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University of Michigan, TCAUP Structures II Slide 1/27 Architecture 324 Structures II Deflection of Structural Members Slope and Elastic Curve Deflection Limits Diagrams by Parts Symmetrical Loading Asymmetrical Loading Deflection Equations and Superposition MAISIE GWYNNE, 2013 - USC University of Michigan, TCAUP Structures II Slide 2/27 Deformation Based on Hooke’s Law and the definition of Young’s Modulus, E And so,

Deflection of Structural Members - · PDF filebut zero slope and deflection. • Free end has maximum slope and deflection, but zero moment. • Slope is either downward (-) or upward

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Page 1: Deflection of Structural Members - · PDF filebut zero slope and deflection. • Free end has maximum slope and deflection, but zero moment. • Slope is either downward (-) or upward

University of Michigan, TCAUP Structures II Slide 1/27

Architecture 324Structures II

Deflection of Structural Members

• Slope and Elastic Curve• Deflection Limits• Diagrams by Parts• Symmetrical Loading• Asymmetrical Loading• Deflection Equations and Superposition

MAISIE GWYNNE, 2013 - USC

University of Michigan, TCAUP Structures II Slide 2/27

Deformation

Based on Hooke’s Law and the definition of Young’s Modulus, E

And so,

Page 2: Deflection of Structural Members - · PDF filebut zero slope and deflection. • Free end has maximum slope and deflection, but zero moment. • Slope is either downward (-) or upward

University of Michigan, TCAUP Structures II Slide 3/27

Deflection

Axial fiber deformation in flexure results in normal (vertical) deflection.

The change in lengths, top and bottom, results in the material straining. For a simple span with downward loading, the top is compressed and the bottom stretched.

The material strains result in corresponding stresses. By Hooke‘s Law, these stresses are proportional to the strains which are proportional to the change in length of the radial arcs of the beam “fibers“.

University of Michigan, TCAUP Structures II Slide 4/27

Deflection

• Deflection is the distance that a beam bends

from its original horizontal position, when

subjected to loads.

• The compressive and tensile forces above and

below the neutral axis, result in a shortening

(above n.a.) and lengthening (below n.a.) of the

longitudinal fibers of a simple beam, resulting in

a curvature which deflects from the original

position.

Axial Stiffness

Flexural Stiffness

Page 3: Deflection of Structural Members - · PDF filebut zero slope and deflection. • Free end has maximum slope and deflection, but zero moment. • Slope is either downward (-) or upward

University of Michigan, TCAUP Structures II Slide 5/27

Slope

• The curved shape of a deflected beam is

called the elastic curve

• The angle of a tangent to the elastic

curve is called the slope, and is

measured in radians.

• Slope is influenced by the stiffness of

the member:

– material stiffness E, the modulus of

elasticity

– sectional stiffness I, the moment of

inertia,

– as well as the length of the beam, L

University of Michigan, TCAUP Structures II Slide 6/27

Deflection Limits (serviceability)

• Various guidelines have been

derived, based on usage, to

determine maximum allowable

deflection limits.

• Typically, a floor system with a LL

deflection in excess of L/360 will

feel bouncy or crack plaster.

• Flat roofs require a minimum

slope of ¼” / ft to avoid ponding.

“Ponding” refers to the retention

of water due solely to deflection of

relatively flat roofs. Progressive

deflection due to progressively

more impounded water can lead

to collapse. International Building Code - 2006

roof pondingfrom IRC, Josh 2014

Page 4: Deflection of Structural Members - · PDF filebut zero slope and deflection. • Free end has maximum slope and deflection, but zero moment. • Slope is either downward (-) or upward

University of Michigan, TCAUP Structures II Slide 7/27

Relationships of Forces and DeformationsThere is a series of relationships involving forces and deformations along a beam, which can be useful in analysis. Using either the deflection or load as a starting point, the following characteristics can be discovered by taking successive derivatives or integrals of the beam equations.

University of Michigan, TCAUP Structures II Slide 8/27

Slope and Deflection in Symmetrically Loaded Beams

• Maximum slope occurs at the ends of the beam

• A point of zero slope occurs at the center line.

This is the point of maximum deflection.

• Moment is positive for gravity loads.

• Shear and slope have balanced + and - areas.

• Deflection is negative for gravity loads.

Page 5: Deflection of Structural Members - · PDF filebut zero slope and deflection. • Free end has maximum slope and deflection, but zero moment. • Slope is either downward (-) or upward

University of Michigan, TCAUP Structures II Slide 9/27

Cantilever Beams

• One end fixed. One end free

• Fixed end has maximum moment,

but zero slope and deflection.

• Free end has maximum slope and

deflection, but zero moment.

• Slope is either downward (-) or

upward (+) depending on which

end is fixed.

• Shear sign also depends of which

end is fixed.

• Moment is always negative for

gravity loads.

