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• 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
• 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
•  y  x   x
• E
•  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.
• to plane DC is:
• P n = Pcos 
• 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
•  y  x   x
• E
•  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