1  Stress and Strain Page 1 of 34  Stress and Strain [5.1] Internal Stress of Solids [5.2] Design of Simple Connections (will not be covered in class) [5.3] Deformation and Strain [5.4] Hooke s Law and Poisson s Ratio
2  Stress and Strain Page 2 of 34 [5.1] Internal Stress of Solids CONCEPT OF STRESS 1) Stress is defined as internal force per unit area at the cut section of a member Q: Is stress constant value in a member? A: The expression of stress (state of stress) depends on how you cut the member Two types of stress components: Normal & Shear 2) Normal stress is a component perpendicular to the cut section (stress in the normal direction) 3) Shear stress is a component parallel to the cut section (stress in the shear direction) 4) The stress components change as the direction of the cut section changes This is called the state of stress transformation : Normal Stress : Shear Stress
3  Stress and Strain Page 3 of 34 GENERAL STATE OF STRESS 1) If we can cut out a small piece of element from a body, it represents the state of stress acting around a chosen point in the body 2) 2-D Cartesian state of stress consists of three stress components: x y xy (Normal stress in x direction) (Normal stress in y direction) (Shear stress on x y plane) 3) 3-D Cartesian state of stress consists of six stress components: x xy,, y yz,, z xz (Normal stresses in x, y, and z directions) (Shear stress on x y, y z, x z planes)
4  Stress and Strain Page 4 of 34 4) Note that the following relations apply for shear stresses: = xy yx = yz zy = xz zx UNITS OF STRESS 1) International Standard or SI system: N/m 2 (newtons per square meters) = Pa 2) U.S. Customary or Foot-Pound-Second system: lb/in 2 (pounds per square inches) = psi 1000 lb = 1 kip kip/in 2 (kilopounds per square inches) = ksi
5  Stress and Strain Page 5 of 34 AVERAGE NORMAL STRESS 1) Assuming that the normal stress is uniformly distributed across the cross section of an axially loaded bar, the average normal stress can be given as: = P A [Average Normal Stress] avg where, avg: Average normal stress at a point on the cross section P : Internal resultant normal force developed on the cross section A : Cross-sectional area 2) The normal stress can be either tensile (+) or compressive ( )
6  Stress and Strain Page 6 of 34 AVERAGE SHEAR STRESS 1) Assuming that the shear stress is uniformly distributed across the cross section of a shear surface, the average shear stress can be given as: =V A [Average Shear Stress] avg where, avg : Average shear stress at a point on the cross section V : Internal resultant shear force developed on the cross section A : Cross-sectional area 2) The shear stress can be either single or double shear SINGLE AND DOUBLE SHEAR 1) Single shear connections are called lap joints, and the resultant internal shear force is equal to the external load applied: V = F
7  Stress and Strain Page 7 of 34 2) Double shear connections are called double lap joints, and the resultant internal shear force is equal to the half of the external load applied: V = F 2 Q: Is it really reasonable to assume that stress is uniformly distributed across the cross section? A: In case of shear stress, it is NOT true (details covered later) Shear failure occurs at the centerline of the member (wooden beam) Typical shear failure of wooden beam
8  Stress and Strain Page 8 of 34 CLASS EXAMPLE The 50-lb lamp is supported by two steel rods connected by a ring at A. Determine which rod is subjected to the greater average normal stress and compute its value. Take. The diameter of each rod is given in the figure. = 60
9  Stress and Strain Page 9 of 34 Name: Student ID: HOMEWORK The 80-kg lamp is supported by two rods AB and BC as shown in the figure. If AB has a diameter of 10 mm and BC has a diameter of 8 mm, determine which rod is subjected to the greater average normal stress.
10  Stress and Strain Page 10 of 34 CLASS EXAMPLE The pins on the frame at D and E each have a diameter of 0.25 in. If these pins are subjected to single shear, determine the average shear stress in each pin.
11  Stress and Strain Page 11 of 34 Name: Student ID: HOMEWORK R1-3 The control arm is subjected to the loading shown in the figure. Determine (to the nearest ¼ in.) the required diameter of the steel pin at C if the allowable shear stress for the steel is 8 ksi. Note that, in the figure, the pin is subjected to double shear.
