Use Hooke’s law, Eq. 14–32, to develop the strain tranformation equations, Eqs. 14–19 and 14–20, from the stress tranformation equations, Eqs. 14–1 and 14–2.
Stress Components Acting along x′, y′ Axes
Normal and Shear Strain Components
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Chapter 14 Solutions
Statics and Mechanics of Materials Plus Mastering Engineering with Pearson eText - Access Card Package (5th Edition)
- A single strain gage, placed in the vertical plane on the outer surface and at an angle 60° to the axis of the pipe, gives a reading at point A of PA = -250(10-6). Determine the principal strains in the pipe at this point. The pipe has an outer diameter of 1 in. and an inner diameter of 0.6 in. and is made of C86100 bronze.arrow_forwardThe 45° strain rosette is mounted on a surface of the bracket . The bracket is made from steel with Esteel = 120 GPa and poison ratio, v = 0.28. The following readings are obtained for each gauge under loadings:Ɛa = [ X+Y ] (10-6)Ɛb = -200(10-6)Ɛc = -180(10-6) x=100, y=100 (a) Estimate the in-plane principal strains and the angle associated with the principal strains, and (b) Calculate the principal stress associated with the principal strains in (a).arrow_forwardA force P and a force Q, applied via a nut, are acting on the arm attached to the end of a shaft made of steel. The strain gauge readings on point A of the shaft show the following deformation values: ε1 = 630x10-6, ε2 = 600x10-6, and ε3 = - 189x10-6. Determine the magnitudes of the applied forces P and Q (E = 200 GPa, υ = 0.3).arrow_forward
- The pipe assembly shown is subjected to a force F = 7 kN and P = 14 kN. It is made of steel with Sy = 250 MPa. Determine the safety factor at point H using the maximum shear stress theory. Select one: a. NH = 6.427 b. NH = 3.570 c. NH = 5.355 d. NH = 4.590arrow_forwardThe ratio of direct stress to the volumetric strain is known as Bulk Modulus Select one: True Falsearrow_forwardThe material is subjected to biaxial loading producing uniform normal stress x and y as shown. The strains are Strainx=−0.00065 and Strainy=−0.00040. Use E=30×106 psi and v=0.30. Determine the following: (a) stress at x. Indicate tension or compression. Use 2 decimal places. (b) stress at y. Indicate tension or compression. Use 2 decimal places. (c) Change in the thickness of the material. Indicate elongation or contraction. Use 5 decimal places and scientific notation of ×10−3(Example: _ . _ _ _ _ _ ×10−3)arrow_forward
- The 60° strain rosette is mounted on the surface of the bracket. The following readings are obtained for each gage: Pa = -780(10-6), Pb = 400(10-6), and Pc = 500(10-6). Determine (a) the principal strains and (b) the maximumin-plane shear strain and associated average normal strain. In each case show the deformed element due to these strains.arrow_forwardHooke's law is used to correlate true stress and true strain in the plastic region of the stress/strain curve. true or falsearrow_forwardThe pipe assembly shown is subjected to a force F = 400 N. The pipe has an inner diameter of 20 mm and an outer diameter of 30 mm. It is made of steel with Sy = 250 MPa. Determine the safety factor at point A using the maximum shear stress theory. Select one: a. NA = 1.843 b. NA = 3.224 c. NA = 2.580 d. NA = 4.299arrow_forward
- The 45° strain rosette is mounted on a steel shaft. The following readings are obtained from each gage: Pa = 800(10-6), Pb = 520(10-6), Pc = -450(10-6). Determine the in-plane principal strains.arrow_forwardThe state of plane strain at a point has strain components of Px = -400(10-6), P y = 200(10-6), and gxy = 150(10-6), Fig. a. Determine the maximum in-plane shear strain and the absolute maximum shear strain.arrow_forwardThe state of strain at the point on the spanner wrench has components of Px = 260(10-6), P y = 320(10-6), and gxy = 180(10-6). Use the strain transformation equations to determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case specify the orientation of the element and show how the strains deform the element within the x–y plane.arrow_forward
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