Solution Found!
The state of strain at the point on the pin leaf has
Chapter 10, Problem 10-3(choose chapter or problem)
The state of strain at the point on the pin leaf has components of \(\epsilon_{x}=200\left(10^{-6}\right), \quad \epsilon_{y}=180\left(10^{-6}\right)\), and \(\gamma_{x y}=-300\left(10^{-6}\right)\). Use the strain transformation equations and determine the equivalent in-plane strains on an element oriented at an angle of \(\theta=60^{\circ}\) counterclockwise from the original position. Sketch the deformed element due to these strains within the x–y plane.
Questions & Answers
QUESTION:
The state of strain at the point on the pin leaf has components of \(\epsilon_{x}=200\left(10^{-6}\right), \quad \epsilon_{y}=180\left(10^{-6}\right)\), and \(\gamma_{x y}=-300\left(10^{-6}\right)\). Use the strain transformation equations and determine the equivalent in-plane strains on an element oriented at an angle of \(\theta=60^{\circ}\) counterclockwise from the original position. Sketch the deformed element due to these strains within the x–y plane.
ANSWER:
Problem 10-3The state of strain at the point on the pin leaf has components of x = 200(10-6 ), y =180(10-6 ), and xy = -300(10-6 ).Use the strain transformation equations and determine the equivalent in-plane strains on anelement oriented at an angle of u = 60 counterclockwise from the original position. Sketch thedeformed element due to these strains within the xy plane. Step-by-step solution Step 1 of 4 ^ 1Write the relation to find the in plane strain along x by using the following relation. x y x y xyX1= 2 + 2 cos2 + 2 sin2Here, are the plane strains along x and y direction is the shear strain, and is the plane X, y xyangle.Substitute 200 x 10 for , 180 x 10 for , -300 x 10 for is the shear strain, and is the X y xyplane angle. 200 + 180 200180 (300)X1= [( ) + ( ) cos2(60°) + sin2(60°) ]x 10-6 2 2 2 =[190 + 10cos120° - 150sin120°) x 10 -6 -6 = 55.1 x 10Therefore, plane strain along x is 55.1 x 10 . -6