02) a) For isotropic material and a complex stress state explain yield criteria 16] and yield curve. 02) a) For isotropic material and a complex stress state explain yield criteria 16] and yield...
(2D Yield Criteria, 16 points) The stress distribution of three wind turbine blades were investigated, and found there are thee critical place, A, B and C that are under a fairly high stress state. The 2D principal stresses at these spots are given below. The uni-axial yield strength (Ơy) ofwind turbine material is 100 MPa. Note that this is purely 2D problem. So, just use the 2D yield boundaries in the lecture note to solve this problem (a) Draw the...
2. [5 pts] Show that for an isotropic linear elastic material subjected to a biaxial stress state, the relationship between stress and strain is such that: Of = (x+vey)}; Oy = (byt veze; O Txy = 207vs Yxy 1-12 1-12 1 Ιαν
For an isotropic material, (a) Calculate the components of the strain tensor and the stress tensor for the following set of given displacements for an isotropic material: Uj = - X1 , U2 = -V – X2, U3 = -V – X3 , E E E where o is a constant. (b) Check the equilibrium equations to see if they are satisfied for zero body forces. (c) Show the edge tractions on a diagram of the body0 S XL SL,0...
An applied stress fluctuates between -33 MPa and 6 MPa. The yield stress of the material is 41 MPa and the endurance limit is 27 MPa. What is the factor of safety according to the ASME-elliptic fatigue failure criteria?
The figure shows the stress-strain curve for a material. The scale of the stress axis is set by s = 330, in units of 106 N/m^2. What are (a) the Young's modulus and (b) the approximate yield strength for this material? (a) Number Units (b) Number Units
Failure criterion The yield stress of a material is 40 MPa. This material is subjected to the maximum normal stress and the minimum normal stressz--0.350. Determine the magnitude of 01 MPa that will cause yielding according to the maximum shear stress criterion (rounding to two decimal places).
Stress-Strain Curve 39· Material Type (l ot J) 40 Modulus of Elasticity 41. Stress Axis 42. Strain Axis G K l-ductile J-brittle 43.Yield Point 4Fracture Point 45. Ultimate Strength Bonus 2: If the force is removed when the material is at point F, what will happen to the material?
Maximum stress that a material can resists is called yield stress. True or False?
Identify the material. that fits the stress-strain
curve.
A)
MILD STEEL
Yield Strength: 30 kN, Ultimate Strength: 40 kN, Young Modulus:
30/0.7 kN/mm = 43 kN/mm
B)
ALUMINUM
Yield Strength: 20 kN, Ultimate Strength: 26 kN, Young Modulus:
20 kN/mm
C)
CAST IRON
Yield Strength: N/A, Ultimate Strength: 20 kN, Young Modulus:
20/0.4 kN/mm = 50 kN/mm
D)
COPPER
Yield Strength: 30 kN, Ultimate Strength: 35 kN, Young Modulus:
30 kN/mm
Instructions CLI Tesile TesTeR Test Specimens Material - Mild...
6. The flow curve equation for a material is given as follow. Plot the engineering stress-strain curve for the material. Label the plot properly. 0 = 552.7 €0.2 MPa [Hint: Transform the flow curve equation to engineering stress-strain equation]