Uploaded on

 

  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Be the first to comment
    Be the first to like this
No Downloads

Views

Total Views
458
On Slideshare
0
From Embeds
0
Number of Embeds
0

Actions

Shares
Downloads
14
Comments
0
Likes
0

Embeds 0

No embeds

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
    No notes for slide

Transcript

  • 1.
    • III .) Stress Considerations
    • A.) Poisson’s Effect - Uniaxial Load
    • Poisson's’ Ratio (  ) is a known
    • material constant that tells you how much strain occurs in a direction
    • perpendicular to an applied load.
    •  = -  y
    •  x
  • 2. x Original Shape Shape After Loading L y,O y P x P x L x,O L x,F L y,F
  • 3.
    • We already know how to find strains and deformations in the direction parallel to the load axis.
    •  x =  x where  x = P x / A x
    •  E
    •  x =  x L x.O
  • 4.
    • To find the strains and deformations in a direction perpendicular to the loading.
    •  = -  y /  x   .25 (for steel)
    •   .33 (for aluminum)
    •  y = -  x
    •  y = -  x because  x =  x
    • E E
    •  y =  y L yO
  • 5.
    • Example: Find the deformation in the x & y-directions of the steel bar shown due to the 20 kip Uniaxial load.
    1” 2.5” 12” 20 k 20 k y z x
  • 6.
    •  x = P x / A x = 20 k____ = 8.0 ksi (<  pl )
    • ( 2.5”)(1.0”)
    •  x =  x = (8.0 ksi)__ = 2.67x10 -4 in/in
    •  E  30,000 ksi
    •  x =  x L x.O = (2.67x10 -4 in/in)(12.0”)
    •  x = 3.20x10 -3 in. (longer)
  • 7.
    •  y = -  x = - (0.25)(8.0 ksi)
    •  E  30,000 ksi
    •  y = - 6.67x10 -5 in/in
    •  y =  y L y.O = (- 6.67x10 -5 in/in)(2.5”)
    •  y = -1.67x10 -4 in. (shorter)
  • 8.
    • B.) Poisson’s Effect - Biaxial Loads
    P x P x P x P x UNIAXIAL BIAXIAL P y P y
  • 9.
    •  The strain in the y-direction now has two parts:
    •  y1 = -  x is caused by P x
    • E 
    •  y2 =  y is caused by P y
    • E
    •  y =  y -  x  is the total strain in E E  the y-direction due to biaxial loading
  • 10.
    •  Similarly,  The strain in the x-direction is found to be.
    •  x =  x -  y  the total strain in the
    • E E x-direction due to
    • biaxial loading
  • 11.
    • Example: Find the deformation in the x & y-directions of the of the same steel bar if a load in the y-direction is added.
    1” 2.5” 12” 20 k 20 k y z x 60 k 60 k
  • 12.
    • Note:  pl = 34 ksi
    •  x = P x / A x = 20 k = 8.0 ksi (<  pl )
    • ( 2.5”)(1.0”)
    •  y = P y / A y = 60 k = 5.0 ksi (<  pl )
    • ( 1.0”)(12.0”)
  • 13.
    •  x =  x -  y = (8.0 ksi) - (0.25)(5.0 ksi) 
    • E E 30,000 ksi 30,000 ksi
    •  x = 2.25x10 -4 in/in
    •  x =  x L x.O = (2.25x10 -4 in/in)(12.0”)
    •  x = 2.70x10 -3 in. (longer)
    • (  x was 3.20x10 -3 in. without y-dir. load)
  • 14.
    •  y =  y -  x = 5.0 ksi - (0.25)(8.0 ksi)
    •  E  E  30,000 ksi 30,000 ksi
    •  y = 1.00x10 -4 in/in
    •  y =  y L y.O = ( 1.00x10 -4 in/in)(2.5”)
    •  y = 2.50x10 -4 in. (longer)
    • (  y was -1.67x10 -4 in. before adding the y-load)
  • 15.
    • The bar got longer and wider
    • (and thinner)
    P x P y P x P y
  • 16.
    •  z = -  y -  x = - (0.25)(5.0 ksi) - (0.25)(8.0 ksi)
    •  E  E  30,000 ksi 30,000 ksi
    •  z = - 1.08x10 -4 in/in
    •  z =  z L z.O = ( -1.08x10 -4 in/in)(1.0”)
    •  z = -1.08x10 -4 in. (thinner)
  • 17.
    • B.) Thermal Effects
    • 1.) Thermal Expansion
    • -most materials expand when heated
    • -the amount of deformation caused
    • by a temperature change is
    • proportional to the change in
    • temperature.
  • 18.
    • That is, if raising the temperature 10 o F
