Tìm thấy 20+ kết quả cho từ khóa "Mechanics of materials"
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The gage length of the specimen was 2 in. We can find the tangent modulus by finding the slope of the tangent at F,. (3.4) where V is the volume of the body. (3.7) Equation (3.7) reflects that the strain energy density is equal to the area of the triangle underneath the stress–strain curve in the linear region. The area of the triangle OAA 1 can be calculated as shown in Equation (E1) and equated to modulus of resilience.. 3.2 THE LOGIC OF THE MECHANICS OF MATERIALS.
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In Example 5.1 the shear strain in a bar was related to the rotation of the disc that was attached to it. Figure 5.15 The logic of the mechanics of materials.. The shear strain of interest to us is the measure of the angle change between the axial direction and the tangent to the circle in Figure 5.16. Equation (5.3) shows that the shear strain is a linear function of the radial coordinate ρ and reaches the maximum value γ max at the outer surface (ρ = ρ max = R), as shown in Figure 5.18a.
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Figure 11.8 Importance of buckled modes.. Figure 11.9 Elastic supports on columns of a water tank.. 11.2.1 Effects of End Conditions. Equation (11.9) is applicable only to simply supported columns. (11.11) where L eff is the effective length of the column. TABLE 11.1 Buckling of columns with different supports Case 1. Figure 11.10 Failure envelopes for Euler columns. 4 into Equation (11.9), we obtain the critical buckling load,.
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For homogenous materials all external and internal axial forces must pass through the centroids of the cross-section and all centroids must lie on a straight line.. Assumption 9 The external (hence internal) axial force does not change with x between x 1 and x 2. Two options for determining internal axial force N. Positive u is in the positive x-direction.. C4.1 Determine the internal axial forces in segments AB, BC, and CD by making imaginary cuts and drawing free body diagrams..
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What is the predicted service life of the beam?. If the beam has the com- posite cross section shown in Figure P6.104, determine the maximum bending normal stress in each of the three materials. But in Figure 6.48b the relative sliding is pre- vented by the shear resistance of the glue—that is, the shear stress in the glue. Because of the bending load P , a normal stress distribution across the cross section will develop as shown in Figure 6.49b..
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For isotropic materials, the principal directions for strains are the same as principal directions for stresses.. August 2012 9-17. C9.5 In a thin body (plane stress) the stresses in the x-y plane are as shown on the stress element. Determine: (a) the principal strains and the principal angle one at the point. (b) the maximum shear strain at the point.. Strain gages are bonded to a free surface, i.e., the strains are in a state of plane stress and not plane strain..
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The raising of the yield point with increasing strain is called strain hardening.. The sudden decrease in the area of cross-section after ultimate stress is called necking.. E Young’s Modulus or Modulus of Elasticity. G is called the Shear Modulus of Elasticity or the Modulus of Rigidity. E = Modulus of Elasticity E s = Secant Modulus at B E t = Tangent Modulus at B. August 2012 3-5. The Modulus of Elasticity of E = 70 GPa and a Poisson’s ratio of ν = 0.25.
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Weight of the Clothes Imaginary cut. Weight of the Clothes. Imaginary cut. along the possible path of the edge of the ring.. Pull of the hand. Pull of the hand τ. of the Clothes. Shear stress in pins. Visualizing the surface on which stress acts is very important.. August 2012 1-6. C1.2 The device shown in Fig. C1.2 is used for determining the shear strength of the wood. The dimensions of the wood block are 6 in x 8 in x 1.5 in.
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Weight of the Clothes Imaginary cut. Weight of the Clothes. Pull of the hand. Pull of the hand τ. of the Clothes. The diameter of the pin is 1 in. The area of the pin is . Figure P1.1. Figure P1.2. Figure P1.5. Figure P1.8. Board Figure P1.11. Figure P1.12. (Figure P1.17). Figure P1.14. Figure P1.15. Figure P1.16. Figure P1.17. An approximation of the monument geometry is shown in Figure P1.19. Figure P1.18. 17 m Figure P1.19.