12  Stress and Strain Page 12 of 34 [5.2] Design of Simple Connections ALLOWABLE STRESS 1) The engineering designs need information of allowable stress for a given material in order to avoid structural failure 2) It is important to choose an allowable stress that restricts the applied external loading to the one, which is less than the load that the material can support: Allowable Load < Failure Load 3) There are many factors to be considered to determine the allowable stress Q: What are those factors? A: Weight of the structure, material costs, magnitude of the loading, type of the loading, vibration fatigue, impact resistance, etc. 4) The factor of safety (FS) is defined as the ratio of the failure load and the allowable load: FS = Failure Load / Allowable Load = F fail F allow 5) If the load applied to the member is linearly related to the stress developed within the member ( = F A, =V A): FS = = fail allow fail allow
13  Stress and Strain Page 13 of 34 DESIGN OF SIMPLE CONNECTIONS 1) The required cross-sectional area of a member under tension or compression can be given from the normal stress equation ( ): = P A A = P [Required Cross-Sectional Area] allow 2) The required cross-sectional area of a connector subjected to shear can be given from the shear stress equation ( =V A): A = V [Required Cross-Sectional Area] allow
14  Stress and Strain Page 14 of 34 3) A normal stress that is produced by the compression of one surface against another is called a bearing stress ( support the bearing stress can also be given from the normal stress equation: A = P [Required Cross-Sectional Area] ( b ) allow b ), and the required cross-sectional area to 4) The required length of a shaft to resist shear stress caused by axial load can also be given from the shear stress equation ( =V A): P l = [Required Shaft Length] allow d
15  Stress and Strain Page 15 of 34 CLASS EXAMPLE The joint is fastened together using two bolts (double shear). Determine the required diameter of the bolts if the failure shear stress for the bolts is fail = 350 MPa. Use a factor of safety for shear, F.S. = 2.5.
16  Stress and Strain Page 16 of 34 Name: Student ID: HOMEWORK R1-4 The suspender rod is supported at its end by a fixed-connected circular disk. If the rod passes through a 40-mmdiameter hole, as shown in the figure, determine the minimum required diameter of the rod and the minimum thickness of the disk needed to support the 20 kn load. The allowable normal stress for the rod is 60 MPa, and the allowable shear stress for the disk is 35 MPa.
17  Stress and Strain Page 17 of 34 CLASS EXAMPLE The rods AB and CD are made of steel having failure tensile stress of fail 510 MPa =. Using a factor of safety, F.S. = 1.75, for tension, determine their smallest diameter so that they can support the load shown. The beam is assumed to be pin connected at A and C.
18  Stress and Strain Page 18 of 34 Name: Student ID: HOMEWORK Member AC shown in the figure is subjected to a vertical force of 3 kn. Determine the position x of this force so that the average compressible stress at C is equal to the average tensile stress in the tie rod AB. The rod has a crosssectional area of 400 mm 2 and the contact area at C is 650 mm 2.
19  Stress and Strain Page 19 of 34 [5.3] Deformation and Strain DEFORMATION OF A DEFORMABLE BODY 1) Whenever an external force is applied to a deformable body, it will change both its size and shape: it is called as deformation Q: How do you measure the deformation? A: There are two types of measurements to quantify the deformation a) Changes made in length: how much it was stretched in length b) Changes made in angle: how much it was distorted in angle CONCEPT OF STRAIN 1) The strain is a quantity used to measure the intensity of deformation Q: Remember, what was stress? A: Intensity of internal force 2) There are two types of strains (just like two types of stresses): : Normal strain : Shear strain 3) Normal strain quantifies axial elongation or contraction of a member (strain in the normal direction) 4) Shear strain quantifies angular distortion of the member (strain in the shear direction)
20  Stress and Strain Page 20 of 34 AVERAGE NORMAL AND SHEAR STRAIN 1) The average normal strain of an axially loaded bar is defined as: = N L avg [Average Normal Strain] where, avg N : Average normal strain over the entire length : Total axial elongation L : Total length of the bar L L 2) The average shear strain of a plane element of a member is defined as: = S L avg [Average Shear Strain] where, avg : Average shear strain on a plane S : Total Angular distortion L : Length of the element S L /2
21  Stress and Strain Page 21 of 34 3) An alternate expression for shear strain is: avg = 2 ' [Angular Shear Strain] 4) Units of strains are given as: a) Normal strain: mm/mm or in/in b) Shear strain: radians GENERAL STATE OF STRAIN 1) If we can cut out a small piece of element from a body, it represents the state of strain around a chosen point in the body (just like state of stress) 2) 2-D Cartesian state of strain consists of three strain components: x y xy (Normal strain in x direction) (Normal strain in y direction) (Shear strain on x y plane)
22  Stress and Strain Page 22 of 34 CLASS EXAMPLE The rigid beam is supported by a pin at A and wires BD and CE. If the maximum allowable normal strain in each wire is max = mm/mm, determine the maximum vertical displacement of the load P.
