University of Florida/Egm3520/s13.team1.r3

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Team 1 : Report 3


Problem R3.1[edit]

PB-10.1, Sec. 10

Find the normal and shear stresses on the inclined facet in these triangles, with thickness t, angle , vertical edge dy, and given normal stress and shear stress .
Are the stresses depending t and dy ?
Equilibrium of triangle, normal and shear stresses on inclined plane in bar under tension and shaft under torsion.

Contents taken from the notes of Dr. Loc Vu-Quoc

R3.1 Solution[edit]


Given(s):

Triangle 1
Find relationship between and normal stress


Find relationship between and shearing stress

Triangle 2

Problem R3.2[edit]

Problem Number 3.2, p.154

(a) Determine the torque T that causes a maximum shearing stress of 45 MPa in the hollow cylindrical steel shaft shown.
(b) Determine the maximum shearing stress caused by the same torque T in solid cylindrical shaft of the same cross-sectional area.

Contents taken from Mechanics of Materials: 6th Edition textbook. Authors: F.P BEER, E.R. JOHNSTON, J.T. DEWOLF AND D.F. MAZUREK ISBN:9780077565664

R3.2 Solution[edit]

Given(s):

(outer radius)
(inner radius)
 : radius of solid cylinder
 : cross-sectional area of solid cylinder
 : cross-sectional area of hollow cylinder
  • subscripts h and s identifies which cylinders are under consideration, whether hollow or solid, respectively


(a) The relation between torque and shearing stress is

Solve the definite integral for the centroidal polar moment of inertia

The torque experienced by the hollow cylindrical steel shaft is

(b) Given that both the hollow and solid cylinders have the same cross-sectional area , we can find the radius of the solid cylinder

Solve for the centroidal polar moment of inertia for the solid cylinder

Using and solve for

Problem R3.3[edit]

Problem Number 3.4, p.154

Knowing that the internal diameter of the hollow shaft shown d = 1.2 in., determine the maximum shearing stress caused by a torque of magnitude T = 7.5 kip*in.

Contents taken from Mechanics of Materials: 6th Edition textbook. Authors: F.P BEER, E.R. JOHNSTON, J.T. DEWOLF AND D.F. MAZUREK ISBN:9780077565664

R3.3 Solution[edit]

Given(s): Internal diameter = Outer diameter =

Use equation 3.9 from the text book to find maximum torque in a hollow shaft:

Where

applied torque
as given by equation 3.11 from the textbook
(Inner radius)
(Outer radius)

Solving for yeilds

Unit conversion:

Problem R3.4[edit]

Problem Number 3.7, p.155

The solid spindle AB has a diameter ds = 1.5 in. and is made of a steel with an allowable shearing stress of 12 ksi, while sleeve CD is made of a brass with an allowable shearing stress of 7 ksi. Determine the largest torque T that can be applied at A.
Figure P3.7

Contents taken from Mechanics of Materials: 6th Edition textbook. Authors: F.P BEER, E.R. JOHNSTON, J.T. DEWOLF AND D.F. MAZUREK ISBN:9780077565664

R3.4 Solution[edit]

For solid spindle AB


For sleeve CD


The smallest is the allowable so the allowable is the torque about the solid spindle AB which is 7.952 kip-in

Problem R3.5[edit]

Problem Number 3.9, p.155

The torques shown are exerted on pulleys A and B. Knowing that both shafts are solid, determine the maximum shearing stress in (a) in shaft AB, (b) in shaft BC.

Contents taken from Mechanics of Materials: 6th Edition textbook. Authors: F.P BEER, E.R. JOHNSTON, J.T. DEWOLF AND D.F. MAZUREK ISBN:9780077565664

R3.5 Solution[edit]

We begin this problem by drawing a free body diagram of the system.

Using the right hand rule we can figure out the direction of the torque on the elements.

By analyzing disk A:

and from disk B

Now that all the torques have been calculated we must calculate the maximum sheer stress in each of the shafts.