University of Michigan, TCAUP Structures II Slide 10/27

Methods to Calculate Deflection

Integrationcan use to derive equations

Diagramssymmetric load cases

Diagrams (by parts)asymmetric load cases

Diagrams (by shifting baseline)asymmetric load cases

Equationssingle load cases

Superposition of Equationsmultiple load cases

Page 6: Deflection of Structural Members - · PDF filebut zero slope and deflection. • Free end has maximum slope and deflection, but zero moment. • Slope is either downward (-) or upward

University of Michigan, TCAUP Structures II Slide 11/27

Deflection by Integration

Load

Shear

Moment

University of Michigan, TCAUP Structures II Slide 12/27

Deflection by Integration

Slope

Deflection

Page 7: Deflection of Structural Members - · PDF filebut zero slope and deflection. • Free end has maximum slope and deflection, but zero moment. • Slope is either downward (-) or upward

University of Michigan, TCAUP Structures II Slide 13/27

Deflection by Diagrams

Load

Shear

Moment

Slope (EI)

Deflection (EI)

University of Michigan, TCAUP Structures II Slide 14/27

Table D-24 I. Engel

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University of Michigan, TCAUP Structures II Slide 15/27

Diagrams by Parts marks vertex which must be present for area equations to be valid.

University of Michigan, TCAUP Structures II Slide 16/27

Diagrams by Base Line ShiftAsymmetrically Loaded Beams

• The value of the slope at each of the

endpoints is different.

• The exact location of zero slope (and

maximum deflection) is unknown.

• Start out by assuming a location of zero slope

(Choose a location with a known dimension

from the loading diagram)

• With the arbitrary location of zero slope, the

areas above and below the base line (“A” and

“B”) are unequal

• Adjust the base line up or down by D distance

in order to equate areas “A” and “B”. Shifting

the base line will remove an area “a” from “A”

and add an area “b” to “B”

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University of Michigan, TCAUP Structures II Slide 17/27

Asymmetrically Loaded Beams (continued)

• Compute distance D with the equation:

so,

• With the vertical shift of the base line, a

horizontal shift occurs in the position of zero

slope.

• The new position of zero slope will be the

actual location of maximum deflection.

• Compute the area under the slope diagram

between the endpoint and the new position of

zero slope in order to compute the magnitude

of the deflection.

University of Michigan, TCAUP Structures II Slide 18/27

Example: Asymmetrical Loading – Diagram method

• Solve end reactions

• Construct shear diagram

• Construct moment diagram

• Choose point of θEI = 0Sum moment areas to each side to find θEI end values

• Calculate distance D

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University of Michigan, TCAUP Structures II Slide 19/27

Example: Asymmetrical Loading Diagram method

• Shift the baseline and find point of crossing. Find new areas

• Check that areas A and B balance. Any difference is just round-off error. (average the two to reduce error)

University of Michigan, TCAUP Structures II Slide 20/27

Example: Asymmetrical Loading Diagram method

• Calculate the deflection in inches for a given member.

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University of Michigan, TCAUP Structures II Slide 21/27

Deflection: by Superposition of Equations

• Deflection can be determined by the use

of equations for specific loading

conditions

• See posted pages for more examples. A

good source is the AISC Steel Manual.

• By “superposition” equations can be

added for combination load cases. Care

should be taken that added equations all

give deflection at the same point, e.g.

the center line.

• Note that if length and load (w) is entered

in feet, a conversion factor of 1728 in3/ft3

must be applied in order to compute

deflection in inches.

University of Michigan, TCAUP Structures II Slide 22/27

Example: Equations Method – By Superposition

• To determine the total deflection of the

beam for the given loading condition,

begin by breaking up the loading diagram

into parts, one part for each load case.

• Compute the total deflection by

superimposing the deflections from

each of the individual loading conditions.

In this example, use the equation for a

mid-span point load and the equation for

a uniform distributed load.

Page 12: Deflection of Structural Members - · PDF filebut zero slope and deflection. • Free end has maximum slope and deflection, but zero moment. • Slope is either downward (-) or upward

University of Michigan, TCAUP Structures II Slide 23/27

Example: Equations Method

• For a W18x55 with an

• E modulus of 30000 ksi

• moment of inertia of 890 in4

• Using an allowable deflection limit of

L / 240.

• Check deflection

University of Michigan, TCAUP Structures II Slide 24/27

Example: Asymmetrical Loading – Superposition of Equations

Standard equations provide values of shear, moment and deflection at points along a beam.

Cases can be superposed or overlaid to obtain combined values.

To find the point of combined maximum deflection, the derivative of the combined deflection equation can be solved for 0. This gives the point with slope = 0 which is a max/min on the deflection curve.

Steel Construction ManualAISC 1989

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University of Michigan, TCAUP Structures II Slide 25/27

Example (same as above): Asymmetrical Loading – Superposition

Deflection equations for cases 5 + 8

Input actual dimensions

Reduce and write deflection equation in terms of x

Differentiate dy/dx to get the slopeequation

Set slope equation = 0

Solve for xThis will be the point of ∆max

University of Michigan, TCAUP Structures II Slide 26/27

Deflection equations (5 + 8)

Input beam distances as before and reduce terms

Solve deflection for x=16.5’

Solve for specific section and material by dividing by EI of the section

Example (same as above): Asymmetrical Loading – Superposition

Page 14: Deflection of Structural Members - · PDF filebut zero slope and deflection. • Free end has maximum slope and deflection, but zero moment. • Slope is either downward (-) or upward

University of Michigan, TCAUP Structures II Slide 27/27

Estimate: Asymmetrical Loading – Superposition of Equations

Or as an estimate…

It is also possible to estimate the deflection location and value without the more exact calculation of x.

If an equation for D max is given, use that (conservative).

Otherwise guess x near mid-span.

e.g. for example above

C.L. = 18 ft

D = 7344 k-ft3 = 2.15”

Steel Construction ManualAISC 1989