    • lengthens a part by 0.10”,
    • then raising the temperature 20 o F will
    • lengthen the same part by 0.20”.
    L 100 o F 110 o F 120 o F 0.10” 0.10”
  • 19.
    •  T =  L (  T)
    •  T =  Change in length caused by a temp change (in)
    •  coefficient of thermal expansion, in/in
    • o F
    • L= Original length of member (in)
    •  T= Change in temperature ( o F)
  • 20.
    • Example:
    • Mackinac Bridge is 5 miles long ( + ).
    • What is the change in length from 3 p.m. when the temp.= 80 o F to midnight when temp.= 50 o F. (A36 Steel)
  • 21.
    •  T =  L (  T)
    •  steel  6.5 X 10 -6 in/in (App G)
    • o F
    • L= (5mi)(5280 ft/mi)(12in/ft)
    • =316,800in
    •  T= 50 o F - 80 o F= -30 o F
    •  T = (6.5x10 -6 in/in/ o F)(316,800in)(-30 o F)
    • = - 61.8 in = - 5.15 ft .
  • 22.
    • 2.) Thermal Stress
    • If you restrain a member when you change it’s temperature, internal stresses will be produced which are called thermal stresses.
    • Thermal stress is determined by picturing a member that is expanded when heated and determining the force (P) which would be required to bring in back to its original length.
  • 23. L  T =  L(  T) P P  PL AE L 60 o F 100 o F 100 o F
  • 24.
    • PL =  L(  T)
    • AE
    • P =  T)E
    • A
    •  T = -  T)E
    •  = Thermal stress in a restrained member due to a change in temperature (psi). ( for  T  PL )
    • E= Modulus of Elasticity (psi)
  • 25.
    • C.) Composite Members (SKIP)
    • Examples:
    • Concrete slab reinforced with steel
    • Carbon fiber and plastic snow skis
    • Kevlar and fiberglass boats
    • Concrete slab on a metal deck
    • .
    concrete corrugated metal deck
  • 26.
    • 1.) Relationship Between Load and Stress
    • We know:
    • a.) E A = E B because two different
    • materials.
    • Example: E A = 30,000 ksi (steel)
    • E B = 3,000 ksi (conc)
  • 27.
    • b.)  A =  B - because the two materials
    • move (deform) together
    • c.)  A =  B
    •  A =  B   - because  = 
    •  E A E B E
    •   A = E A  B = n  B , where n = E A
    • E B E B
  • 28.
    • We also know:
    • d.) P = P A + P B because each material
    • carries a part of the load
    • P =  A A A +  B A B because P =  A
    • P = (n  B )A A +  B A B because  A = n  B
    • P =  B (nA A + A B )
    • (or)
    • P =  A (A A + A B /n) because  B =  A /n
  • 29.
    • 2.)To find the stresses due to a known
    • axial load (P) in a member (A) made of two known materials (E A , E B ).
    • a.) first find  B from:
    • P =  B (nA A + A B )
    • b.) then find  A from:
    •  A = n  B
  • 30.
    • 3.)To find the allowable axial load (P) on a member (A A , A B ) made of two known
    • materials (E A , E B ,  A,all ,  B,all ).
    • a.) find  A from:
    •  A = n  B,all
    • b.) if  A <  A,all , then find P all from:
    • P all =  B,all (nA A + A B )
    • c.) if  A >  A,all , then find P all from:
    • P all =  A,all (A A + A B /n)
  • 31.
    •  D.) Stress Concentration:
    • 1.) At a section with a hole in we previously assumed the stress is uniformly distributed across the net section.
    •  P
    • A net
    • A net = A - A hole
     avg
  • 32.
    • 2). The actual distribution is not uniform, but is much higher than  avg near the hole and less than  avg away from the hole.
    •  t,max = stress next to the hole
    •  t,max = k(P )
    • A net
    • k = stress concentration factor
     t,max  avg
  • 33.
    • 3.) Stress Concentration Factor (k)
    • k is determined from empirical curves.
    • (see Fig.11-12, p.285)
    • k is based on the ratio r/d:
