ft. 8 ft. 10 ft.

30 kips 30 kips

216 KN216 KN 216 KN216 KN

1.8 KN/m1.8 KN/m

3.0 m3.0 m3.6 m3.6 m 2.4 m2.4 m209.7 kN209.7 kN 238.5 kN238.5 kN

754.9 kN.m754.9 kN.m 715.5 kN.m715.5 kN.m

68402 Slide # 62


• Step II. Determine Lb, Cb, Mu, and Mu/Cb for all spans.

Span Lb

(m)

Cb Mu

(kN-m)

Mu/Cb

(kN-m)

AB 3.6 1.67 754.9 452.8

BC 2.4 1.0(assume)

754.9 754.9

CD 3.0 1.67 715.5 429.2

It is important to note that it is possible to have different Lb and

Cb values for different laterally unsupported spans of the same

beam.

Cb – Table 3-1

68402 Slide # 63

Ex. 4.5 – Beam Design• Step III. Design the beam and check all laterally unsupported spans

Assume that span BC is the controlling span because it has the largest Mu/Cb although the corresponding Lb is the smallest.

Required Zx = 754.9*106/(0.9*344) = 2438x103 mm3

After few trials select W21 x 68 from section property Table.

Lp = 1.94 m Lr = 5.73 m (From Tables)

For all members Lp < Lb < Lr

Check the selected section for spans AB, BC, and CDSpan Lb (m) Cb bMn for Cb value bMp limit

AB 3.6 1.67 1125.5 kN.m 811.8 kN.m

BC 2.4 1.0 773.6

CD 3 1.67 1208.7 kN.m 811.8 kN.m

pr

pbrppbn LL

LLMMMCM )(

68402 Slide # 64

Ex. 4.5 – Beam DesignThus, for span AB, bMn = 811.8 kN.m > Mu - OK!

For span BC, bMn = 773.6 kN.m > Mu - OK!

For span CD, bMn = 811.8 kN.m > Mu - OK!

• Step IV. Check for local buckling

Bf / 2Tf = 6.04; Corresponding p = 0.38 (E/Fy)0.5 = 9.19

Therefore compact flange

H/Tw = 43.6; Corresponding p = 3.76 (E/Fy)0.5 = 90.55

Therefore compact web

Compact section. - OK!

• This example demonstrates the method for designing beams with several laterally unsupported spans with different Lb and Cb values.

68402 Slide # 65

Ex. 4.6 – Beam Design Design the simply-supported beam shown below. The

uniformly distributed dead load is equal to 15 kN/m and the uniformly distributed live load is equal to 30 kN/m. A concentrated live load equal to 40 kN acts at the mid-span. Lateral supports are provided at the end reactions and at the mid-span. Use Grade 50 steel.

wD = 1.0 kips/ft.

12 ft. 12 ft.

10 kips

A

BC

wL = 2.0 kips/ft.

30 ft.

wsw = 0.1 kips/ft.


12 ft. 8 ft. 10 ft.

30 kips 30 kips40 kN15 kN/m30 kN/m

3.6 m 3.6 m

68402 Slide # 66

Ex. 4.6 – Beam Design• Step I. Assume the self-weight and calculate the factored design loads.

Let, wsw = 1.5 kN/m

wD = 15 + 1.5 = 16.5 kN/m

wL = 30 kN/m

wu = 1.2 wD + 1.6 wL = 67.8 kN/m

Pu = 1.6 x 40 = 64 kNThe reactions and the bending moment diagram for the factored loads are shown below.

62.24 kips 62.24 kips

wu = 4.52 kips/ft.

12 ft. 12 ft.

16 kips

B

x M(x) = 62.24 x - 4.52 x2/2

64 kN

67.8 kN/m

3.6 m 3.6 m276.1 kN 276.1 kN

M(x) = 276.1(x) + 67.8(x)2/2

68402 Slide # 67

• Step II. Calculate Lb and Cb for the laterally unsupported spans.

Since this is a symmetric problem, need to consider only span AB

Lb =3.6 m, M(x) = 276.1 x – 67.8 x2/2

Therefore,

MA = M(x = 0.9 m) = 221 kN.m - quarter-point along Lb = 3.6 m

MB = M(x = 1.8 m) = 387 kN.m - half-point along Lb = 3.6 m

MC = M(x = 2.7 m) = 498 kN.m - three-quarter point along Lb= 3.6 m

Mmax = M(x = 3.6 m) = 554.6 kN.m - maximum moment along Lb =3.6 m

Therefore, Cb = 1.36


68402 Slide # 68


• Step III. Design the beam section

Mu = Mmax = 554.6 kN.m

Lb = 3.6 m, Cb = 1.36

Required Zx = 554.6*106/(0.9*344) = 1791x103 mm3

After few trials, select W21 x 57 steel section

Lp = 1.46 m Lr = 4.37 m

Lp < Lb < Lr

bMn = 699 kN.m > bMp = 639.3 kN.m

bMn= 639.3 >Mu OK

68402 Slide # 69


• Step V. Check for local buckling.

