Explanation The force F y produces a moment in negative z –direction. Calculate the moment about A due to force F y . M z = F y ⋅ l BC (I) Here, the length of the rod B C is l B C . The force F z produces a moment M y in the negative y-direction and the distance between the force and centre of axis for this moment is the length of rod AB + length of handle. The force F x produces a moment M y in y-direction and the distance between the force and centre of axis for this moment is the length of rod BC . Write the resultant of the moment in y-axis for the element. M y = F z ⋅ ( l BC ) − F x ⋅ ( l ) (II) Here, the length of the rod BC is l BC , and the length of the rod AB + length of handle is l . The force F y produces a moment M z in z-direction and the distance between the force and centre of axis for this moment is the length of rod (AB+BC). Write the total moment for the element. M t = M y 2 + M z 2 (III) Write the expression for bending stress in x-direction. σ ( bx ) = M t ⋅ c I (IV) Here, the moment on the bar is M t , the moment of inertia is I ,and the distance of bending force from the centre of axis is c . The maximum bending stress is at outer surface or radius that is at c = d 2 . Write the expression for moment of inertia. I = π ⋅ d 4 64 Here, the diameter of the bar is d . Substitute d 2 for c , and ( π ⋅ d 4 64 ) for I in the Equation (IV). σ ( bx ) = M t ⋅ ( d 2 ) ( π ⋅ d 4 64 ) = 32 M t π ⋅ d 3 Write the expression for axial stress on the element. σ ( ax ) = F x A (V) Here, the cross-sectional area of the bar is A . Write the expression for cross-sectional area of the bar. A = π d 2 4 Substitute π d 2 4 for A in the Equation (V). σ ( ax ) = F x ( π d 2 4 ) Write the total stress on the element in x-direction. σ x = σ ( ax ) + σ ( bx ) (VI) Substitute F x ( π d 2 4 ) for σ ( ax ) , and ( 32 M t π ⋅ d 3 ) for σ ( bx ) in the Equation (VI). σ ( ax ) = F x ( π d 2 4 ) + ( 32 M t π ⋅ d 3 ) (VII) Write the expression for torsional stress. τ x z = T × R J (VIII) Here, the radius of the bar is R , and the polar moment of inertia is J . Write the polar moment of inertia for the bar. J = π ⋅ d 4 32 Substitute ( π ⋅ d 4 32 ) for J , and d 2 for R in the Equation (VIII). τ x z = T × ( d 2 ) ( π ⋅ d 4 32 ) = 16 T π ⋅ d 3 (IX) Write the torque equation on element. T = F ⋅ z (X) Here, the force on the element is F and the distance between the axis of rotation and the point of application of force is z . The distance between the axis of rotation and the point of application of force is the length of the rod BC . Write the Neubar constant for the stress element due to bending. a = 0.246 − 3.08 ( 10 − 3 ) S u t + 1.51 ( 10 − 5 ) ( S u t ) 2 − 2.67 ( 10 − 8 ) ( S u t ) 3 (XI) Here, the minimum tensile strength for the element is S u t . Write the expression for notch sensitivity for the element due to bending. q = 1 1 + a r (XII) Here, the notch radius is r . Write the expression for fatigue-stress concentration factor due to bending. K f = 1 + q ( K t − 1 ) (XIII) Here, the stress concentration factor for bending is K t . Write the Neubar constant for the stress element due to torsion. a = 0.190 − 2.51 ( 10 − 3 ) S u t + 1.35 ( 10 − 5 ) ( S u t ) 2 − 2.67 ( 10 − 8 ) ( S u t ) 3 (XIV) Write the expression for notch sensitivity for the element due to torsion. q s = 1 1 + a r (XV) Write the expression for fatigue-stress concentration factor due to torsion. K f s = 1 + q s ( K t s − 1 ) (XVI) Here, the stress concentration factor for torsion is K t s . Write the expression for von Mises stress for the stress element. σ ′ a = { [ ( K f ) bending ( σ a ) bending ] 2 + 3 [ ( K f s ) torsion ( τ a ) torsion ] 2 } 1 2 (XVII) Here, the fatigue stress-concentration factor for bending is ( K f ) bending , the fatigue stress-concentration factor for axial load is ( K f ) axial , the axial bending stress is ( σ a ) bending , the fatigue stress-concentration factor for torsion is ( K f s ) torsion and the torsional stress is ( τ a ) torsion . The mid-range stress is same as the alternating stress. σ ′ m = σ ′ a Write the expression for localized yielding. n y = S y σ ′ m + σ ′ a (XVIII) Here, the ultimate yielding stress is S y . Write the expression for endurance limit of the rotating-beam specimen. S ′ e = 0.5 S u t (XIX) Here, the minimum tensile strength is S ′ e . Write the expression for surface factor. k a = a S u t b (XX) Here, the constant are a and b . Write the expression for effective diameter. d e = 0.37 d (XXI) Write the expression for size factor. k b = 0.879 d − 0.107 (XXII) Write the endurance limit at the critical location. S e = k a ⋅ k b ⋅ S ′ e (XXIII) Write the modified Goodman expression. 1 n f = σ ′ a S e + σ ′ m S u t (XXIV) Here, the factor of safety is n f . Conclusion: Substitute 100 lbf for F z , 8 in for l BC , ( 75 lbf ) for F x , and 5 in for l in the Equation (II). M y = − ( 100 lbf ) ⋅ ( 8 in ) − ( 75 lbf ) ⋅ ( 5 in ) = ( − 800 − 375 ) lbf ⋅ in = − 1175 lbf ⋅ in Substitute ( − 200 lbf ) for F y , and 8 in for l BC in the Equation (I)

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ISBN: 9781260531374

Author: BUDYNAS

Publisher: MCG

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Axial Load

When a bar of the uniform cross-section is acted by a load, such that the load acts along the longitudinal axis of the bar, such kinds of loadings are known as axial loadings. In general terms, a load is an external force acting on a component. An axial …

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Solution Summary: The author calculates the minimum factor of safety for fatigue based on infinite life. The force F_y produces a moment in negative z-direction.

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