aect360 lecture 11

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Lecture 11 - Page 1 of 12

Lecture 11 Beam Deflections, Combined stresses

Beam Deflections

All beams deflect (or sag) under load. Even the strongest, most substantial

beam imaginable will deflect under its own weight. Under normal conditions, theactual amount of deflection in floor beams is generally un-noticeable, see figurebelow:

A) Actual Beam Def lections:

The actual beam deflections can be calculated using the following formulae:

Case 1 Uniform Load on simply-supported beam:

EIwL

3845

4

max = occurs at midspan

L

w

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Case 2 Concentrated load at midspan:

EI

PL

48

3


Case 3 - Two equal concentrated loads at 1/3 points:

EI

PL

2.28

3

max=

occurs at midspan

Case 4 Three equal concentrated loads at points:

EI

PL

1.20

3


L

P

L

P P

L

P P P

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Case 5 Cantilever with uniform load:

EI

wL

8

4

max = occurs at the free end

Case 6 Cantilever with concentrated load at free end:

EI

PL

3

3

max = occurs at the free end

B) Allowable Beam Deflections:

Studies have shown that excessive deflection in beams causes undesirableeffects such as cracked ceilings and floors as well as vibration. Building codestypically specify the maximum allowable deflection so as to avoid theseproblems. The actual deflections are compared against the allowable deflectionsin another check of structural adequacy, sometimes referred to as serviceabilitychecking. It is possible that a lightly-loaded beam having a relatively long spanmay be adequate in a stress analysis, but fail due to deflection.

The International Building Code, (IBC) dictates allowable deflections, and isshown below.

L

w

L

P

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Example 1GIVEN: A simply-supported W16x26 steel beam carries a total uniform DL+LLload of 1300 PLF (including beam weight).REQUIRED:

1) Determine the maximum actual deflection.

2) If the allowable DL+LL deflection is L/360, determine if the beam isacceptable.

The actual mid-span deflection is determined by the formula:

EI

wL

384

54

max =

)301)(000,000,29(384

)/"12'24)(/"12

1300(5

4

4

maxinPSI

ftxft

PLF

=

max = 1.11 in.

The allowable deflection is determined as:

360

Lallow =

360

)"12'24( xallow =

allow= 0.80 in.

Since the actual deflection of 1.11 is more than the allowable, it isunacceptable.

24-0

w = 1300 PLF

From beamproperties

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Example 2GIVEN: A simply-supported W30x99 steel girder carries a uniform load of 1100PLF (including beam weight) as well as 2 point loads of 14 kips each acting atthe 1/3 points.REQUIRED: Determine the maximum actual mid-span deflection.

Since the loads acting on the girder are symmetric, the maximum deflectionwill occur at the mid-span. This means that the total deflection can be calculatedas the sum of the deflections from the point loads plus the deflection from theuniform load:

unifpototal += int

)384

5()

2.28(

43

EI

wL

EI

PLtotal +=

))3990)(000,29(384

)"12'27)(/"12

1.1(5

())3990)(000,29(2.28

)"12'27)(16((

4

4

4

3

inKSI

xft

KLF

inKSI

xkipstotal +=

total = 0.17 + 0.11

total = 0.28

w =1.1 KLF

27-0

16 kips 16 kips

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Combined Stresses

Many structural elements are subject to two or more types of stressessimultaneously. For example, a column that carries vertical compressiveloads may also be subject to lateral wind loads creating bending moments.

Or, a spandrel beam carrying bending moment may also experiencetorsion. This phenomenon is called combined stresses.

Designing for combined stresses involves determining the sum of theratios of the actual stresses to the allowable stresses not exceeding 1.0.In other words, the sum of the actual stresses cannot exceed the sum ofthe respective allowable stresses.

0.1tressesAllowableS

ssesActualStre

So, if a member were subject to compressive stress and bending stresssimultaneously:

0.1+b

b

c

c

F

f

F

f

where: fc = actual compressive stressFc = allowable compressive stressfb = actual bending stressFb = allowable bending stress

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Example 3GIVEN: A pair of W8x15 steel columns are used to support the 9000 lb. sign asshown in the figure below. Additionally, a lateral wind force of 21 PSF strikes thesign creating moment in the columns. The allowable bending stress is 24 KSIand the allowable compressive stress is 8 KSI.

REQUIRED: Determine if the W8x15 columns are adequate based uponcombined stresses.

Determine the actual compressive stress, fc on each column:

Area

Loadfc =

244.4

)9(2

1

in

kipsfc =

fc = 1.01 KSI

From textbook Appendix

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Determine the actual bending stress, fb on each column:

The maximum moment acting on the column occurs at the base of thecolumn at point A. Taking moments at point A:

MA = 189 PLF(8)(14)= 21,168 ft-lb

Determine the maximum bending stress, fb

as follows:

x

bS

Mf =

38.11

)/"12)(21168(

in

ftlbftfb

=

fb= 21,527 PSI= 21.5 KSI

A

8-0

10-0

Side view of column

w = 9(21 PSF)= 189 PLF

(18-0) = 9

Moment arm = 14-0

Equivalent pointload acting onsign

From textbook Appendix

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Checking combined stresses:

0.1+b

b

c

c

F

f

F

f

????0.124

5.21

8

01.1+

KSI

KSI

KSI

KSI

0.13 + 0.90 = 1.03

Column is UNACCEPTABLE since 1.03 > 1.0

Example 4

GIVEN: An 18 deep x 6 wide glued-laminated wood beam is attached to theglued-laminated 8 x 8 wood column eccentrically as shown below. The beamthus creates a moment on the column. Assume the allowable bending stress forthe column is 1600 PSI and the allowable axial (compressive) stress is 850 PSI.REQUIRED: Determine if the column is acceptable based upon combinedbending & compressive stresses.

Moment

8

9000 lbs.Eccentricity = 6

Side View F.B.D.

Beam

Column

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Step 1 Determine actual axial stress on column:

Actual Axial stress =Area

Load

"8"89000Lbs=

Actual Axial stress = 140.6 PSI

Step 2 Determine actual bending stress on column:

Actual Bending stress =S

M

Where: M = Applied moment

= Pe

= (9000 Lbs.)(6.5)

= 58,500 lb-in

S = Section modulus

=

6

2bh

=6

)"8)("8(2

= 85.33 in3

Actual Bending stress =3

33.85

500,58

in

inlb

Actual Bending stress = 685.6 PSI

h = 8

b = 8

Column cross-section

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Step 3 Determine combined stress:

0.1+b

b

c

c

F

f

F

f

????0.11600

6.685

850

6.140+

PSI

PSI

PSI

PSI

0.17 + 0.43 = 0.6

Column is ACCEPTABLE since 0.6 < 1.0

aect360 lecture 11

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