Transcript Class 13 14 15 CIVE 2110 Combined Stress

Classes #13, 14, 15
Civil Engineering Materials – CIVE 2110
Combined Stress
Fall 2010
Dr. Gupta
Dr. Pickett
1
Combined Stresses
Assume:
Linear Stress-Strain relationship
Elastic Stress-Strain relationship
Homogeneous material
Isotropic material
Small deformations
Stress determined far away from
points of stress concentrations
(Saint-Venant principle)
Combined Stresses
Procedure:
Draw free body diagram.
Obtain external reactions.
Cut a cross section, draw free body diagram.
Draw force components acting through centroid.
Compute Moment loads about centroidal axis.
Compute Normal stresses associated with each load.
Compute resultant Normal Force.
Compute resultant Shear Force.
Compute resultant Bending Moments.
Compute resultant Torsional Moments.
Combine resultants (Normal, Shear, Moments) from all
loads.
-Example:
Combined Stress
# 8.6
-Pg. 451-452
-Hibbeler, 7th edition
Combined Stress
-Example:
# 8.6
-Pg. 451-452
-Hibbeler, 7th edition
-Problem:
Combined Stress
# 8-43, 8-44
-Pg.
458
-Hibbeler, 7th edition
Remember:
for Shear Stress
 
 
Areas and Centroids,
Mechanics of Materials, 2nd ed,
Timoshenko, p. 727
Stress Transformation
General State of Stress:
- 3 dimensional
Six stresses  x , y , z , xy, xz , yz
Remember:
 xy   yx
 xz   zx
 yz   zy
 
 
Stress Transformation
General State of Stress:
- 3 dimensional
Six stresses  x , y , z , xy, xz , yz
Plane Stress
- 2 dimensional
Three stresses  x , y , xy
Remember:
 xy   yx
 xz   zx
 yz   zy
 
 
Stress
Transformation
Plane Stress
2 dimensional
Stress Components are:
 x   xx  Norm al Stress on X face in X Direction
 y   yy  Norm al Stress on Y face in Y Direction
 xy   yx
 xy  Shear Stress on X face in Y
Direction
 yx  Shear Stress on Y face in X
Direction
+ = CCW, upward
on right face
 
 
Plane Stress
Transformation
State of Plane Stress
at a POINT
May need to be determined
In various
ORIENTATIONS, .
 
+ = CCW, upward on
right face
 
Plane Stress
Transformation
Must determine:  x '
 y'  x' y'
To represent the same stress as:
x
 y  xy
Must transform:
Stress – magnitude
- direction
Area – magnitude
- direction
+ = CCW, upward
on right face
 
 
Steps for
Plane Stress
Transformation
To determine  x' and  x' y '
acting on X’ face,
:
- Draw free body diagram
at orientation .
- Apply equilibrium equations:
ΣFx’=0 and ΣFy’=0
by multiplying stresses
Ax = (ΔA)CosΔ
on each face
by the area of each face.
Ay = (ΔA)SinΔ
Steps for
Plane Stress
Transformation
To determine  y '
acting on Y’ face,
:
- Draw free body diagram
at orientation .
- Apply equilibrium equations:
ΣFx’=0 and ΣFy’=0
by multiplying stresses
on each face
by the area of each face.
- Remember:  x ' y '   y ' x '
Ax = (ΔA)SinΔ
Ay = (ΔA)CosΔ
-Problem:
Plane Stress Transformation
-Pg.
# 9-6, 9-9, 9-60
484
-Hibbeler, 7th edition
Equations
Plane Stress
Transformation
A simpler method,
General Equations:
- Draw free body diagram
at orientation .
- Apply equilibrium equations:
ΣFx’=0 and ΣFy’=0
by multiplying stresses
on each face
by the area of each face.
- Sign Convention:
+ = Normal Stress = Tension
+ = CCW, upward on
+ = Shear Stress = CCW, Upward on right face right face
+ =  = CCW from + X axis
 
 x' y '   y 'x'
Equations
Plane Stress
Transformation
- Draw free body diagram
at orientation .
- Apply equilibrium equations:
ΣFx’=0 and ΣFy’=0
by multiplying stresses
on each face
by the area of each face.
- Sign Convention:
+ = Normal Stress = Tension
+ = Shear Stress =
+ = CCW, Upward on right face,
+ = CCW, upward
+ =  = CCW from + X axis
 
on right face
 x' y '   y 'x'
Equations
Plane Stress
Transformation
Fx '  0
0   x ' A
  x ACos Cos


  ACos Sin
  ASin Sin
  xy ASin  Cos
xy
y
-
Equations
Plane Stress
Transformation
factor out A
 x '   x Cos2  2 xy SinCos   y Sin2
 1  Cos2 
 Sin2 
 1  Cos2 
 x'   x 
  2 xy 
  y

2
2


 2 


 y  y Cos2
 x  xCos2
 x' 

  xy Sin2 

2
2
2
2
  x  y    x  y 
  
Cos2   xy Sin2
 x '  
 2   2 
-
Equations
Plane Stress
Transformation
Fy '  0


0   x ' y ' A
  x ACos Sin


  ACos Cos
  ASin Cos
  xy ASin  Sin
xy
y
-
Equations
Plane Stress
Transformation
factor out A
 x ' y '   xyCos2   y SinCos   xy Sin2   xCosSin
 x' y'
 x' y'
 x' y'
 1  Cos2 
 1  Cos2 
 Sin2 
 Sin2 
  xy 
   xy 
  y
  x

2
2




 2 
 2 
 1  Cos2  1  Cos2 
 Sin2 


  xy 
   y  x 

2


 2 
  x  y 
 Sin2
  xyCos2  
 2 
-
Equations
Plane Stress
Transformation
for  y '
set     90
note: Cos2  Cos2  90  Cos2  180  Cos2
note: Sin2  Sin2  90  Sin2  180   Sin2
  x  y    x  y 
  
Cos2   xy Sin2
previously  x '  
 2   2 
 x  y   x  y 
  
Cos2   xy Sin2
consequently  y '  
 2   2 
-
Equations of
Plane Stress Transformation
The equations for the transformation of
Plane Stress are:
  x  y    x  y 
  
Cos2   xy Sin2
 x '  
 2   2 
  x  y    x  y 
  
Cos2   xy Sin2
 y '  
 2   2 
 x' y'
  x  y 
Sin2
  xyCos2  
 2 
-Problem:
Plane Stress Transformation
-Pg.
# 9-6, 9-9, 9-60
484
-Hibbeler, 7th edition