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The motor on the left provides a torque of 5500 N•m to the gear shaft shown. The motor runs machines connected to gears B, C and D requiring torques of 3000 N•m, 1500 N•m and 1000 N•m respectively. A, B, C and D are spaced 2000 mm apart and the diameter of the shaft is 75 mm. The shear modulus of the shaft material is 80x109 N/m2. Determine: Minimum diameter required (in millimeters) if the shearing stress in the shaft is limited to 100x106 N/m2 Rotation of D with respect to A Note: Draw the Free Body Diagram, Compute for all of the necessary elements, Include the units/dimensions

The motor on the left provides a torque of 5500 N•m to the gear shaft shown. The motor runs machines connected to gears B, C and D requiring torques of 3000 N•m, 1500 N•m and 1000 N•m respectively. A, B, C and D are spaced 2000 mm apart and the diameter of the shaft is 75 mm. The shear modulus of the shaft material is 80x109 N/m2. Determine: Minimum diameter required (in millimeters) if the shearing stress in the shaft is limited to 100x106 N/m2 Rotation of D with respect to A Note: Draw the Free Body Diagram, Compute for all of the necessary elements, Include the units/dimensions

Automotive Technology: A Systems Approach (MindTap Course List)
Automotive Technology: A Systems Approach (MindTap Course List)
6th Edition
ISBN:9781133612315
Author:Jack Erjavec, Rob Thompson
Publisher:Jack Erjavec, Rob Thompson
Chapter11: Lower End Theory And Service
Section: Chapter Questions
Problem 11RQ: What is the purpose of a thrust main bearing?
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The motor on the left provides a torque of 5500 N•m to the gear shaft shown. The motor runs machines connected to gears B, C and D requiring torques of 3000 N•m, 1500 N•m and 1000 N•m respectively. A, B, C and D are spaced 2000 mm apart and the diameter of the shaft is 75 mm. The shear modulus of the shaft material is 80x109 N/m2.

Determine:

  • Minimum diameter required (in millimeters) if the shearing stress in the shaft is limited to 100x106 N/m2
  • Rotation of D with respect to A

Note: Draw the Free Body Diagram, Compute for all of the necessary elements, Include the units/dimensions, use the proper formula and round-off all the answers and final answers to 3 decimal places.

