Description
Basic Principles of Structures : Week 7
Strain  measure of deformation
strain= change in length
original length ALL inches, feet Lot inches, feet
dimensionless
Shudy Sou
normal strain :
fr
Exongitudinal  All
F
AW

Elatitudinal 
AW
latitudinal
nal =
* positive strain means it got longer,
negative strain means it got shorter
Example 59 find the strain in the cable
12"
KI
Elong
at We also discuss several other topics like What is the meaning of ethnocentrism in anthropology?
L = 1122+12 = 12.04 L = 24.08 Lo 24' AL= 0.08
Eling Element Ew7,0033
Example 5.7 calculate the strain developed
p=100 kip ISI SF.036" ?
long
shortened, so it is
5045)
d=6"
negative
Evorga 20.036" Enoga .003
IZ
shear strain:
tito
Intel Don't forget about the age old question of What is the meaning of conditional proofs?
We also discuss several other topics like What is the meaning of performance production measures in measuring motor performance?
8( gamma) = shear strain 8 is measured in radians 
1 circle = 2th radians
radians = degrees & ZT
 360
to
original angle = 90° = Il
Example:
determine the average normal strain along AC and the shear strain at E relative to the x, y axes
Ac
Ac = 150 +1502 = 212.132 mm
150
MM
150 mm
150mm zum
150 mm We also discuss several other topics like Where was martin luther king's childhood home?
mo  AC
Ac' = J150271523 = 213.551 mm
150
mm
152 mm
E = AL = 213.551 212.132
To 212.132 E = .0067 (normal strain) shear strain:
tan' a 150
75 mm 8= 45.379°
O'= 20 o'= 90.7599
76mm
mm
152 mm
radians  degrees x 270
360
90.759 x 270 = 1.584 rad
360
8= 7  o
8 = 11.584 Don't forget about the age old question of Why is oxygen so important to most animals?
18=
.0132
rad
Ssans
ultimate
Stress strain curve
yield strength plastic behavior elastic behavior plastic behavior
o breaking
linear
stress will tensile strength strength
loading into
eventually decrease elastic behavior
elastic region
as strain increases
Strain
will not cause loading into plastic
permanent
region will cause deformation permanerit deformation A: “strongesť  can withstand the most stress B: brittle has little to no plastic region Don't forget about the age old question of What is optimistic time in pert?
C: ductile  ex steel strain
stress
Hooke's Law
: relationship between stress and strain
* only in the elastic region where there is a LINEAR
relationship between stress and strain
oEE
strain
stress
elastic modulusia constant for any given material
Example :
axial load (normal force) = 4 Kip (kilopounds) yield stress = 51.5 ksi (kilopounds per square inch) determine the crosssectional area.
If the member is 3 feet long and elongates by 0.02 inches, what load was applied ? E14*10% ksi
ve
Å
57.5=4
A .07 in
Lo= 3 ft = 36 in
keep units
E = AL = .025 555*104 (strain) Lo 36
AL = .02 in Keren en
sistent
o=EE
0 =(14*103)(5.55*104)= 7.78 ksi
o
7.18
P=.545 kip]
Example:
If the wire has a diameter of 0.2 inches, determine how much it stretches when the distributed load acts
on the strut.
AL=?
200 16 let
wire has an elastic modulus of 29x109 ksi
q ft
find force acting through AB :
R=900#
KAB 1 300
resultant force from distibuted load: R Ź(9)(200) = 900 #
point of application : 2 / L = 2 (9)
= 6 ft
from small end
q ft
Ozm = 9003 + (ABsin30). 90
AB= 600 #
find AL:
P= 600 # A= r2 = π(0.2) = .03 in
0=600  20000 psi o
= 20 ksi
:03
OE€ E= AL
20 = ( 29x10°) E= 6.9x104
cos3023
АР.
L 30°
JB
e necin 29 ft
AB = 10.39 ft
9 ft Bar 6.9x104=AL IAL= .007 ft
10.39 ft
Poisson's Ratio
force
force
7 ore I ho

to

The
Two types of strain Elongitudinal  AL
same direction as the force
Elatitudinal =
ah
In the elastic region the ratio
{lat will be constant
Elong
This ratio is
Poisson's Ratio
M =
Elat
(mu)
Elong
 negative because one of the two values will be getting smaller
Poisson's Ratio will always be between 0 and 0.5
300N
+ 15mm
Example:
determine the change in the length and diameter of the rod if an axial load of 300 N is applied. E= 2.70 GPa v=0.4
belastic 100 spoissons ratio modulus
. * Pascal = N/m2 –
OI >300 N .
should convert
mm tom 200 mm
o= 300 = 1.698x10° N/m2 or Pa
776.0075) ds 15 mm r= 7.5 mm
= .0075 m O E Elomp 1.699 *10* = ( 2.70x409) Eling Elang = 6.294104
= AU
Evega 4L
6.29~104 =
DL= 1.26*10*mor
Z
m
126 mm
0.42  {cat
als will al
Elat
2.5*104
Poisson's Ratio
6.29x1054
mm
2.5 x 10
=
.
25410*= Ad
Este and
Ad = –.0377 mm]
15 mm
Factor of Safety =
failure stress Allowed stress
If Ophil  1000 Pa and diameter = 15 mm, what force can be applied?
oh
1000+
P
P
.177 N]
76.0075)
Use factor of safety of 2 :
2=
O fail
Fallow = 500 Pa
O allow
z= 1000 Pa
Tallow 500 Pa = p
(.0075)
och
P=.028 N
Axial Forces
oE Elong P = E.AL
alt
AL  PL
AE