Samantha Ramirez, MSE Stress The intensity of the internal force acting on a specific plane (area) passing through a point. Δ ΔA Δ z Δ 1 2 ΔA Δ x Δ y ΔA is an infinitesimal size area with a uniform force acting on it. 1
Stress Equations Normal Stress (σ) is the intensity of force, or force per unit area, acting normal to ΔA. z σ z = lim A 0 A Shear Stress (τ) is the intensity of force, or force per unit area, acting parallel to ΔA. Subscripts irst letter is the orientation of force normal to ΔA Second letter is the orientation of force causing shear x τ zx = lim A 0 A y τ zy = lim A 0 A σ x, σ y, σ z, τ xy, τ xz, τ yx, τ yz, τ zx, τ zy Stress States τ xy = τ yx, τ xz = τ zx, τ yz = τ zy σ x, σ y, σ z, τ xy, τ xz, τ yz Volume Element Units N/m 2 (Pa) 1000 N/m 2 (kpa) lb f /in 2 (psi) 1000 lb f /in 2 (ksi) 2
Saint-Venant s Principle The stress and strain at points in a body sufficiently away from the region of load application will be the same as the stress and strain produced by any applied loadings that have the same statically equivalent resultant. Deformation Deformation is defined as the changes to a body s shape and size when a load is applied. Deformation is specified by normal and shear strains and occurs because of loads and temperature. 3
Average Normal Strain (ε) The elongation or contraction of a line segment per unit length ε avg = s s = L s s Strain depends on geometry of formation Stress and strain are independent of each other Average Shear Strain (γ) The change in angle between two lines that were originally perpendicular γ nt = π 2 lim θ B A along n C A along t 4
Tensile Test P is measured using a load cell σ avg = P A δ is elongation measured using an extensometer (optical or mechanical) ε avg = δ L o A strain gauge can be used to measure strain directly Stress-Strain Diagram The most important result from the tension test is the stress-strain diagram (σ-ε diagram). Nominal or engineering stress and strain are used to create a σ-ε diagram. σ = A ε = L L o 5
Stress-Strain Diagram Properties Proportional Limit The upper stress limit in the linear elastic region of a σ-ε diagram Elastic Limit If the load is removed before reaching the elastic limit, the specimen will return to its original shape If the load passes the elastic limit, the specimen will permanently deform Yield Strength The stress that causes a material to breakdown and deform permanently When yield strengths is not well defined, the offset yield strength can be calculated A line parallel to the initial straight-line portion of the stress-strain curve at 0.2% strain (0.002 in/in). Ultimate Tensile Strength The maximum stress the material will reach through testing Necking occurs (visibly see permanent deformation) racture Stress The stress when the material breaks Hooke s Law The proportional limit was observed by Robert Hooke in 1676 E = σ ε E: Young s Modulus, the constant of proportionality Young s Modulus or the modulus of elasticity is a mechanical property of a material that indicates stiffness. : Slope of the straight-line portion of the stress-strain curve σ ε Theoretical Moduli Steel: E=29 Msi or 200 GPa Aluminum: E=10 Msi or 69 Gpa The modulus of elasticity can only be used if a materials has linear elastic behavior and is being subjected to stresses below the proportional limit. If a material has yielded, the modulus of elasticity cannot be used. 6
Poisson s Ratio A deformable body subjected to a normal tensile load not only elongates but also contracts laterally. ε long = δ L ε lat = δ r ν = ε lat ε long Shear Stress- Shear Strain Diagram Hooke s Law for Shear G = τ γ G: Shear modulus of elasticity or modulus of rigidity τ: Shear stress γ: Shear strain Relation of modulus of elasticity and rigidity G = E 2(1+ν) or steel, E=29000 ksi G=11000 ksi ν=0.32 7
Uniaxial Tensile Test Axially Loaded Bar (Uniaxial Tensile Test) Assumptions Homogeneous material Isotropic material Bar remains straight and cross-section flat P is applied along the centroid axis 8
Average Normal Stress න d = න σda P = σ Avg A σ Avg = P A Example 1 The built-up shaft consists of a pipe AB and solid rod BC. The pipe has an inner diameter of 20 mm and an outer diameter of 28 mm. The rod has a diameter of 12 mm. Determine the average normal stress at D and E and represent the stress on a volume element located at each of these points. 9
