EMA 3702 Mechanics & Materials Science (Mechanics of Materials) Chapter 2 Stress & Strain  Axial Loading


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1 MA 3702 Mechanics & Materials Science (Mechanics of Materials) Chapter 2 Stress & Strain  Axial Loading
2 MA 3702 Mechanics & Materials Science Zhe Cheng (2018) 2 Stress & Strain  Axial Loading Statics vs. Mechanics of Materials Statics  Deals with undeformable bodies (rigid bodies) Mechanics of Materials  Deals with practical, deformable bodies Need to calculate the stress and deformation (relative and absolute) of a body under various loading (stress) states Compute forces and related information (stress/strain) for certain statically indeterminate problems
3 Poisson 's Ratio When a material is subject to normal stress (e.g., tensile) in one direction, it subject to deformation in other (transverse) directions. Define Poisson s Ratio v lateral strain axial strain For isotropic material: y x z x Within elastic limit x x y z x
4 Multiaxial Loading Axial loading from more than one directions? Assumptions: ach effect is linear The deformation is small and does not change the overall condition of the body. Principle of Superposition: The combined effect (strain) = individual effect (strain)
5 Generalized Hooke's Law Normal strain along each direction (x, y, z) comes from three contributions: x y z x y z x y z x y z
6 MA 3702 Mechanics & Materials Science Zhe Cheng (2018) 2 Stress & Strain  Axial Loading Class xercise A 500 mm long, 8 mm diameter rod (i.e., crosssection area of ~50 mm 2 ) made of homogeneous rod is subject to 10 kn axial tensile load. The length increases by 0.5 mm while the diameter shrinks by 2 μm. Please calculate elastic modulus and Poisson s ratio. lastic modulus 3 P / A PL 1010 N 500mm 2 / L A 50mm 0.5mm Pa 200GPa N m 2 Poison ratio v y x mm / 8mm 0.5mm / 500mm
7 Shearing Strain (1) For general loading condition, shearing stresses are present, the angle between neighboring surfaces of cube unit will change: From a cube oblique parallelepiped
8 Shearing Strain (2) The change in angle due to shearing stress (from π/2) is defined as shearing strain xy in radians Sign convention for shearing strain: If shear causes reduction in that angle, it is positive; otherwise, negative
9 Hooke s Law for Shearing Within elastic (proportional) limit, xy G xy yz G yz xz G xz G is Shear Modulus or Modulus of Rigidity
10 Generalized Hooke s Law x y xy x x x z G xy yz G yz xz xz G Relationship between, G, and ν G y y y z z z 2 1
11 MA 3702 Mechanics & Materials Science Zhe Cheng (2018) 2 Stress & Strain  Axial Loading Class xercise A 2 inch thick 2 8 in 2 rectangular block with shear modulus G = 100 ksi is bonded between the rigid ground and one rigid top plate, as illustrated. The top plate is subjected to shear force P. Knowing the top plate moved 0.04 inch horizontally. Please calculate the average shearing strain and the shear force P. Average shearing strain arctan In case one wants to know the angle in deg, arctan in 0.02 / in 0.04 in γ P Shear force P P A G A in lb / in lb
12 StressStrain Distribution: Ideal vs. Actual Idealized condition: Uniform stress distribution In reality: Nonuniform distribution ave P A Local stress changes in complex way
13 SaintVenant's Principle The stress distribution far away from the region where the load is applied is NOT related to the mode of application. Implication: for axial loading, uniform stress if it is far away from loading points MA 3702 Mechanics & Materials Science Zhe Cheng (2018) 2 Stress & Strain  Axial Loading
14 Stress Concentrations (1) Stress rises at locations where geometric discontinuity exists Stress Concentration Factor K max ave The sharper the crack tip (or smaller r), the higher K
15 Stress Concentrations (2) Stress rises at locations where geometric discontinuity exists K max ave The sharper the fillet (smaller r), the higher K The larger the dimension ratio (D/d), the higher K
16 Plastic Deformation Idealized elastoplastic behavior: After linear elastic deformation, plastic deformation is characterized by no change in stress (i.e., flat line) until fracture or rupture. y Y Rupture A
17 Stress Concentration Involving Plastic Deformation As load P increases When max < Y lastic deformation K max ave P ave A max K A When max = Y Load for yielding: When ave = Y (Ultimate) load for fracture: Yielding occurs P Y P U Y K Y A Fracture occurs (after certain deformation) A
18 Cube with unit length subject to multiaxial stress of x, y, z The total volume changes from 1 to υ: Dilation & Bulk Modulus (1) Neglecting the high order terms yields: Relative change in volume, also called dilation, e z y x e 1 From the relationships between ε i and σ i, we have z y x e 2 1 z y x 1 ) )(1 )(1 1 ( z y x
19 Dilation & Bulk Modulus (2) Dilation e 1 2 x y Special case: Hydrostatic pressure or x = y = z = p z e 31 2 p Define as bulk modulus or modulus of compression: e p
20 Dilation & Bulk Modulus (3) Dilation e p Practically, when applying pressure, volume should shrink or dilation e should be negative must be positive Therefore, (12ν) > 0 0 < ν < ½ For boundary conditions ν = 0 3 e 3 p ν = ½ e 0 Incompressible solid
21 Residual Stresses After the applied load is removed, some stresses may still remain inside the material xample: solder a bar over a hole on a metal plate During cooling, metal bar gradually hardens and (tensile) stress develops within the bar, which may even exceeds yield strength!
