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ME 422 Problem Set 3 Name A stationary circular section shaft has bearings (simple supports) at x=0.4 (section B ) and x=1.4 (section G). The bearing at G provides axial support for the shaft. The shaft carries a variety of transverse forces, axial forces, and axial torques. Vertical reactions at the supports are calculated to be R_(B)=350N and RG_(c)= 150N downward. (a) Sketch internal load diagrams (transverse shear force V(x), bending moment M(x), axial torque T(x), and axial force F(x) ) for the bar. (b) Five potential critical points P1-P5 are indicated on the shaft. Using your internal load diagrams and the stress elements provided, indicate the stress components (bending normal stress, sigma _(M), axial normal stress, sigma _(A), transverse shear stress, au _(V), torsional shear stress, au _(T) ) acting at the points. Show the proper directions for the stress components, but do not calculate numerical values. Do not show a stress component if its value is zero. P1 and P2 are on the top or bottom fibers of the bar. P3 and P4, and P5 are at the neutral axis. Point P5 is on the "hidden" side of the bar. A B c ME 422 Problem Set 3 Name A stationary circular section shaft has bearings (simple supports) at x=0.4 (section B) and x=1.4 (section G). The bearing at G provides axial support for the shaft. The shaft carries a variety of transverse forces, axial forces, and axial torques. Vertical reactions at the supports are calculated to be R = 350 N and Ro = 150 N downward. (a) Sketch internal load diagrams (transverse shear force V(x), bending moment M(x), axial torque T(x),and axial force Fxfor the bar. (b) Five potential critical points P1-P5 are indicated on the shaft. Using your internal load diagrams and the stress elements provided, indicate the stress components (bending normal stress, om, axial normal stress, transverse shear stress, tv, torsional shear stress, r) acting at the points. Show the proper directions for the stress components, but do not calculate numerical values. Do not show a stress component if its value is zero. P1 and P2 are on the top or bottom fibers of the bar.P3 and P4, and P5 are at the neutral axis. Point P5 is on the"hidden"side of the bar M=90N-m |300N T=50N-m To=80N-m 100N Tp= P2 Show P3 150N Sample a 200N P1 250N RB=350N Rg=150N 0.4 m 0.2m 0.2m 02 02 02m .. P5 (EC) Spring 2024

          ME 422
Problem Set 3
Name
A stationary circular section shaft has bearings (simple supports) at x=0.4 (section B ) and x=1.4 (section G). The bearing at G provides axial support for the shaft. The shaft carries a variety of transverse forces, axial forces, and axial torques. Vertical reactions at the supports are calculated to be R_(B)=350N and RG_(c)= 150N downward. (a) Sketch internal load diagrams (transverse shear force V(x), bending moment M(x), axial torque T(x), and axial force F(x) ) for the bar.
(b) Five potential critical points P1-P5 are indicated on the shaft. Using your internal load diagrams and the stress elements provided, indicate the stress components (bending normal stress, sigma _(M), axial normal stress, sigma _(A), transverse shear stress, 	au _(V), torsional shear stress, 	au _(T) ) acting at the points. Show the proper directions for the stress components, but do not calculate numerical values. Do not show a stress component if its value is zero. P1 and P2 are on the top or bottom fibers of the bar. P3 and P4, and P5 are at the neutral axis. Point P5 is on the "hidden" side of the bar.
A
B
c
ME 422 Problem Set 3 Name A stationary circular section shaft has bearings (simple supports) at x=0.4 (section B) and x=1.4 (section G). The bearing at G provides axial support for the shaft. The shaft carries a variety of transverse forces, axial forces, and axial torques. Vertical reactions at the supports are calculated to be R = 350 N and Ro = 150 N downward. (a) Sketch internal load diagrams (transverse shear force V(x), bending moment M(x), axial torque T(x),and axial force Fxfor the bar. (b) Five potential critical points P1-P5 are indicated on the shaft. Using your internal load diagrams and the stress elements provided, indicate the stress components (bending normal stress, om, axial normal stress,  transverse shear stress, tv, torsional shear stress, r) acting at the points. Show the proper directions for the stress components, but do not calculate numerical values. Do not show a stress component if its value is zero. P1 and P2 are on the top or bottom fibers of the bar.P3 and P4, and P5 are at the neutral axis. Point P5 is on the"hidden"side of the bar M=90N-m |300N T=50N-m To=80N-m 100N Tp= P2 Show P3 150N Sample a 200N P1 250N RB=350N Rg=150N 0.4 m 0.2m 0.2m 02 02 02m ..
P5 (EC)
Spring 2024
        
