jurg conzett – traversina bridge mom ent loadi ng

Jurg Conzett – Traversina Bridge

Moment

Loading

Riccardo Morandi – Santa Barbara Power Station

Materials Review

Stress-Strain curve

allow

Myf f

I

allow

Myf f

I

= Modulus of Elasticity = E

fy

Stress-Strain curve

Comparison of materials

Modulus of Elasticity (E)Yield Stress (fy)

compressionbending tensionMaterial

Steel

Wood

Concrete

Glass

29,000 ksi

1700 ksi

3100 ksi

10,000 ksi

36 ksi

1.0 ksi

0.5 ksi

24 ksi

36 ksi

0.7 ksi

0.3 ksi

24 ksi

36 ksi

1.5 ksi

3 ksi

145 ksi

Comparison of materials

Modulus of Elasticity (E) compressionbending tensionMaterial

Steel

Wood

Concrete

Glass

17

1

2

6

36

1

0.5

24

50

1

0.5

34

24

1

2

97

Yield Stress (fy)

Allowable Stress Design

Make sure that materials do not reach their yield stress by providing a factor of safety (FOS).

Factor of Safety

Steel: 0.6

Factor of Safety

Steel: 0.6

Allowable flexural stress = factor of safety x yield stress

Fb = 0.6 x fy

Factor of Safety

Steel: 0.6

Allowable flexural stress (Fb)= factor of safety x yield stress

Fb = 0.6 x fy

Fb = 0.6 x 36 ksi

Fb = 21.6 ksi

Moment = bending stress (fb) x SECTION MODULUS

What is section modulus?

Moment = bending stress x SECTION MODULUS


Property of the cross sectional shape.

Moment = bending stress x SECTION MODULUS


Property of the cross sectional shape.

Where do you find it?

Look it up in the tables OR calculate it

Section Modulus = S = b h2

6

b

hneutral axis

b

h

Deflection

Deflection

the measured amount a member moves depends upon:

• Rigidity or stiffness of the material

• Property of the cross sectional shape

• Length of beam

• Load on beam

Deflection

• Rigidity or stiffness of the material

Modulus of Elasticity (E)


Moment of Inertia (I)

Moment of Inertia



Look it up in tables OR calculate it

Moment of inertia = I = b h3 12

b

hneutral axis

b

h

14”

Area = 14 in2

I = 485 in4

Area = 14 in2

I = 229 in4

Area = 14 in2

I = 1.2 in4

14”

14”

P

L

P

L

Rx Ry M

Deflection = P L3 3 E I

P

L

P

L

Rx Ry M

Deflection = w L4 8 E I

w

L

L

Rx Ry

w

M

Deflection = 5 w L4 384 E I

w

L

LRx Ry

w

Ry

Deflection = P L3

48 E I

P

L

L

Rx Ry Ry

P

Moment of Inertia



Look it up in tables OR calculate it

Bigger Moment of Inertia, smaller deflection

STRUCTURAL ANALYSIS :

Determining Structural Capacity

From Structural Analysis we have developed an understanding of all :

Actions - Applied forces such as dead load, live load, wind load, seismic load.

Reactions - Forces generated at the boundary conditions that maintain equilibrium.

Internal forces - Axial, shear and moment (P V M) in each structural element.

Determination of Structural Capacity is based on each element’s ability to perform under the applied actions, consequent reactions and internal forces without :

Yielding - material deforming plastically (tension and/or stocky compression).

Buckling - phenomenon of compression when a slender element loses stability.

Deflecting Excessively - elastic defection that may cause damage to attached materials/finishes – bouncy floors.

TENSILE YIELDING and ALLOWABLE STRESS :

deformation

stre

ss

FY = yield stress

Elastic Range

Plastic Range

deformation

stre

ss

FY

fA

Force on the spring generates an axial stress and elastic deformation

P1

(fA = P/Area of Section)

deformation

stre

ss

FY

When Force is removed, the spring elastically returns to its original shape

deformation

stre

ss

FY

fA

A Larger Force may generate an axial stress sufficient to cause plastic deformation

P2

deformation

stre

ss

FY

fA

When the larger force is removed, the plastic deformation remains (permanent offset)

deformation

stre

ss

FY

fT

To be certain that the tension stress never reaches the yield stress, Set an ALLOWABLE TENSILE STRESS :

FTension = 0.60 FY

If using A36 Steel : FY = 36 ksi

Allowable Tensile Stress (FT ):

FT = 0.60 FY = 0.60 (36 ksi) = 21.6 ksi


Allowable Tensile Stress FT:

FT = 0.60 FY = 0.60 (36 ksi) = 21.6 ksi

P = 5,000 lb or 5 kips

fA = P/Area (actual axial stress fA = P/A)

P force

Aarea

fA stress


Allowable Tensile Stress :