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/in. 2.14 The average normal strain in bar A due to the application of force P in Figure P2.13 was found to be –6000 μ in./in. 2.16 The average normal strain in bar A due to the application of force P in Figure P2.15 was found to be –5000 μ in./in. 2.17 The average normal strain in bar F due to the application of force P , in Figure P2.15 was found to be -2000 μ in./in. Figure P2.13 A. Figure P2.15 F. Figure P2.18.
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Calculate relative rotation of each shaft segment in terms of the reac- tion torque of the left (or right) wall. C5.8 Two hollow aluminum (G = 10,000 ksi) shafts are securely fastened to a solid aluminum shaft and loaded as shown Fig. Point E is on the inner surface of the shaft. C5.8, Determine (a) the rotation of section at C with respect to rotation the wall at A. August 2012 5-15. C5.9 Under the action of the applied couple the section B of the two tubes shown Fig.
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C2.1 Due to the application of the forces in Fig. C2.1, the displace- ment of the rigid plates in the x direction were observed as given below.. 1.8 mm u C = 0.7 mm u D = 3.7 mm. 1000 μ rad is equal to a strain 0.001 rad. C2.2 A thin triangular plate ABC forms a right angle at point A.. Determine the average shear strain at point A.. C2.2 δ A = 0.008 in. Strain at a point. 0.500 μmm u D = 0.250 μmm
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We obtain the extension of the bar as displacement of point B.. Figure P4.7 C. (c) The axial rigidity of the bar is EA = 8000 kips. Determine the movement of the section at D with respect to the section at A.. (c) The axial rigidity of the bar is EA = 80,000 kN. Determine the movement of the section at C.. (c) The axial rigidity of the bar is EA = 2000 kips. Determine the movement of the section at B.. (c) The axial rigidity of the bar is EA = 50,000 kN. Figure P4.8. Figure P4.9. Figure P4.10.
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Determine the maximum intensity of the distributed load w.. The maximum shear stress at a point is the radius of the biggest circle.
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The area moment of inertia I zz and the first moment Q z of the area A s shown in Figure 9.12a are. Figure 9.11 Beam and loading in Example 9.5. The arrow of rotation along side the shear strains corresponds to the rotation of the line on which the point lies, as shown in Figure 9.13.. Step 3: Draw the horizontal axis to represent the normal strain, with extensions (E) to the right and contractions (C) to the left, as shown in Figure 9.14a.
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The coordinates of the point on Mohr’s Circle are the normal and shear stress on the plane represented by the point.. σ nn σ xx + σ yy. 2 + τ nt 2 σ xx – σ yy. Construction of Mohr’s Circle. The rotation arrow next to the shear stresses corresponds to the rotation of the cube caused by the set of shear stress on planes V and H.. Draw the vertical axis with clockwise direction of shear stress up and counterclockwise direction of rotation down..
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of the elastic curve.
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C6.1 Due to the action of the external moment M ext and force P, the rigid plate shown in Fig. C6.1 was observed to rotate by 2 o from the ver- tical plane in the direction of the moment. Both bars have an area of cross-section of A = 1/2 in 2 and a modulus of elasticity of E = 30,000 ksi. ε 1 = 2000 μ in./in.. Above equations are independent of material model as these equa- tions represents static equivalency between the normal stress on the entire cross-section and the internal moment..
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C7.2 In terms of w, L, E, and I, determine (a) the equation of the elastic curve. (b) the deflection at x = L.. Class Problem 7.2. C7.3 Write the boundary value problem for determining the deflec- tion of the beam at any point x. The internal moments are:. August 2012 7-8. Class Problem 7.3. v 1 and v 2 represents the deflection in segment AB and BC. For the beams shown, identify all the conditions from the table needed to solve for the deflection v(x) at any point on the beam.
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The complexity of finding the state of stress under combined loading can be simplified by first determining the state of stress due to individ- ual loading.. In order to use subscripts to determine the direction (sign) of stress components, a local x, y, z coordinate system can be established for a structural member such that the x direction is normal to cross-section, i.e., the x-direction is along the axis of the structural member..