Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress

s = Average Stress (N/mm2 or MPa) Axial Stress F= Axial Force (Newtons, N) A = Cross-Sectional Area Perpendicular to “F” (mm2) E = Young’s Modulus of Material, MPa L = Original Length of Component, mm s = Average Stress (N/mm2 or MPa) D = Total Deformation (mm) AE = “Axial Stiffness of Component”

Direct Shear Stress taverage = Avearge Shear Stress (MPa) P = Shear Load A = Area of Material Resisting “P”

Examples of Direct Shear Stress Bolted Joint with Two Shear Planes. P = 50 KN D = 13 mm tavg = ? Area of bolt (Ab) = p D2 / 4 = p (13)2 / 4 = 132.7 mm2 A resisting shear = 2 Ab tavg = P / 2Ab = 50000 N/ 2(132.7) mm2 = 188.4 MPa

Direct Shear II Fillet Weld 13 13 150 175 Fillet Weld 9.2 Find the load P, such that the stress in the weld does not exceed the allowable stress limit of 80 MPa.

tavg = P / Aw = 80 MPa Solution: Aw = Throat x Total Length = (9.2)(175)(2) = 3217 mm2 P / 3217 = 80 MPa P = (3217)(80) N = 257386 N = 257 kN

BENDING Before C After T Neutral Plane

Displacement in Beam Curvature of Beam = y x v M = Moment; EI = Bending stiffness of Beam

Bending Stress smax (C) y cc s ct smax (T)

Source of Internal Moment Fc d Ft Fc = Ft M= Ft d

Equivalent FBD M = Fd F d F

Bending Stiffness Ixx = bh3 / 12 (mm4) Iyy= hb3 / 12 (mm4) EI = Bending Stiffness E = Young’s Modulus (Material Dependant) I = Moment of Inertia (2nd Moment of Area) y Ixx = bh3 / 12 (mm4) h Iyy= hb3 / 12 (mm4) y b

Moment of Inertia IXX = Ix’x’ + A y2 Parallel Axis Theorem A If X-X is the neutral axis: S A y = 0

Try it! Locate the Neutral Axis and find the Moment of Inertia for the “T” section shown below. Consider the XX axis, all dimensions are mm. 300 Ans: Ct = 200 mm Ixx = 2.50x108 mm4 50 X X 250 Ct 60

Ok... Neutral Axis 300 50 X X Ct/2 250 Ct 60

Mech 422 – Stress and Strain Analysis 300 Ok... 50 75 X X 75 250 200 60 Mech 422 – Stress and Strain Analysis

Determine the Bending Stiffness of beams with this cross section made of: 1) Steel, E=203 000 MPa 2) Aluminum Alloy, E= 72 000 MPa 3) Glass Reinforced Polyester, E = 30 000 MPa

Determine the Bending Stiffness of beams with this cross section made of: 1) Steel, E=203 000 MPa 2) Aluminum Alloy, E= 72 000 MPa 3) Glass Reinforced Polyester, E = 30 000 MPa Ans: 1) EI = 5.08 x 1013 Nmm2 2) EI = 1.80 x 1013 Nmm2 3) EI = 0.75 x 1013 Nmm2 In the ratio 1 : 0.36 : 0.15

Bending Stress: 300 kN M max = 200 KN.m s tension = 200x106 N.mm (200) mm / 2.50 x108 mm4 = 160 MPa MAX s compression = 200x106 N.mm (100) mm / 2.50 x108 mm4 = 80 MPa

Bending Shear A V

Bending Shear A V V S Fy = 0

Bending Shear V A V V V S Fy = 0 S Fx = 0 S M = 0

Shear Stress - Bending t avg = V / A t max = 1.5 V / A t avg t max For a Rectangle: t max y t max = 1.5 V / A t avg = V / A A = b h b h t avg Max Shear Stress is at N.A.

t y = V Q / I b Shear in Bending Q = 1st Moment of Area In General: b @ top of web t y = V Q / I b X X Q = 1st Moment of Area b @ N.A. b = width of the X-section at the plane of interest.

1st Moment of Area Qy = A d A=zw w Consider all of the X -section above (or below) the plane of interest. z d y NA A=zw Qy = A d

Find The 1st Moment of Area at the Neutral Axis. (dimensions are mm.) 300 50 X X 250 Ct 60

Q = S Ay The 1st Moment of Area at the Neutral Axis: (dimensions are mm.) Q = S Ay 300 = (300)(50)(75) + (60)(50)(25) = 1.2 x 106 mm3 Note: At the N.A. b=60 mm 50 75 X X 25 250 200 60

Shear Stress: t y = V Q / I b t max = (200x103 N) (1.2x106 mm3) 300 kN 100 kN 200kN t y = V Q / I b V max = 2000 KN t max = (200x103 N) (1.2x106 mm3) (2.50x108 mm4 (60) mm t avg = V/Aweb = (200x103 N) / (60)(300) mm2 = 11.1 MPa = 16 MPa

Bending Shear The “Maximum” Stress Distributions in the beam are: 300 Compression 80 MPa tmax=16 MPa 50 N.A. X X 250 60 160 MPa Tension

Torsion r T = Torque, G = Shear Modulus of Elasticity L = Length of Shaft, J = Polar Moment of Inertia q is in Radians!

Shear Stress Distribution tmax r D

Polar Moment of Inertia Do Di

Shear Modulus, G E = Young’s Modulus v = Poisson’s Ratio Example: Steel E = 203 000 MPa, v = 0.3 G = 78 000 MPa

Shear Stress-Strain Curve Shear Stress, t Shear Modulus, G Shear Strain, g

500 mm Torque 150 mm 50 mm 25 KN

T = 25x103 N (150 mm) = 3.75x106 N.mm L = 500 mm = p (50)4 / 32 = 6.14x105 mm4 3.75x106 N.mm (500) mm = = 3.9x10-2 = 2.2o 6.14x105 mm4 (78000 N/mm2) 3.75x106 N.mm (25 mm) = = 152.7 MPa 6.14x105 mm4 FS = 200/152.7 = 1.31

Superposition Assume the beam in our example is made of steel with a yield stress of 350 MPa. If it is subjected to an additional Axial Tension of 5000 kN along it N.A., will it yield ? 300 kN F = 5000 KN 200 kN 100 kN

Result: Axial Net + Bending = Tension Tension Compression 80 MPa F/A = 167 MPa 80 MPa 83 MPa (Tension) Tension N.A. + = Tension < 350 MPa No Yield! 160 MPa 327 MPa Tension

Rule for Adding Stresses: Like stresses at a point acting in the same direction and on the same plane can be added algebraically. You can’t add a shear stress to a tensile/compressive stress. You can’t add a stress in one location to one at another. The effects of combined shear and tension/compression are covered later in this course.