FT = 0.60 FY = 0.60 (36 ksi) = 21.6 ksi


fA = P/Area

FT = Pmax /AreaRequiredPmax

Areq

FT stress



FT = 0.60 FY = 0.60 (36 ksi) = 21.6 ksi


fA = P/Area

FT = Pmax /AreaRequired

AreaRequired = Pmax/FT

Areq

Pmax

FT stress



FT = 0.60 FY = 0.60 (36 ksi) = 21.6 ksi


fA = P/Area

FT = Pmax /AreaRequired

AreaRequired = Pmax/FT = 5k / 21.6 ksi

= .25 in2

Areq

5k

21.6 ksi

FLEXURAL YIELDING and ALLOWABLE BENDING STRESS :

deformation

stre

ss

FY = yield stress

Elastic Range

Plastic Range

fb = M/S

S = Section Modulus

deformation

stre

ss

FY

fb

Force on the BEAM generates an bending stress (tension and compression) and elastic deformation

(fb = Mmax/S)

P1

deformation

stre

ss

FY

When Force is removed, the BEAM elastically returns to its original shape

deformation

stre

ss

FY

fb

A Larger Force may generate an bending stress sufficient to cause plastic deformation

P2

deformation

stre

ss

FY

fb

When the larger force is removed, the plastic deformation remains.

deformation

stre

ss

FY

Fb

To be certain that the bending stress never reaches the yield stress, Set an ALLOWABLE BENDING STRESS :

Fbending = 0.60 FY


Allowable Bending Stress (Fb) :

Fb = 0.60 FY = 0.60 (36 ksi) = 21.6 ksi


Allowable Bending Stress :

Fb = 0.60 FY = 0.60 (36 ksi) = 21.6 ksi

Mmax = 316 k-ft

Mmax = 316 k-ft (12 in / ft) = 3792 k-in



Fb = 0.60 FY = 0.60 (36 ksi) = 21.6 ksi

Mmax = 316 k-ft

Mmax = 316 k-ft (12 in / ft) = 3792 k-in

fb = M/S (actual bending stress fb = M/S)



Fb = 0.60 FY = 0.60 (36 ksi) = 21.6 ksi

Mmax = 316 k-ft

Mmax = 316 k-ft (12 in / ft) = 3792 k-in

fb = M/S

Fb = Mmax / SRequired



Fb = 0.60 FY = 0.60 (36 ksi) = 21.6 ksi

Mmax = 316 k-ft

Mmax = 316 k-ft (12 in / ft) = 3792 k-in

fb = M/S


SRequired = Mmax / Fb



Fb = 0.60 FY = 0.60 (36 ksi) = 21.6 ksi

Mmax = 316 k-ft

Mmax = 316 k-ft (12 in / ft) = 3792 k-in

fb = M/S


SRequired = Mmax / Fb = 3792 k-in / 21.6 ksi = 176 in3


Mmax = 316 k-ft

Mmax = 316 k-ft (12 in / ft) = 3792 k-in


Fb = 0.60 FY = 0.60 (36 ksi) = 21.6 ksi

fb = M/S


SRequired = Mmax/Fb = 3792 k-in / 21.6 ksi = 176 in3

Use W24x76 : SX-X = 176in3

BUCKLING and ALLOWABLE COMPRESSION STRESS :

PC Buckling is a compressive phenomenon that depends on :

1. ‘unbraced length’ of the compression element:(k x l)

2. shape of the section:(radius of gyration ryy)

3. Allowable Material compressive stress:(Fc)

‘unbraced length’ (kxl) depends upon the boundary conditions of an element

l

The radius of gyration (ryy) is a property of a members cross section.

It measures the distance from the neutral axis a member’s area may be considered to be acting

I = Ar2

r = (I/A)0.5

(I = moment of inertia)

Allowable Compression Stress Fc depends on ‘kl/r’

k = 1.0

l = 15 ft = 180 in

assume ryy = 3.0 in.**

kl/r = 60

Fc = 17.4 ksi

** we must always come back and verify this assumption **


Pmax = 240 kips (typ. read this from your P diagram]

Allowable Compression Stress (Fc) :

FC = 17.4 ksi

fC = P/Area

FC = Pmax/AreaRequired

AreaRequired = Pmax/FC = 240k / 17.4 ksi = 13.8 in2

ryy = 3.02 inW12x65 A = 19.1 in2


Pmax = 240 kips

Allowable Compression Stress :

FC = 17.4 ksi

fC = P/Area

FC = Pmax/AreaRequired


Use W12x65 Area = 19.1 in2

check actual stress: fC = P/A

fC = 240 kips / 19.1 in2 = 12.6 ksi OK!