-Bearing
Transcribed Image Text:-Bearing
Mechanics of Deformable Bodies SHEAR AND МОМENT EQUATIONS AND DIAGRAMS TORSION (Torsional Shearing Stress, t) SHEAR AND MOMENT EQUATIONS (Procedure) 1. Compute the support reactions of the beam. 2. Divide the beam into segments so that the loading within each segment is continuous. 3. ON EACH BEAM SEGMENT, introduce an imaginary cutting plane within the segment, located at a distance “x" from the left end of the beam, that cuts the beam into parts. Draw an FBD for the part of the beam lying to the left. At the cut section, show V (Shear) and M (Bending moment) acting in their positive directions. 5. Determine the equations for V and M from the equilibrium equations obtainable from the FBDS. These expressions, which are usually functions of x, are the shear force and bending moment equation for the segment. 6. Do the preceding steps for all segments until you obtain the shear and moment equation for all segments. 4. TP Tr Tmax= T - torsional shearing stress Tmax - maximum shear stress (N/m²) T- twisting moment/ torque (Nm, lb-in, lb-ft) p - distance from the center of the shaft J- polar moment of inertia of the section (mm², in*) r - radius of the shaft (m, mm, in,ft) SHEAR AND MOMENT DIAGRAM For solid cylindrical shaft: Circular section If the shear-force diagram is positive and looks like this ... J = D4 32 Vp Ď J= = = 0.098 Linear Linear 16T 32 Constant (first order) (first order) Tmar = (zero order) For hollow cylindrical shaft: Tube section ... then the bending-moment diagram looks like this: J = (Dª – dª) 32 J= 16TD Quadratic (second order) Mp Linear Tmaz = (first order) 7(Dª – dª) Quadratie (second order) M. M. D- outside diameter (mm) d - inside diameter (mm) Slope becomes more positive Slope becomes less positive Constant slope (an upward ramp) (a hill that (an arch) gets steeper) ANGLE OF TWIST Constant (zero order) TL in radians JG Linear (first order) Linear VE (first order) ... then the bending-moment diagram looks like this: M ME T- torque (Nmm) L- length of shaft (mm) G- shear modulus (MPa) J- polar moment of inertia (mm4) Quadratic (second order) Linear (first order) Quadratic (second order) M. M Slope becomes more negative Slope becomes less negative Constant slope (a downward ramp) (a waterfall) (a valley) E G = 2(1+v) Equation Load Diagram Shear-Force Diagram V Bending-Moment Diagram M Rule 1: Concentrated loads create discontinuities in the shcar-force diagram. [Equation (7.5) Slepe -V Positive jump Slepe = M. E - modulus of elasticity v - poisson's ratio AV- P. Rule 2: The change in shear force is equal to the area under the distributed-load curve. [Equation (7.3)] POWER TRANSMITTED BY THE SHAFT Av-v, -V, - dc My V. - V, - ada A shaft rotating with a constant angular velocity o (in radians per second) is being acted by a twisting moment Т. Rule 3: The slope of the V diagram is equal to the intensity of the distrihuted Iw. [Equation (7.1)) Slepe - P= To = 2nfT Rule 4: The change in bending moment is equal to the area under the shear-force diagram. [Equation (7.4)) T- torque (N;m, ft:lb) 0 - angular velocity of the shaft (rad/s) f- number of revolutions per second/ 1 cycle P- power (watts, N'm) r - radius M, - M, = 5."V dt Rule 5: The slope of the M diagram is oqual to the intensity of the shear force V. [Equation (7.2)] Slepe - V, Slope - Siope V f: 1 Hz (Hertz) = 1 cycle = 2nrad = W= 2nrf Rule 6: Concentrated moments create discontinuities in the bending-moment disgram. [Equation (7.6)] No elliet shear force V hending moent AM --M.
Transcribed Image Text:Mechanics of Deformable Bodies SHEAR AND МОМENT EQUATIONS AND DIAGRAMS TORSION (Torsional Shearing Stress, t) SHEAR AND MOMENT EQUATIONS (Procedure) 1. Compute the support reactions of the beam. 2. Divide the beam into segments so that the loading within each segment is continuous. 3. ON EACH BEAM SEGMENT, introduce an imaginary cutting plane within the segment, located at a distance “x" from the left end of the beam, that cuts the beam into parts. Draw an FBD for the part of the beam lying to the left. At the cut section, show V (Shear) and M (Bending moment) acting in their positive directions. 5. Determine the equations for V and M from the equilibrium equations obtainable from the FBDS. These expressions, which are usually functions of x, are the shear force and bending moment equation for the segment. 6. Do the preceding steps for all segments until you obtain the shear and moment equation for all segments. 4. TP Tr Tmax= T - torsional shearing stress Tmax - maximum shear stress (N/m²) T- twisting moment/ torque (Nm, lb-in, lb-ft) p - distance from the center of the shaft J- polar moment of inertia of the section (mm², in*) r - radius of the shaft (m, mm, in,ft) SHEAR AND MOMENT DIAGRAM For solid cylindrical shaft: Circular section If the shear-force diagram is positive and looks like this ... J = D4 32 Vp Ď J= = = 0.098 Linear Linear 16T 32 Constant (first order) (first order) Tmar = (zero order) For hollow cylindrical shaft: Tube section ... then the bending-moment diagram looks like this: J = (Dª – dª) 32 J= 16TD Quadratic (second order) Mp Linear Tmaz = (first order) 7(Dª – dª) Quadratie (second order) M. M. D- outside diameter (mm) d - inside diameter (mm) Slope becomes more positive Slope becomes less positive Constant slope (an upward ramp) (a hill that (an arch) gets steeper) ANGLE OF TWIST Constant (zero order) TL in radians JG Linear (first order) Linear VE (first order) ... then the bending-moment diagram looks like this: M ME T- torque (Nmm) L- length of shaft (mm) G- shear modulus (MPa) J- polar moment of inertia (mm4) Quadratic (second order) Linear (first order) Quadratic (second order) M. M Slope becomes more negative Slope becomes less negative Constant slope (a downward ramp) (a waterfall) (a valley) E G = 2(1+v) Equation Load Diagram Shear-Force Diagram V Bending-Moment Diagram M Rule 1: Concentrated loads create discontinuities in the shcar-force diagram. [Equation (7.5) Slepe -V Positive jump Slepe = M. E - modulus of elasticity v - poisson's ratio AV- P. Rule 2: The change in shear force is equal to the area under the distributed-load curve. [Equation (7.3)] POWER TRANSMITTED BY THE SHAFT Av-v, -V, - dc My V. - V, - ada A shaft rotating with a constant angular velocity o (in radians per second) is being acted by a twisting moment Т. Rule 3: The slope of the V diagram is equal to the intensity of the distrihuted Iw. [Equation (7.1)) Slepe - P= To = 2nfT Rule 4: The change in bending moment is equal to the area under the shear-force diagram. [Equation (7.4)) T- torque (N;m, ft:lb) 0 - angular velocity of the shaft (rad/s) f- number of revolutions per second/ 1 cycle P- power (watts, N'm) r - radius M, - M, = 5."V dt Rule 5: The slope of the M diagram is oqual to the intensity of the shear force V. [Equation (7.2)] Slepe - V, Slope - Siope V f: 1 Hz (Hertz) = 1 cycle = 2nrad = W= 2nrf Rule 6: Concentrated moments create discontinuities in the bending-moment disgram. [Equation (7.6)] No elliet shear force V hending moent AM --M.
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