ENGR 225 Section 1.3 – 1.6. Internal Loadings Resultant Force and Moment.
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Transcript of ENGR 225 Section 1.3 – 1.6. Internal Loadings Resultant Force and Moment.
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- ENGR 225 Section 1.3 1.6
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- Internal Loadings
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- Resultant Force and Moment
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- Coplanar Loadings
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- Stress Derivation A very small finite force F is acting on an associated area A. Replace this force with its directional components, F x, F y, and F z. Take the limit as A approaches zero of the quotient of the component forces and the areas This quotient is called, stress. It represents the intensity of the force on a specified plane.
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- Normal Stress, z The intensity of force per unit area acting normal to A.
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- Shear Stress, zx zy The intensity of force per unit area acting tangent to A.
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- Normal Stress in an Axially Loaded Bar
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- Normal Stress Distribution dF = dA dF = dA Resultant Internal force P = dA
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- Average Normal Stress Distribution If we assume is constant throughout the area or Averaged over the cross sectional area, then P = dA
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- Assumptions in using relation for average normal stress 1. Uniform cross section throughout the length 2. Uniform deformation. 3.Applied load is along Centroidal axis 4.Homogenous and Isotropic material 5.Only axial load applied. 6.Weight of the bar is neglected. Uniaxial Stress Tensile or Compressive
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- Example 1.7: The 80 kg lamp is supported by two rods AB and BC. If AB has a diameter of 10 mm and BC has a diameter of 8 mm determine the average normal stress in each rod.
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- Lecture Example: Determine the stress on the floor of a 135 lb. woman standing still wearing the following shoes that she purchased on sale. Assume that her weight is equally distributed between the heels and toes of her feet.
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- Average Shear Stress
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- Single Shear
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- Double Shear
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- Lecture Example: Determine the average normal and average shear stress developed in the wood fibers that are oriented along the section a-a.
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- Lecture Example: Determine the average compressive stress along the smooth areas of contact defined by AB and BC.
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- Lecture Example: Determine the average shear stress along the horizontal plane defined by EDB.
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- Complementary Property of Shear Stress
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- zy yz zy yz Force balance gives zy = zy Moment balance about x axis gives zy = yz Hence zy = zy = yz = yz Two- dimensional case y z
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- Allowable Stress Failure Load : Material testing To ensure safe working : Allowable Load < Failure Load F.S. = F failure / F allowable Factor of Safety : accounts for Unknown natural factors Errors in manufacturing and assembly Errors in load estimation Material weathering
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- Design of Simple Connections If = P/A holds F.S. = fail / allow Hence A = P / allow
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- Rods A and B are made of steel having failure stress 510 MPa. Use Factor of Safety 1.75, determine the smallest diameter to support the given load. Beam is assumed to be pin connected at A and C.
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- Determine the intensity w of the maximum distributed load that can be supported by the hanger assembly so that an allowable shear stress of 13.5 ksi is not exceeded in the 0.4 in diameter bolts at A and B and an allowable tensile stress of 22 ksi is not exceeded in the 0.5 in diameter rod AB.