Example 2 If the turnbuckle is subjected to an axial force of P=900 lb, determine the average normal stress developed in section a-a and in each bolt shank at B and C. Each bolt shank has a diameter of 0.5 in. 10
Average Shear Stress in Connections Direct Shear Shear Stress is the force per unit area that acts in a plane of the cross-sectional area. τ Avg = V A Single Shear Situation #1 M N V Situation #2 V Situation #3 Pin 11
Double Shear Situation #4 Situation #5 Pin 2 2 2 2 2 Example 3 If the joint is subjected to an axial force of P=9 kn, determine the average shear stress developed in each of the 6-mm diameter bolts between the plates and the members and along each of the four shaded shear planes. 12
actor of Safety actor of Safety (S) is a ratio of the failure load, fail, divided by the allowable load, allow. (S>1) S = ail Allow If the load applied to the member is linearly related to the stress developed within the member, then S = N ail N Allow = σ ail σ Allow S = V ail V Allow = τ ail τ Allow Example 4 The pin is made of a material having a failure shear stress of 100 MPa. Determine the minimum required diameter of the pin to the nearest mm. Apply a factor of safety of 2.5 against shear failure. 13
Example 5 Determine the maximum vertical force P that can be applied to the bell crank so that the average normal stress developed in the 10 mm diameter rod, CD, and the average shear stress developed in the 6 mm diameter double sheared pin B not exceed 175 MPa and 75 MPa, respectively. Example 6 To the nearest 1/16, determine the required thickness of member BC and the diameter of the pins at A and B if the allowable normal stress for member BC is σ Allow =29 ksi and the allowable shear stress for the pins is τ Allow =10 ksi. 14
Normal Stress Concentrations Holes and sharp transitions at a cross section will create stress concentrations. In engineering practice, only the maximum stress at these sections must be known. The maximum normal stress occurs at the smallest cross-sectional area. Stress Concentration actor A stress concentration factor K is used to determine the maximum stress at sections where the cross-sectional area changes. K = σ max σ avg K is found from graphs in handbooks of stress analysis σ avg is the average normal stress at the smallest cross section 15
Stress Concentration actor Tables Example 7 Determine the maximum normal stress developed in the bar when it is subjected to a tension of P=2 kip. 16
Thin-Walled Pressure Vessels Cylindrical or spherical vessels are commonly used in industry to serve boilers or tanks. A thin wall refers to a vessel having an inner radius to wall thickness ratio of 10 or more r t 10. We will assume a uniform or constant stress distribution throughout the thickness because it is thin. Pressure vessels are subjected to loadings in all directions. Cylindrical Vessels Circumferential or Hoop Stress (σ 1 ) Longitudinal or Axial Stress (σ 2 ) σ 1 = pr t σ 2 = pr 2t 17
Spherical Vessels σ 2 = pr 2t σ 1,σ 2 : the normal stress in the hoop and longitudinal directions, respectively. Each is assumed to be constant throughout the wall of the cylinder, and each subjects the material to tension. p: the internal gauge pressure developed by the contained gas r: the inner radius of the cylinder t: the wall thickness (r/t 10) Example 8 The tank of the air compressor is subjected to an internal pressure of 90 psi. If the internal diameter of the tank is 22 in, and the wall thickness is 0.25 in, determine the stress components acting at point A. Draw a volume element of the material atthis point, andshow the results on the element. 18
Example 9 The A-36 steel band is 2 in wide and is secured around the smooth rigid cylinder. If the bolts are tightened so that the tension in them is 400 lb, determine the normal stress in the band, the pressure exerted on the cylinder, and the distance half the band stretches. Example 10 A pressure-vessel head is fabricated by gluing the circular plate to the end of the vessel as shown. If the vessel sustains an internal pressure of 450 kpa, determine the average shear stress in the glue and the state of stress in the wall of the vessel. 19