BUCKLING and ALLOWABLE COMPRESSION STRESS :

Allowable Compression Stress depends on slenderness ratio = kl/r

Slenderness Ratio = kl/r

k = coefficient which accounts for buckling shape

for our project gravity columns, k=1.0

for moment frames see deformed shape


l = unbraced length (inches)

rx > ry


r = radius of gyration (inches)

typical use ry (weak direction)

Allowable Compression Stress (Fc)

slenderness ratio = kl/r

assume r = 2 in., k = 1.0

lcolumn = 180 in

kl/r = 90

use Table C-36 to

determine Fc = 14.2 ksi

Pmax Column 2 = 238 kips (assume columns are continuous from foundation to roof, total length 30 feet)

FC = 14.2 ksi



FC = 14.2 ksi



fC = Pmax/Area = 238 k / 19.1 in2 = 12.5 ksi


FC = 14.2 ksi




check ry for W12x65 and verify FC


FC = 14.2 ksi





ry (W12x65) = 3.02

kl/r = (1.0)(180 in)/3.02in = 60


FC = 14.2 ksi





ry (W12x65) = 3.02

kl/r = (1.0)(180 in)/3.02in = 60, using Table C-36

Fc = 17.4 ksi


FC = 14.2 ksi





ry (W12x65) = 3.02

kl/r = (1.0)(180 in)/3.02in = 60, using Table C-36

Fc = 17.4 ksi > fc , therefore ok

Column 2, Efficiency Check: W12x65

fC = 12.5 ksi (actual stress fc = P/A)

FC = 17.4 ksi [allowable stress from chart C-36]

fC/FC < 1.0


fC = 12.5 ksi

FC = 17.4 ksi

fC/FC = 12.5 ksi/17.4 ksi = 0.72 < 1.0


fC = 12.5 ksi

FC = 17.4 ksi

fC/FC = 12.5 ksi/17.4 ksi = 0.72 < 1.0

(72% of capacity is used)

ALLOWABLE BENDING + COMPRESSION:

80 kips

40 kips 40 kips

200 kips200 kips

- compression

- co

mpre

ssio

n

+ t

ensi

on

- 40 kips

+ 2

00 k

ips

- 2

00 k

ips

80 kips 80 kips

900 k-ft

900 k-ft900 k-ft

Axial Diagram

Moment Diagram

fa=P/Area

fb=M/S

Axial Stress (fa)

Bending Stress (fb)

Combined Stress (fa+fb)

+

+ =

=

Axial Stress (fa)

Bending Stress (fb)

+

+

To be certain that the combined stress (bending + axial) never reaches the yield stress, use the INTERACTION EQUATION

fb/Fb + fa/Fa < 1.0

Mmax = 900 k-ft Pmax = 200 kips

Assume 50% capacity of bending (fb)

Mmax = 900 k-ft Pmax = 200 kips


50% Fb = (0.5)(21.6 ksi) = 10.8 ksi

SREQ = Mmax/50%Fb = 900k-ft (12in/1ft) / 10.8ksi

SREQ = 1000in3

TRY W36x260, Sx-x = 953 in3 A = 76.5 in2 ry-y = 3.78 in


fb = Mmax/S = 900 k-ft (12in/1ft) / 953in3 = 11.3 ksi



Fb = 21.6 ksi

fb/Fb = 11.3ksi/21.6ksi = 0.52



Fb = 21.6 ksi

fb/Fb = 11.3ksi/21.6ksi = 0.52

fc = Pmax/Area = 200 kips/76.5 in2 = 2.6 ksi

Slenderness ratio: k=2.0 l = 180 in

kl/r = (2.0)(180 in)/3.78 in = 95



Fb = 21.6 ksi

fb/Fb = 11.3ksi/21.6ksi = 0.52



kl/r = (2.0)(180 in)/3.78 in = 95, using Table C-36

Fc = 13.6 ksi

fc/Fc = 2.6ksi/13.6ksi = 0.19



Fb = 21.6 ksi

fb/Fb = 11.3ksi/21.6ksi = 0.52




Fc = 13.6 ksi

fc/Fc = 2.6ksi/13.6ksi = 0.19

fb/Fb + fc/Fc = 0.71 < 1.0, therefore ok


70% Fb = (0.7)(21.6 ksi) = 15.1 ksi

SREQ = Mmax/70%Fb = 900 k-ft (12in/1ft) / 15.1 ksi

SREQ = 720 in3


fb = Mmax/S = 900 k-ft (12 in/1 ft) / 684 in3 = 14.2 ksi



Fb = 21.6 ksi

fb/Fb = 14.2 ksi/21.6 ksi = 0.73



Fb = 21.6 ksi

fb/Fb = 14.2 ksi/21.6 ksi = 0.73



kl/r = (2.0)(180 in)/3.56 in = 101



Fb = 21.6 ksi

fb/Fb = 14.2 ksi/21.6 ksi = 0.73




Fc = 12.85 ksi

fc/Fc = 3.4 ksi/12.85 ksi = 0.26



Fb = 21.6 ksi

fb/Fb = 14.2 ksi/21.6 ksi = 0.73




Fc = 12.85 ksi

fc/Fc = 3.4 ksi/12.85 ksi = 0.26

fb/Fb + fc/Fc = 0.99 < 1.0, therefore ok

jurg conzett – traversina bridge mom ent loadi ng

Documents

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allowable flexural stress

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cross sectional shapewhere

e i p lp lrx ry mdeflection

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material property