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    311ch9 311ch9 Presentation Transcript

      • I) Stresses and Strains
      • In statics we assumed rigid bodies
      • In strength of materials, we acknowledge that bodies are deformable , not rigid.
      • We will study the stresses applied forces produce in a body and the accompanying strains .
      • A.) Axial Tensile and Compressive
      • Stresses
      • Consider a 2” x 4” piece of wood with
      • force P applied at each end.
      800 lb 800 lb 2” 4” A B
      • Anywhere you cut this bar across its section, in order to satisfy equilibrium , the 800 lb force must act on that section.
      •  F x = 0 = - 800 lb + P A = 0
      • P A = 800 lb
      P A 800 lb A B
      • We assume that the force is distributed evenly throughout the section so that an equal portion of the 800 lb force acts on each square inch of the cross-section
      800 lb 2” 4” 1” 1”
      • Since we have 8 square makes, the amount of force on each square inch is:
      • 800 lb = 100 lb = 100 psi
      • 8in 2 in 2
      • Which is the definition of stress :
      •  = P
      • A
      •  = stress = unit stress
      • = average stress
      • = engineering stress
      • P = applied force
      • A= constant cross-sectional area over which the stress develops
      •  t = Tensile Stress (produced by
      • Tensile Forces)
      •  c = Compressive Stress (produced by
      • Compressive Forces)
      • Normal Stress = stress acting on a plane perpendicular to (or normal to) the line of action of the applied force (as in our example).
      • we will discuss stresses on inclined planes in week 4.
      • B. TENSILE AND COMPRESSIVE
      • STRESSES - Application
      • 1.) Analyze the capacity of existing
      • member
      • P (all) =   (all) A
      • A= cross-sectional area, perpendicular to the direction of the force
      • P (all) = axial load capacity (max allowable axial load)
      • = amount of load the member can
      • safely carry
      •  (all) = allowable axial stress=amount of stress which is judged acceptable for the given material
      • Example: 2 x 4 wood with tensile force applied. Find the axial load capacity of the 2 x 4.
      • A = 3.5” x 1.5” = 5.25in 2 ,  (all) = 400 psi
      •  (all) = 400 psi
      • P t(all) =  (all) A=(400 psi)(5.25in 2 ) = 2,100 lb
      • 2. Design a member to support a known
      • load.
      • A (req) = P_
      •  (all)
      • A (req) = required cross-sectional area of the axially loaded member being designed.
      • P = applied axial load
      •  (all) = allowable axial stress
      • Example: Find what diameter steel rod is required to support a 2100# tension force.
      • P = 2100 lb
      •  (all) = 24,000 psi (= lb/in 2 )
      • A (req) = P = 2,100 lb = .0875 in 2
      •  (all) 24,000 lb/in 2
      diameter = d 2100 lb 2100 lb
      • A (req) = .0875 in 2 =  (r) 2 =  (d req ) 2
      • 4
      • (d req ) 2 = .0875 in 2 (4) = 0.1114 in 2
      •  
      •  d req =(0.1114 in 2 ) 1/2 = .333” = 3/8”
      • 
      • Therefore a 3/8” diameter steel rod supports as much tensile force as 2” x 4”.
      • Note : App. A,B,C*,D* give cross-sectional areas for different steel shapes. E gives cross-sectional areas for lumber.
      • 3.) Cautions concerning design and
      • analysis:
      • a.) Some materials have a higher allowable compressive stress than tensile stress (  t(all) <  c(all) )
      • Examples: Wood, concrete, cast iron
      • (refer to App. F, G, 722, 723)
      • b.) The length of a member effects how much compressive force it can sustain. Will discuss in depth in Week 14 (columns).
      • C all < T all
      • For now, we will assume compressive members are short enough to ignore effect of length.
      • c). Holes for bolts etc. must be subtracted from the gross cross-sectional area to obtain the net cross-sectional area (A net ).
      • d.) Design Manuals have been published by Steel (AISC), Wood (AITC, NDS), Concrete (ACI), and Aluminum associations.
      • e.) Assuming even force distribution over the cross-sectional area is usually valid, but not always .
      • C. STRESS ON NET AREA
      • Ex. 2 x 4 with bolt holes for 2-1/2” bolts
      • Determine allowable tensile load
      9/16” dia. holes 3 1/2” 1 1/2” T T
      • T all =  t,all A net
      •  t(all) = 400 psi
      • A net = A - A holes
      • = (3.5” x 1.5”) - (2)(9/16”)(1.5”)
      • = 3.56 in 2
      • T all = (400 psi)(3.56 in 2 ) = 1425 lb
      • Note : 3.5” and 1.5” are “dressed dimensions” (see APP.E)
      • 2 x 4 is actually 1.5” x 3.5”
      • 2 x 8 is actually 1.5” x 7.25”
      • D.) Bearing Stress (  p )
      • -A contact pressure between separate bodies.
      • -A type of compressive stress
      • Example: Wood rafter in a building
      2 X 4 2 X 8 2,000 lb
    •  p = P A bear P = 2000 lb A bear = 1 1/2” x 71/4” = 10.88 in 2  p = 2,000 lb = 184 psi 10.88 in 2
      • E.) Shear Stress
      • Normal stress (tensile and compressive) acts in a direction perpendicular to the surface on which it acts. It is produced by a force whose line of action is perpendicular to the surface on which the stress is produced.
      • Shear stress acts in a direction parallel to the surface on which it acts. It is produced by a force whose line of action is parallel to the surface on which the stress is produced.