Bf / 2Tf = 7.87; Corresponding p = 0.38 (E/Fy)0.5 = 9.192

Therefore, compact flange

l = h/tw = 50.0; Corresponding p = 3.76 (E/Fy)0.5 = 90.55

Therefore, compact web

Compact section. - OK!

• This example demonstrates the calculation of Cb and the iterative design method.

68402 Slide # 70

Shear Capacity The shear capacity of the beam is

nvu VV vwyn CAFV 6.0

For I-shaped sections, the factors Cv and v are functions of the shear buckling of the web and thus the ration h/tw

yw F

Ethif 24.2/

0.1vC

representing the case of no web instability.

Aw = dtw

9.0v

0.1v

68402 Slide # 71

Shear Capacity

9.0v

y

vw

y

v

F

Ekth

F

Ek37.1/10.1

yw

vv

Fth

EkC

2/

51.1

The second case represents inelastic web buckling

y

vw F

Ekth 37.1/

y

vw F

Ekth 10.1/ 0.1vC

The last case represents elastic web buckling

For all other doubly and singly sym. sections and channels except round HSS

9.0vw

yvv th

FEkC

10.1

The first case represents the case of no web instability.

9.0v

9.0v

68402 Slide # 72

Shear Capacity

• Aw = dtw

• The web plate buckling coefficient, kv, is given

• For unstiffened webs with h/tw <260, kv = 5 except for the stem of tee shapes, kv =1.2

• For stiffened webs

• a = clear distance between transverse stiffeners

• h = for rolled shapes, the clear distance between flanges less the fillet of corner radii.

2

2

2600.3,5

55

w

v

thhaorha

hak

68402 Slide # 73

Beam Bearing Plates Design of a beam bearing plate would require checking:

t

B N

1- Web Yielding and Web crippling to determine N

2- Bearing capacity to determine B

3- Plate moment capacity to determine t

AISC Specifications: Chapter K

68402 Slide # 74

Beam Bearing Plates When a bearing plate is used at beam end, two limit states shall be

considered

1. Web Yielding

This represents yield of the web at the vicinity of the flange due to excessive loading

kN wyn tFNkR )5.2(

CASE 1: At Support

R

R

wyn tFNkR )5(

CASE 2: Interior Load

nRR 0.1

68402 Slide # 75

Beam Bearing Plates

Web crippling represent the possible buckling of the web

2. Web Crippling

nRR 75.0

w

fy

f

wwn t

tFE

t

t

d

NtR

5.1

2 314.0

CASE 1: At Support

CASE 2: Interior Load

2.0d

N

w

fy

f

wwn t

tFE

t

t

d

NtR

5.1

2 2.04

14.0 2.0d

N

w

fy

f

wwn t

tFE

t

t

d

NtR

5.1

2 318.0

68402 Slide # 76

Ex. 4.7 – Beam Design Check the beam shown in the figure below for:

• Shear capacity.

• Web yielding.

• Web crippling.

• Assume the width of the bearing plate is 100 mm. Use Grade 50 steel.

wD = 1.0 kips/ft.

12 ft. 12 ft.

10 kips

A

BC

wL = 2.0 kips/ft.

40 kN10 kN/m25 kN/m

2 m 2 m

W16x26

68402 Slide # 77

Ex. 4.7 – Beam Design• Step I. The section used from Example 4.6 is W21x57.

The self-weight wsw = 0.83 kN/m

wD = 10 + 0.38 = 10.38 kN/m

wL = 25 kN/m

wu = 1.2 wD + 1.6 wL = 52.5 kN/m

Pu = 1.6 x 40 = 64 kNThe reactions and the bending moment diagram for the factored loads are shown below.

62.24 kips 62.24 kips

wu = 4.52 kips/ft.

12 ft. 12 ft.

16 kips

B

x M(x) = 62.24 x - 4.52 x2/2

64 kN

52.5 kN/m

2 m 2 m137 kN 137 kN

Vu = 137 kN

68402 Slide # 78


• Step II. h/tw = 56.8

Assume unstiffened web, h/tw <260, kv = 5

Assume unstiffened web, h/tw <260, kv = 5

Vn = 0.9*(0.6Fy)*d*tw*Cv

Vn = 0.9*(0.6x344)*399*6.4x10-3 = 474.4 kN> Vu

8.5654344

20000024.224.2

wy t

h

F

E

9.01

8.563.59344

200000510.110.1

v

wy

v

C

t

h

F

Ek

68402 Slide # 79


• Step III. Web yielding critical is support

k = 19 mm

R = (2.5k + N)*Fy*tw

R = 1x(2.5x19 + 100)x344x6.4/1000 = 324.7 kN

R > reaction = 137 kN OK

• Step IV. Web crippling critical is support

d = 399 mm tw = 6.4 mm tf = 8.8 mm

N/d = 100/399 = 0.25 > 0.2

w

fy

f

wwn t

tFE

t

t

d

NtR

5.1

2 2.04

14.0

68402 Slide # 80


R = 179 kN > reaction = 137 kN OK

kNRn 7.238104.6

8.8344200000

8.8

4.62.0

399

100414.64.0 3

5.12

68402Slide # 1 Design of Beams for Flexure Monther Dwaikat Assistant Professor Department of...

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