Download pdf - 06 Stress Strain

8/13/2019 06 Stress Strain

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Stress strain relations Prof Schierle 1

Stre

ss

Strain

Stress/ StrainRelations


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Stress / strain test

= /

Pull bar to cause stress and strain

Record load P

Compute stress f

f = P/A

A = cross section area

Record strainL

Compute unit strain = L / L L = unstressed length

Plot measure points on stress strain graph Draw line through plotted points


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Stress / Strain RelationsHookes Law for isotropicmaterial

(material of equal properties in any direction)

Stress causes strain deformation

Stress/strain relations are visualized

by a spring as substitute for a rod

to amplify stress / strain relations

1 Elongation under tension

2 Shortening under compression3 Stress / strain graph

P Applied load

L Unstressed length

L Elongation/shortening under load Unit strain =L/L

f Stress f = P/A (A = cross section area)

E Elastic modulus E = f /

(defines stiffness of material)


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Stress / Strain

1 Test load (increasing from 1 to 5 k)

2 Stress-strain graph

For each load the strain is recorded

A line through recorded points

defines stress / strain relations

Slope defines Elastic Modulus E

E = f /

f = P / A = stress

P = applied load

A = test bar cross-section area

= L/L = unit strainL = elongation / shortening

L = unstressed length


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5 Elastic material (rubber)

Resumes initial length after load

6 Plastic material (clay)

Remains deformed after load

S = permanent set

7 Brittle material (concrete)

8 Ductile material (steel)

Stress / strain graphs

1 Test loads (1-5)

2 Stress / strain graph

3 Linear material

Strain increase linear with stress4 Nonlinear material

Strain increase nonlinear with stress

Stress strain relations Copyright Prof Schierle 2014 5


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1 Poissons ratio (shrinks / expands material

normal to applied load)

2 Creep = deformation over time (critical in

concrete)C = Creep

T = Time

3 Elastic/plastic material4 Steel (idealized graph)

5 Steel (mild steel)

6 High strength steel

7 Concrete

8 Wood (under tensile stress)


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Stress / Strain

3 Elastic / plastic material

E Elastic range

P Plastic range

Slope of decreasing stress,

parallel to increasing stress,causes permanent deformation

4 Steel graph (idealized)

A Proportional limit

B Elastic limitC Yield point

CD Yield plateau

E Ultimate strength

F Breaking point



Mild steel (Fy = 36 ksi yield stress)(Mild steel is increasingly replaced by

high-strength steel of 50 ksi yield strength)

Allowable stress for Fy = 36 ksi

Type of stress Allowable stresses

Axial stress Fa = 21.6 ksi (0.6 Fy)Bending stress Fb = 21.6 ksi (0.6 Fy)

Shear stress Fv = 14.4 ksi (0.4 Fy)

Allowable stress for Fy = 50 ksiType of stress Allowable stresses

Axial stress Fa = 30 ksi (0.6 Fy)

Bending stress Fb = 30 ksi (0.6 Fy)

Shear stress Fv = 20 ksi (0.4 Fy)



1,5001.5Masonry

3,0003Concrete

30,00030Steel

1,4001.4Wood

Stiffness E (ksi)

Elastic modulus

Strength F (ksi)Material

Typical strength F vs. stiffness E

Axial deflection LDerivation of formula

f =P/A =L / L

E = f/

E =(P/A) / (L/L)

E = P L / (AL)

L = PL / AE



Examples

Elevator cables

Assume

4 cables each, 60% metallic area

Breaking strength Fy = 210 ksi

Allowable stress (210 ksi / {3x4}) Fa = 17.5 ksiElastic Modulus E=16,000 ksi

Cable length each L = 700

Load P = 8 k

Metallic areaAm = 4 x .6 r2= 4 x .6 0.252 Am = 0.47 in2

Stress

f = P / A = 8 / 0.47 f = 17 ksi

17 < 17.5, ok

Elongation

L = PL / AE = 8k x 700x12 / (0.47x16000) L = 8.9

~

70%metallic

~

60%metallic



Cable elevator

Suspended on 4 to 8 wire ropeseach alone strong enough to

support the elevator

Safety breaks block elevator

if all cables break

Air cushion slows free-fallin case of failure

Shock absorber cushionsimpact in case of failure



Settlement effect on curtain wall

Assume

Steel structure, aluminum curtain wall

2-story mullion, length L = 30x12 L = 360

Column stress increase during construction f = 15 ksi

Elastic modulus (steel) E = 29,000 ksi

Column shortening due to stress

L = PL / AE

Since f = P/ A

L = f L/E = 15 ksi x 360 / 29000 L = 0.19

Note:To allow settlement without stress in mullion,

provide minimum mullion expansion joints

(also verify joint width for thermal expansion)



Suspended structure settlementAssume:

Steel columns, strand hangers

10 stories @ 14 = 10x14x12 L = 1680Average column stress f = 18 ksi

Average strand stress f = 60 ksi

Elastic modulus (steel) E = 29,000 ksi

Elastic modulus (strand) E = 22,000 ksi

Column strain

L = PL/AE

Since f = P/A L = f L/EL = 18 ksi x 1680/29000 L = 1.0

Strand strain

L = 60ksi x 1680 / 22000 L = 4.6

Differential settlement L = 5.6Note:

To reduce differential settlement:

Prestress strands to reduceL by half and adjust

floor levels for DL and partial LL



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