    • r = the radius of the hole
    • d = the net width of the member
  • 34.
    • 3.) Stress Concentration Factor (k)
  • 35.
    • 4.) When to consider stress concentration in design.
    • a.) If a member is ductile and subjected to fairly static loads , it is assumed plastic deformation will provide a uniform stress distribution, use  avg in design.
    • b.) If the member is made of a brittle
    • material or is subjected to repetitive loads , you must consider the higher stress(  t,max ) which occurs near the hole.
  • 36.
    • 4.) When to consider stress concentration in design.
    • Repetitive (cyclical) stress Examples:
    • - rotational machinery:
    • - axles, gears, sprockets, drive shafts
    • - others…
    • - number of cycles:
  • 37.
    • III.) Stress Considerations (Review)
    • A.) Poisson’s Ratio
    •  trans /  long  y /  x
    • 1.)Uniaxial Loads
    •  y  x   x
    • E
    • 2.)Biaxial Loads
    •  x   x -  y
    • E E  
  • 38.
    • B.) Thermal Effects
    • 1.) Thermal Expansion
    •  t =  L(  T)
    • 2.) Thermal Stresses
    •  t =  E(  T)
  • 39.
    • C.) Composite Members - SKIPPED
    • n = E A / E B = Modular Ratio
    •  A  = n  B
    • P =  B (nA A + A B )
    • D.) Stress Concentration
    •  t,max  = k  avg  = k(P/A net )
  • 40.
    • E.) Stresses on Inclined Planes
    • 1.) We are interested in finding the
    • stresses on planes other than the plane which is normal to the axial load because failure does not always occur on the normal plane.
    • a.) An axially loaded ductile material will usually fail along a plane which is at a 45 degree angle to the load.
    • b.) An axially loaded brittle material will usually fail along a plane which is normal to the load.
  • 41.
    • If “A” is the area of plane DB,
    • and the thickness is constant,
    • Then the area of plane DC is:
    • A/ cos 
    D B C P P  Area = A Area = A_ cos 
  • 42.
    • The load perpendicular (normal)
    • to plane DC is:
    • P n = Pcos 
    • The load parallel (shear)
    • to plane DC is:
    • P v = Psin 
    P P  n v P v P n C B D
  • 43.
    • The stress perpendicular (normal)
    • to plane DC is:
    •  n = P n__
    • A/cos 
    •  n = Pcos 
    • A/cos 
    •  n = P(cos  
    • A
    P  n  n C B
  • 44.
    • The stress parallel (shear)
    • to plane DC is:
    •  v = P v__
    • A/cos 
    •  v = Psin 
    • A/cos 
    •  v = Psin  cos 
    • A
    •  v = (P/2A)sin2 
    P  n  v C B D
  • 45.
    • The maximum normal stress is when the plane is inclined at 0 degrees:
    •  n = P(cos   P
    • A A
    • The minimum normal stress is when the plane is inclined at 90 degrees:
    •  n = P(cos90   0
    • A
  • 46.
    • The maximum shear stress is when the plane is inclined at 45 degrees:
    •  v = P(sin(2x45))  =  P
    • 2A 2A
    • The minimum shear stress is when the plane is inclined at 0 or 90 degrees:
    •  v = P(sin(2x90)  =  P(sin(2x0)  = 0
    • A A
  • 47.  
  • 48.
    • III.) Stress Considerations (Ch. 11Review)
    • A.) Poisson’s Ratio
    •  trans /  long  y /  x
    • 1.)Uniaxial Loads
    •  y  x   x
    • E
    • 2.)Biaxial Loads
    •  x   x -  y
    • E E  
  • 49.
    • B.) Thermal Effects
    • 1.) Thermal Expansion – when member is free to move
    •  t =  L(  T)
    • 2.) Thermal Stresses – when member is restrained
    •  t =  E(  T)
  • 50.
    • C.) Stress Concentrations  t,max = k  avg = k P A
  • 51.
    • D.) Stresses on Inclined Planes
    •  n = P(cos  
    • A
    •  n,max = P , when  = 0
    • A
    •  v = Psin  cos 
    • A
    •  v,max = P/2A  , when  = 45 
    P   n  v A