Using equation 3.9 from the text book we see that

Where T = torque, c = the radius of the cross section, J = centroidal polar moment of inertia ( for a solid shaft)


By analyzing shaft AB we find:

By analyzing shaft BC we find:

Problem R3.6[edit]

Problem Number 3.17, p.156

The allowable stress is 50 MPa in the brass rod AB and 25 MPa in the aluminum rod BC. Knowing that a torque of magnitude T = 1250 N*m is applied at A, determine the required diameter of (a) rod AB, (b) rod BC.
Figure 3.6-3.7.png

Contents taken from Mechanics of Materials: 6th Edition textbook. Authors: F.P BEER, E.R. JOHNSTON, J.T. DEWOLF AND D.F. MAZUREK ISBN:9780077565664

R3.6 Solution[edit]

Egm3520.s13.team1.scheppegrell.jas (discusscontribs) 16:58, 22 February 2013 (UTC)

Known Values:





As there are no torsional forces being applied between points A and C, it can also be seen that



Formulas: ,



(a) Maximum Allowable Shear Stress in the Rod:

Torque Applied to the Rod:

Radius of the Rod AB:

Substitute Known Values into the Radius Equation:

Simplify and Solve for Radius:

Find Diameter of AB:



(b) Maximum Allowable Shear Stress in the Rod:

Torque Applied to the Rod:

Radius of the Rod BC:

Substitute Known Values into the Radius Equation:

Simplify and Solve for Radius:

Find Diameter of BC:

Problem R3.7[edit]

The solid spindle AB is made of a steel with an allowable shearing stress of 12 ksi, and sleeve CD is made of brass with an allowable shearing stress of 7 ksi. Determine (a) the largest torque T that can be applied at A if the allowable shearing stress is not to be exceeded in sleeve CD, (b) the corresponding required value of the diameter d, of spindle AB.


Contents taken from Mechanics of Materials: 6th Edition textbook. Authors: F.P BEER, E.R. JOHNSTON, J.T. DEWOLF AND D.F. MAZUREK ISBN:9780077565664


R3.7 Solution[edit]

(a) For the sleeve CD



(b) At the solid spindle AB



A

for the sleeve CD

B

at the solid spindle AB


Failed to parse (syntax error): {\displaystyle \displaystyle => \frac{2T}{(\pi \tau)^{1/3}} = (\frac{2*19.213 kip-in}{12 ksi *\pi)^1/3 = 1.0064 in }



Failed to parse (unknown function "\math"): {\displaystyle \displaystyle => 2*1.0064 in = 2.01 in <\math> <math> \displaystyle ds = 2.01 in \sigma }

Problem R3.8[edit]

Problem Number 3.10, p.155

Figure P3.8
In order to reduce the total mass of the assembly of Prob 3.9, a new design is being considered in which the diameter of shaft BC will be smaller. Determine the smallest diameter of shaft BC for which the maximum value of the shearing stress in the assembly will not increase.

Contents taken from Mechanics of Materials: 6th Edition textbook. Authors: F.P BEER, E.R. JOHNSTON, J.T. DEWOLF AND D.F. MAZUREK ISBN:9780077565664

R3.8 Solution[edit]

We begin this problem by drawing a free body diagram of the system.

Using the right hand rule we can figure out the direction of the torque on the elements.

By analyzing disk A:

and from disk B

Now that all the torques have been calculated we must calculate the maximum sheer stress in each of the shafts.

Using equation 3.9 from the text book we see that

Where T = torque, c = the radius of the cross section, J = centroidal polar moment of inertia ( for a solid shaft)


By analyzing shaft AB we find:

By analyzing shaft BC we find:

By comparing the maximum sheer stress in both AB and BC we see that the torque is greater in AB

No we create the variable d' and solve for it using the known values.

Let 2cunknown = dunknown

Therefore the smallest possible diameter for shaft BC is 39.6mm