23  Stress and Strain Page 23 of 34 Name: Student ID: HOMEWORK R1-5 A force acting on the grip of the lever arm shown in the figure, causes the arm to rotate clockwise through an angle of = rad. Determine the average normal strain developed in the wire BC.
24  Stress and Strain Page 24 of 34 CLASS EXAMPLE The rectangular plate is subjected to the deformation shown by the dashed line. Determine the average shear strain of the plate. xy
25  Stress and Strain Page 25 of 34 Name: Student ID: HOMEWORK R1-6 The plate is deformed into the dashed shape shown in the figure. If in this deformed shape horizontal lines on the plate remain horizontal and do not change their length, determine: (a) the average normal strain along the side AB, and (b) the average shear strain in the plate relative to the x and y axes.
26  Stress and Strain Page 26 of 34 [5.4] Hooke s Law and Poisson s Ratio TENSION AND COMPRESSION TEST 1) The mechanical properties of a material need to be determined by experiment 2) One of the most important tests that determines relationship between the average normal stress and average normal strain is tension and compression test 3) Tension and compression test provides a stress-strain diagram for a material HOOKE S LAW 1) The stress-strain diagrams of the most engineering materials exhibit a linear relationship between stress and strain within the elastic region 2) The Hooke s law is defined as: = E [Hooke s Law]
27  Stress and Strain Page 27 of 34 3) The proportionality constant E is a slope of the linear line within the elastic region, and called modulus of elasticity or Young s modulus 4) The unit of modulus of elasticity is the same as stress, and typical values are given as: GPa (SI) ksi (U.S. Customary) POISSON S RATIO 1) When a body is subjected to an axial tensile force, it elongates axially but contracts laterally 2) When a body is subjected to an axial compressive force, it contracts axially but expands laterally 3) From the strain equation, longitudinal strain is defined as: = L long and lateral strain is defined as: = ' r lat 4) The ratio of longitudinal and lateral strains is constant within the elastic range, and called Poisson s ratio: = lat long [Poisson s Ratio]
28  Stress and Strain Page 28 of 34 Q: Why there is a negative sign in Poisson s ratio definition? A: The directions of the strains are opposite: the negative sign makes the Poisson s ratio positive 6) The range of Poisson s ratio is: 0 0.5, and typical values are: 0.25 ~ 0.3 SHEAR STRESS-STRAIN DIAGRAM 1) When a body is subjected to the pure shear stress, it undergoes shear strain 2) The mechanical properties of a material under pure shear stress and strain can be experimentally obtained by torsion test 3) Torsion test provides a shear stress-strain diagram for a material
29  Stress and Strain Page 29 of 34 HOOKE S LAW FOR SHEAR 1) The stress-strain diagrams of the most engineering materials exhibit a linear relationship between shear stress and shear strain within the elastic region 2) The Hooke s law for shear is defined as: = G [Hooke s Law for Shear] 3) The proportionality constant G is a slope of the linear line within the elastic region, and called shear modulus of elasticity or modulus of rigidity 4) Three material constants: modulus pf elasticity (E), Poisson s ratio ( ), and shear modulus of elasticity (G) are related as: E G = [Constitutive Relations] 2(1 + )
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31  Stress and Strain Page 31 of 34 CLASS EXAMPLE A bar having a length of 5 in. and cross-sectional area of 0.7 in 2 is subjected to an axial force of 8000 lb. If the bar stretches in., determine the modulus of elasticity of the material. The material has linear-elastic behavior.
32  Stress and Strain Page 32 of 34 Name: Student ID: HOMEWORK R1-7 A bar made of A-36 steel (Young s modulus 200 GPa and Poisson s ratio 0.32) has the dimension shown in the figure. If an axial force of P = 80 kn is applied to the bar, determine the change in its length and the change in the dimensions of its cross section after applying the load. The material behaves elastically.
33  Stress and Strain Page 33 of 34 CLASS EXAMPLE The plastic rod is made of Kevlar 49 (Poisson s ratio 0.34 and modulus of elasticity 131 GPa) and has a diameter of 10 mm. If an axial load of 80 kn is applied to it, determine the change in its length and the change in diameter.
34  Stress and Strain Page 34 of 34 Name: Student ID: HOMEWORK An aluminum specimen shown in the figure has a diameter of d 0 = 25 mm and a gauge length of L 0 = 250 mm. If a force of 165 kn elongates the gauge length 1.2 mm, determine the modulus of elasticity. Also, determine by how much the diameter of the specimen is to be reduced. Take shear modulus of elasticity of 26 GPa and yield strength of 440 MPa.