Date post: | 04-Jun-2018 |
Category: |
Documents |
Upload: | mdsajidalam |
View: | 243 times |
Download: | 0 times |
of 15
8/13/2019 06 Stress Strain
1/15
Stress strain relations Prof Schierle 1
Stre
ss
Strain
Stress/ StrainRelations
8/13/2019 06 Stress Strain
2/15
Stress strain relations Prof Schierle 2
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
8/13/2019 06 Stress Strain
3/15
Stress strain relations Prof Schierle 3
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)
8/13/2019 06 Stress Strain
4/15
Stress strain relations Prof Schierle 4
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
8/13/2019 06 Stress Strain
5/15
Stress strain relations Prof Schierle 5
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
8/13/2019 06 Stress Strain
6/15Stress strain relations Prof Schierle 6
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)
8/13/2019 06 Stress Strain
7/15
8/13/2019 06 Stress Strain
8/15Stress strain relations Prof Schierle 8
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
8/13/2019 06 Stress Strain
9/15Stress strain relations Prof Schierle 9
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)
8/13/2019 06 Stress Strain
10/15Stress strain relations Prof Schierle 10
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
8/13/2019 06 Stress Strain
11/15Stress strain relations Prof Schierle 11
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
8/13/2019 06 Stress Strain
12/15Stress strain relations Prof Schierle 12
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
8/13/2019 06 Stress Strain
13/15Stress strain relations Prof Schierle 13
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)
8/13/2019 06 Stress Strain
14/15Stress strain relations Prof Schierle 14
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
8/13/2019 06 Stress Strain
15/15Stress strain relations Prof Schierle 15
a p p y e n d