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me 422 problem set 3 name a stationary circular section shaft has bearings simple supports at x04 section b and x14 section g the bearing at g provides axial support for the shaft the shaft  34038

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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ME 422 Problem Set 3 Name A stationary circular section shaft has bearings (simple supports) at x=0.4 (section B ) and x=1.4 (section G). The bearing at G provides axial support for the shaft. The shaft carries a variety of transverse forces, axial forces, and axial torques. Vertical reactions at the supports are calculated to be R_(B)=350N and RG_(c)= 150N downward. (a) Sketch internal load diagrams (transverse shear force V(x), bending moment M(x), axial torque T(x), and axial force F(x) ) for the bar. (b) Five potential critical points P1-P5 are indicated on the shaft. Using your internal load diagrams and the stress elements provided, indicate the stress components (bending normal stress, sigma _(M), axial normal stress, sigma _(A), transverse shear stress, au _(V), torsional shear stress, au _(T) ) acting at the points. Show the proper directions for the stress components, but do not calculate numerical values. Do not show a stress component if its value is zero. P1 and P2 are on the top or bottom fibers of the bar. P3 and P4, and P5 are at the neutral axis. Point P5 is on the "hidden" side of the bar. A B c ME 422 Problem Set 3 Name A stationary circular section shaft has bearings (simple supports) at x=0.4 (section B) and x=1.4 (section G). The bearing at G provides axial support for the shaft. The shaft carries a variety of transverse forces, axial forces, and axial torques. Vertical reactions at the supports are calculated to be R = 350 N and Ro = 150 N downward. (a) Sketch internal load diagrams (transverse shear force V(x), bending moment M(x), axial torque T(x),and axial force Fxfor the bar. (b) Five potential critical points P1-P5 are indicated on the shaft. Using your internal load diagrams and the stress elements provided, indicate the stress components (bending normal stress, om, axial normal stress, transverse shear stress, tv, torsional shear stress, r) acting at the points. Show the proper directions for the stress components, but do not calculate numerical values. Do not show a stress component if its value is zero. P1 and P2 are on the top or bottom fibers of the bar.P3 and P4, and P5 are at the neutral axis. Point P5 is on the"hidden"side of the bar M=90N-m |300N T=50N-m To=80N-m 100N Tp= P2 Show P3 150N Sample a 200N P1 250N RB=350N Rg=150N 0.4 m 0.2m 0.2m 02 02 02m .. P5 (EC) Spring 2024
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Transcript

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00:01 Hi students, here in the question we first calculate the equivalent stiffness using the formula k equivalent equals to 2 u upon del square.
00:12 Putting up the value we have k equivalent equals to having the formula for u and del square.
00:24 Here we have to calculate firstly bending moment m, the bending moment m equals to m e g into x.
00:34 Putting up the value we have m equals to 0 .014 into 9 .8 into 0 .17 that give us m equals to 0 .024 kg meter square per second square.
00:50 Next we calculate the strain energy u using the formula m square upon 2ei.
00:57 Putting up the value we have 0 .024 square upon 2 into 200 into 10 to the power 9 into i that is 1 .005 into 10 to the power minus 7 that give us u equals to 2 .39 into 10 to the power minus 9 joule.
01:18 Next calculating the deflection del we have del equals to m theta e g x square upon 2ei...
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