Notes for Test 1 2017 - University of Michigan · • Shear and slope have balanced + and ... TCAUP...

University of Michigan, TCAUP Structures II Slide 1/30

Notes for Test 12017

These are just some pages selected

from the PowerPoints. Basically these

are pages with definitions and the

problem procedure pages.


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


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


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 with total deflections

greater than L/120 are in danger

of ponding.

International Building Code -2006

roof pondingfrom IRC, Josh 2014


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.


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.


Deflection by Diagrams

Load

Shear

Moment

Slope (EI)

Deflection (EI)


Table D-24 I. Engel


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)


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

Actual Flexure Stress fb

fb = Mc/I = M/S

S = I/c = bd2/6

University of Michigan, TCAUP Wood Structures Slide 11/59

Allowable Flexure Stress Fb’

Fb from tables determined by species and grade

Fb’ = Fb (usage factors)

usage factors for flexure:CD Load Duration FactorCM Moisture FactorCL Beam Stability FactorCF Size FactorCfu Flat UseCr Repetitive Member Factor

Allowable Stress Design by NDSFlexure

Actual Shear Stress fv

fv = VQ / I b = 1.5 V/A

Can use V at d from support as maximum


Allowable Shear Stress Fv’

Fv from tables determined by species and grade

Fv’ = Fv (usage factors)

usage factors for shear:CD Load Duration FactorCM Moisture Factor

Allowable Stress Design by NDSShear

Actual Compression Stress fb

fc = P/A


Allowable Compression Stress Fc’

Fc from tables determined by species and grade

Fc’ = Fc (CM CD Ct CF Ci CP)

Allowable Stress Design by NDSCompression

Analysis Procedure

Given: loading, member size, material and span.

Req’d: Safe or Unsafe

1. Find Max Shear & Moment• Simple case – equations• Complex case - diagrams

2. Determine actual stresses• fb = M/S• fv = 1.5 V/A

3. Determine allowable stresses• Fb and Fv (from NDS)• Fb’ = Fb (usage factors)• Fv’ = Fv (usage factors)

4. Check that actual ≤ allowable• fb ≤ F’b• fv ≤ F’v

5. Check deflection 6. Check bearing (Fb = Reaction/Abearing )


from NDS 2012

Analysis Procedure

Given: member size, material and span.Req’d: Max. Safe Load (capacity)

1. Assume f = F• Maximum actual = allowable stress

2. Solve stress equations for force• M = Fb S• V = 0.66 Fv A

3. Use maximum forces to find loads• Back calculate a load from forces• Assume moment controls

4. Check Shear• Use load found is step 3 to check

shear stress.• If it fails (fv > F’v), then find load based

on shear.

5. Check deflection 6. Check bearing


from NDS 2012

Design Procedure

Given: load, wood, spanReq’d: member size

1. Find Max Shear & Moment• Simple case – equations

• Complex case - diagrams

2. Determine allowable stresses

3. Solve S=M/Fb’

4. Choose a section from Table 1B• Revise DL and Fb’

5. Check shear stress• First for V max (easier)• If that fails try V at d distance

from support.• If the section still fails, choose a new

section with A=1.5V/Fv’

6. Check deflection

7. Check bearing



Failure Modes

FLEXURE AXIAL

Strength

Stability

Serviceability

Deflection Bearing (crushing limit)


Failure Modes

Short Columns – fail by crushing

– fc = Actual compressive stress – A = Cross-sectional area of column (in2)– P = Load on the column – Fc = Allowable compressive stress per codes

Intermediate Columns – crush and buckle

Long Columns – fail by buckling

– E = Modulus of elasticity of the column material (psi)– K = Stiffness (curvature mode) factor– L = Column length between ends (inches)– r = radius of gyration = (I/A)-2 (inches)


Leonhard Euler (1707 – 1783)

Euler Buckling (elastic buckling)

– A = Cross sectional area (in2)– E = Modulus of elasticity of the material (lb/in2)– K = Stiffness (curvature mode) factor– L = Column length between pinned ends (in.)– r = radius of gyration (in.)

portrait by Emanuel Handmann,1753


Slenderness Ratio

Radius of Gyration: a distance from the N.A. to where the concentrated mass of the section would yield the same rotational inertia about the N.A.

– r = Radius of Gyration– I = Moment of Inertia– A = Cross-sectional Area

Slenderness Ratios:

The larger ratio will govern.Try to balance for efficiency

rx = 0.999

ry = 0.433


End Support Conditions

K is a constant based on the end conditionsl is the actual length le is the effective length (curved part)

le = Kl

K= 0.5

K= 2.0

K= 0.7

K= 1.0

Both ends fixed.

One end free, one end fixed.

Both ends pinned.

One end pinned, one end fixed.


Actual Flexure Stress fb

fc = P/A


Analysis of Wood Columns

Data:• Column – size, length• Support conditions• Material properties – Fc , E• Load

Required:• Pass/Fail or margin of safety

1. Calculate slenderness ratio l/dlargest ratio governs. Must be < 50

2. Find adjustment factorsCD CM Ct CF Ci

3. Calculate CP

4. Determine F’c by multiplying the tabulated Fc by all the above factors

5. Calculate the actual stress: fc = P/A6. Compare Allowable and Actual stress.

F’c > fc passes


Analysis of Wood Columns

Data:• Column – size, length• Support conditions• Material properties – Fc , E

Required:• Maximum Load, Pmax

1. Calculate slenderness ratio l/dlargest ratio governs. Must be < 50

2. Find adjustment factorsCD CM Ct CF Ci

3. Calculate CP

4. Determine F’c by multiplying the tabulated Fc by all the above factors

5. Set actual stress = allowable, fc = F’c6. Find the maximum allowable load

Pmax = F’c A

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Notes for Test 1 2017 - University of Michigan · • Shear and slope have balanced + and ... TCAUP...

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