22 MA 3702 Mechanics & Materials Science Zhe Cheng (2018) 2 Stress & Strain  Axial Loading Homework 2.0 Read text book chapter sections 2.1 to 2.7, (you may skip the sample problems) and give an honor statement confirm reading.
23 MA 3702 Mechanics & Materials Science Zhe Cheng (2018) 2 Stress & Strain  Axial Loading Homework 2.1 A plastic wire is pulled by a tensile force of 4 N, and its length increases by 1%. If the plastic has elastic modulus of 4 GPa. Please calculate (a) average normal stress in the wire and (b) the diameter of wire
24 Homework 2.2 MA 3702 Mechanics & Materials Science Zhe Cheng (2018) 2 Stress & Strain  Axial Loading The 4 mm diameter metal wire F has modulus = 200 GPa. If the maximum stress should not be higher than 250 MPa, and the elongation of the wire should be longer than 5 mm. Please calculate the maximum load P can be applied. F D 2 m F 2 m 3 m
25 MA 3702 Mechanics & Materials Science Zhe Cheng (2018) 2 Stress & Strain  Axial Loading Homework 2.3 Members CD and BC have common = psi and crosssection area of 0.75 in 2 and 0.6 in 2, respectively. For the forces shown, determine the elongation of (a) CD and (b) BC. 25 kips D 50 kips 5 ft 5 ft A C B 5 ft
26 Homework 2.4 MA 3702 Mechanics & Materials Science Zhe Cheng (2018) 2 Stress & Strain  Axial Loading A centric load of P=400 kn is applied to the composite block shown through a rigid end plate. The iron plate (G Fe =200 GPa) is sandwiched between two Cu plates (G Cu =80 GPa). Center iron plate is 40 mm thick while the outer Cu plates are 20 mm thick. All plates are 500 mm long and 100 mm wide, Calculate the normal stress in (a) the center Fe core and the outer Cu plates 400 kn 500mm 20 mm 40 mm 20 mm
27 Homework 2.5 MA 3702 Mechanics & Materials Science Zhe Cheng (2018) 2 Stress & Strain  Axial Loading A rigid bar H is supported by two steel wire of 1/16 in diameter (= psi) and a pin and bracket at H. Knowing that the wires were initially tight. Determine (a) the additional tension in each wire when a 220lb load P is applied at. (b) The corresponding deflection (downward displacement) of point. 8 in 200 lb F G H 10 in 10 in 10 in 10 in
28 MA 3702 Mechanics & Materials Science Zhe Cheng (2018) 2 Stress & Strain  Axial Loading Homework 2.6 A tensile force of 2 kn is applied to a test steel ( = 200GPa, ν = 0.30) bar with rectangular crosssection with initial thickness of 2 mm, initial width of 10 mm, and initial length of 100 mm. Please calculate the absolute change in (a) sample length, (b) sample thickness, (c) sample width, and (d) sample crosssection area.
29 MA 3702 Mechanics & Materials Science Zhe Cheng (2018) 2 Stress & Strain  Axial Loading Homework 2.7 Two blocks of rubber with shear modulus G=10 MPa are bonded to rigid supports and a plate AB as illustrated. c = 100 mm and P = 50 kn. Calculate the minimum value for dimension a and b for the rubber blocks if the shearing stress should not exceed 1 MPa and the deflection of the plate is at least 4 mm.
30 MA 3702 Mechanics & Materials Science Zhe Cheng (2018) 2 Stress & Strain  Axial Loading Homework 2.8 Cylindrical rod CD has length of L = 50 inch and a crosssection area of 0.5 in 2. It is made of a material assumed to be elastoplastic with = psi and yield strength Y = 36 ksi. A tensile force P is applied to the rod and then completely removed to give it a permanent deformation (or set) of δ P. Calculate the maximum value of P and maximum amount deformation δ m by which the rod should be stretched if the desired value of δ P is (a) 0.1 in, (b) 0.2 in.
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