    • T T Tension -pulling apart V V Shear -sliding past
      • Example: Nail through boards
      Shear Plane V/5 V/5 V V Shear Plane
      •  v = V
      A  v = shear stress (psi) V = shear force (lb) A= cross-sectional area parallel to the force.
      • Example: Find the shear stress in the 1” diameter bolt shown.
      1” diameter bolt 4,000 lb Steel Plates 4,000 lb Shear Plane
      •  v = V
      • A
      • V = 4,000 lb
      • A =  d 2 =  (1”) 2 = .785 in 2
      • 4 4
      •  v = 4,000 lb = 5,096 psi
      • .785 in 2
      • F.) TENSILE AND COMPRESSIVE
      • STRAINS AND DEFORMATIONS
      • In picking a member size, we not only
      • need to know it can handle the load
      • without failing due to too much stress
      • (i.e. fracture or “break”), we must also
      • be sure it will not deflect (or deform)
      • excessively.
      • Example: Dock with wooden ladder for
      • a footbridge.
      • This is an example of deformation or
      • deflection due to bending stress which
      • we will cover later.
      • Similarly, when a steel rod is in Tension,
      • it will deform, but it is not as noticeable.
      •  = deformation = the amount a body is
      • lengthened by a Tensile force and
      • shortened by a compressive force.
      L  T T
      • To permit comparison with acceptable
      • values, the deformation is usually
      • converted to a unit basis, which is the
      • strain .
      •  =  
      • L
      •  = strain (= unit strain)
      •  = deformation that occurs over length L
      • L = original length of member
      • Example:
      • Example: 3/8” cable, 100’ long stretches
      • 1” before freeing a Jeep which
      • is stuck in the mud.
      • Find - the strain in the cable.
      100’ Jeep Bronco
      •  =  
      • L
      •  = 1”
      • L = 100’ (12”/1) = 1200”
      •  = 1” = 0.0008333 in/in
      • 1200”
      • We’ll come back to see if this is will break the cable.
      • G.) Shear Strain – SKIP !
      • The effect of shear on a body is to cause the body to rotate thru an angle  .
      P P L V V  v 
      • SKIP !
      • tan  =  v _ _
      • L
      •  = shear strain (radians)
      •  v = shear deformation (in)
      • L = Distance over which shear deformation
      • occurs (in)
      • SKIP !
      • Since  is always a very small angle, and
      • for small angles:
      • tan  = 
      • Then:  =  v /L
      • Review of Stress and Strain
      • Axial Stress and Strain
      •  = P
      • A
      •  =  L 
      • Shear Stress and Strain
      •  = V_ 
      • A
      •  =  v /L - SKIP !
      • H.) The Relationship Between Stress
      • and Strain
      • As you apply load to a material, the strain increases constantly (or proportionately) with stress.
      • Example: In a tension test you apply a gradually increasing load to a sample. You can determine the amount of strain (  that occurs in a sample at any given stress level (  .
      •  (ksi)  (in/in)
      • 0 0.000
      • 3 0.001
      • 6 0.002
      • 9 0.003
      • 12 0.004
    • Stress ,  (ksi) Strain ,  in/in x 0.001)           
      • Since the stress is proportional to the strain, ratio of stress to strain is constant .
      •  / 
      •  (ksi)  (in/in)  (ksi)
      • 0 0 0
      • 3 0.001 3000
      • 6 0.002 3000
      • 9 0.003 3000
      • 12 0.004 3000
      • This constant ratio of stress to strain is called the Modulus of Elasticity (E).
      • E =  / 
      • The Modulus of Elasticity is always the same for a given material (see p. 617). We call it a material constant .
      • Knowing E for a given material and :
      • E =  / 
      • 1.) We can find how much stress is in the material if we know the strain:
      •  = E 
      • 2.) We can find how much strain is in the material if we know the stress:
      •  =  E
      • CAUTION !
      • If the tension test continues, the stress will reach a level called the Proportional Limit (  PL ). If the stress is increased above  PL , the strain will increase at a higher rate.
    •  Stress  ), ksi Strain (  ), in/in  PL
      • Strain is directly proportional to stress as long as the stress does not exceed the proportional limit (  PL ) of the material. That is:
      •   = E = constant, ONLY if   PL
      • Anytime you use E to find stress or strain, you must check to make sure   PL .
      • Ex. Given: Previous Truck cable strain
      • Find: Stress in the steel cable
      •  = 1”
      • L = 1200”
      •  = 1” = 0.0008333 in/in
      • 1200”
      •   E ( as long as  PL )
      • 
      • E= 30,000,000 psi (for steel)
      
      •  E  = 30,000,000 psi (.0008333 in/in)
      • = 24,990 psi (pretty high)
      • CHECK: is  <  PL ?
      •  = 24,990 psi <  PL = 34,000 psi (OK)
      • For Shear Stress and Strains:
      • G =  v = Modulus of Rigidity
      • Substituting  = P and  = 
      • A L
      • into E =   Assuming  <  pL )
        • 
        • E = P/A
        •  /L
        • Note: E, G are constant
      • Solving for 
      •  = P
      • L AE
      •  = PL IF ,  <  PL
      • AE
      • Example 2:
      • How much will a 1” diameter by 150” long steel bar stretch if a 1000 lb axial tension load is applied to it ?
      •  = P = 1000 lb = 1,273 psi <  Y  36,000 psi
      • A  (0.5”) 2
      •  =  = __1,273 psi___ = 0.00004244 in/in
      • E 30,000,000 psi
      •  L  0.00004244 in/in)(150”) = 0.006366 ”