DEFORMATION EXAMPLE: OF A STEEL BAR UNDER AXIAL LOADING
External forces applied at centroids of elements
APPLY FORCES TO A STEEL BAR, WHICH IS FIXED ON ONE END. THE BAR IS COMPOSED OF THREE SECTIONS WITH DIFFERENT CROSS-SECTIONAL AREAS AND DIFFERENT LENGTHS.
A
4' 2in2
2' B 4in
2
3' C 2in2 D 20k 80k
HA VA
100k
OBTAIN REACTIVE FORCES AT SUPPORT WITH EQUILIBRIUM CONDITION.
A
HA = 40k V A= 0
B
C
D 20k
100k
80k
Determine PAB , the internal force on cross-section between A and B
A
HA = 40k V A= 0
B
C
D 20k
100k
80k
A
HA = 40k V A= 0
B
C
D 20k
PAB= 40k
100k
(tensile) 80k
Determine PBC , the internal force on cross-section between B and C
A
HA = 40k V A= 0
B
C
D 20k
100k
80k
A
HA = 40k V A= 0
B
C
D
PBC= 60k
100k
20k
(compressive) 80k
Determine PCD , the internal force on cross-section between C and D
A
HA = 40k V A= 0
B
C
D 20k
100k
80k
HA = 40k
A HA = 40k
A B
B C
C D 20k
D 20k
V A= 0
V A= 0 100k
100k
80k
80k PCD= 20k
(tensile)
FORCES AT EACH SEGMENT
A
HA = 40k
B
C
D 20k
A
H A = 40k
B
V A= 0
100k
80k
FAB = 40k PAB=40k
PBC=60k FBC = -60K
B
C
PBC = -60K FBC=60k
F =20k PCD = 20K CD
C
D
PCD=20k 20k
DRAW AXIAL FORCE DIAGRAM
A
HA = 40k V A= 0
B
C
D 20k
100k
80k
40K
0 T
20K
T
A
B
C
C
D
-60K AXIAL FORCE DIAGRAM
DETERMINE DEFORMATION OF EACH SECTION A
H A = 40k
B
B
FAB = 40k
C
(elongates)
FBC = -60K FBC=60k
FBC=60k FBC = -60K (compresses)
F = 20K CD
TOTAL DEFORMATION
C
D
FCD=20k 20k (elongates)
AD AB BC CD
Elongation (+)
TOTAL DEFORMATION OF THE BAR
AD AB BC CD AD
ft to in conversion!
PL PL PL AE AB AE BC AE CD
(40)(4 12) (60)(2 12) (20)(3 12) (2.0)(30 103 ) (4.0)(30 103 (2.0)(30 ) 103 ) (0.032) (0.012) (0.012) elongation 0.032in
INDICATING EXTENSIVE DEFORMATION. (EXTENSION AS OPPOSED TO SHORTENING)
AD
AD 0
SMALL DISPLACEMENT APPROXIMATIONS
a
a
DB
DB=Vertical displacement of B
Assuming point B moves perpendicular to the longitudinal axis of rigid bar AB, then:
DB ~ B
EXAMPLE: Determine the vertical displacement at C
D
deformable bar
P
A
rigid bar
B
BD
6 ft 3 ft
C
DC
6 ft
3 ft
A
B
C
BD
B' C'
DC
A
From Similar Triangles:
6 ft
B
BD
B' A C
AB
BD
AC DC
9 ft
AC DC BD AB
DC
C'
9 ft DC 6 ft BD 1.5 BD
CASE OF UNIAXIAL STRESS
DEFINITION OF NORMAL STRAIN
L
y
x
L'
P
dL'dL L' L dL L
CASE OF UNIAXIAL STRESS
DEFINITION OF NORMAL STRESS
x
dA
P
dP
dP dA
UNIAXIAL STRESS SITUATION
y
y
x
x
x
z
x
z
x
dP P x dA A
y 0
z 0
HOOKE'S LAW FOR UNIAXIAL STRESS SITUATION
P x A
y 0
z 0
x
x
E
y
x
E
z
x
E
y lateral strain z Poisson' s ratio axial strain x x
1 0 2
For engineering materials
Given: A 15 ft long steel member that is subjected to an axial force P=120 k. Determine:
Example Problem 1
a) the change in the length of the member, and b) the change in the dimensions of the cross-section
y
P
z
x
dz= 5 in. CROSS-SECTION
STEEL:
E =29 x 103 ksi
P
A = 15 in2
u = 0.30
Solution: Note the member is uniaxially loaded.
y
z
x
Normal Stress:
P
P 120 k x 8.0 ksi 2 A 15 in y z 0
Normal Strain:
x 8.0 ksi 6 in. x 276 x 10 276in / in 3 E 29x10 ksi in. y z x 0.30276in / in 83in / in
Dimensional Changes:
x x L x 276 x 106 15 ft x 12 in / ft 4.97 x 102 in y yd y z zd z
6 4
83 x 10 3 in. 2.49 x 10 83 x 10 5 in. 4.15 x 10
6
in.
4
in.