Mechanics of materials/Problem set 3

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Problem 3.1 (Pb-10.1 in sec.10.)[edit]

Problem Statement[edit]

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 on t and dy? For each of the above two triangles, deduce the normal and shear stresses for the following angles:

(a) Find the normal and shear stresses on the inclined facet in the triangles pictured below.
(b) Are the stresses depending on t and dy?
(c) Calculate the normal and shear stresses for angles and

Problem 10.1

Given[edit]


Solution[edit]

Part (a)[edit]

Step 1[edit]

Draw free body diagrams for top triangle.


Free Body Diagram



Step 2[edit]

Using equations of equilibrium, we can obtain expressions for the shear force and the normal force on the inclined facet.

(3.1-1)

(3.1-2)

Equation 3.1-1 can be rearranged as so,

(3.1-3)

Which can be substituted back into equation (3.1-2) and solved for

(3.1-4)

And substituted back into equation (3.1-3)

(3.1-5)

Step 3[edit]

Now, using these forces we can solve for the normal stress, , and the shear stress, , on the inclined facet.
The normal and shear stress can be represented as

(3.1-6)

(3.1-7)

Where


Substituting and back into (3.1-6),

(3.1-8)

Substituting and back into (3.1-7),

(3.1-9)



Performing a similar process to the lower triangle, we receive the expressions

(3.1-10)

(3.1-11)

And using the definition of stress, the shear stress and normal stress are

(3.1-12)

(3.1-13)

Part (b)[edit]


From these results, we can see that and are only dependent on and for the upper triangle and and for the lower triangle, not or

Part (c)[edit]


Upper Triangle[edit]

Using equation (3.1-8) and ,

(3.1-14)

(3.1-15)

When ,

(3.1-16)

(3.1-17)


Lower Triangle[edit]


Using equation (3.1-12) and (3.1-13) when ,

(3.1-18)

(3.1-19)

When ,

(3.1-20)

(3.1-21)

Problem 3.2 (P3.2, Beer 2012)[edit]

Problem Statement[edit]

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

3.2.png

Given[edit]

Inner radius of cylinder:
Outer radius of cylinder:

Solution[edit]

Part (a): Determining torque in a hollow cylinder:[edit]

Consider a hollow cylindrical shaft having torque T, with inner radius and the outer radius , causing a maximum shear stress .

From the torsion equation,

(3.2-1)

Calculate polar moment of inertia fro the hollow cylindrical shaft.
Substitute 0.045mm and 0.030mm for

(3.2-2)

(3.2-3)

Calculate the torque for the hollow cylindrical shaft

(3.2-4)

Substitute in values.

(3.2-4)

Therefore, the torque is

 

(3.2-5)

Part (b): Determining the maximum shearing stress in a solid cylinder:[edit]

Consider a solid cylindrical shaft having torque T, with radius c, and polar moment of inertia J.

(3.2-6)

Calculate polar moment of inertia for the solid cylindrical shaft.

(3.2-7)

(3.2-8)

(3.2-9)

Insert all values into equations 3.2-6

(3.2-10)

Therefore, maximum shearing stress required is

 

(3.2-11)

Problem 3.3 ( P3.4, Beer 2012)[edit]

Problem Statement[edit]

Knowing that the internal diameter of the hollow shaft shown is , determine the maximum shearing stress
caused by a torque of magnitude

Torque acting on shaft

Given[edit]

Internal Diameter

(3.4-1)

External Diameter

(3.4-2)

Torque

(3.4-3)

Inner Radius

(3.4-4)

Outer Radius

(3.4-5)

Solution[edit]

Step One:[edit]

The Torsional formula allows the ability ti find the maximum shearing stress:

(3.4-6)

In this case can be represented as and to find the polar moment of inertia:

(3.4-7)

Substituting values into the equation:

(3.4-8)

 

(3.4-9)

Problem 3.4 ( P3.7, Beer 2012)[edit]

Problem Statement[edit]

A solid spindle AB made of steel has a diameter of 1.5 in. and an allowable shear stress of 12 ksi. The sleeve CD around it is made of brass and has an allowable shear stress of 7 ksi.

What is the largest torque that can be applied at point A?

Figure for 3.4

Solution[edit]

From the torsion equation,

(3.4-1)

Rearranging Equation 3.4-1 to solve for the the torque, , gives,

(3.4-2)

Now can be calculated with the given parameters in the problem statement.

Step Two: Calculating and [edit]

The allowable shearing stress of solid spindle AB is

The diameter of the solid spindle AB is

Free Body Diagram of Solid Spindle AB

FBD3.7.png FBDC.png


Radius c is half the diameter



The polar moment of AB, can then be determined with this newly calculated radius

(3.4-4)

The maximum sheer stress is equal to the torque at the radius in this case over the polar moment of inertia[1]



Solving for




The allowable shearing stress of the sleeve CD is is

The diameter of the sleeve CD is

The thickness of the sleeve CD is

Free Body Diagram of Sleeve CD

Fbd3.7 2.png

Radius is equal to half of diameter of sleeve



Radius is equal to the radius of the sleeve minus the thickness of the sleeve



The polar moment of inertia is




The maximum sheer stress is equal to the torque at the radius in this case over the polar moment of inertia



Solving for




Allowable value of torque T is the smaller one of the two

Step Three: Substitute given and calculated values and solve for [edit]

Substituting , , and into Equation 3-4.1 gives,

 

Problem 3.5 ( P3.9, Beer 2012)[edit]

Problem Statement[edit]

Figure for 3.5

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

Given[edit]

Diameter of the shaft AB,
Diameter of the shaft BC,
Acting torque at A,
Acting torque at B,

Solution[edit]

Part (a): Determine maximum shearing stress in shaft AB:[edit]

To solve the problem, the radius for shaft AB and BC must be found. To calculate, divide the given diameter's by 2.

(3.9-1)

(3.9-2)

For shaft AB acting torque is
Therefore, the maximum sheer stress for shaft AB can be given by

(3.9-3)

Substituting the cross sectional area for J

(3.9-4)

Simplifying the equation, we get

(3.9-5)

Now, insert the values

(3.9-6)

The torque in shaft AB is,

 

(3.9-7)

Part (b): Determine maximum shearing stress in shaft BC[edit]

For shaft BC acting torque

(3.9-8)

(3.9-9)

(3.9-9)

Therefore, the maximum sheer stress for shaft BC can be given by

(3.9-10)

The equations then simplifies similarly to equation 3.9-5

(3.9-11)

Now, insert the values

(3.9-11)

Torque in shaft BC is,

 

(3.9-7)

Problem 3.6 (P3.17, Beer2012)[edit]

Problem Statement[edit]

Aluminum and brass rods secured to a wall.


There is a 1250 N*m torque applied at point A. The maximum allowable stress in the brass rod, AB, is 50 MPa and the maximum allowable stress in the aluminum rod, BC, is 25 MPa. Determine the minimum diameter of (a) rod AB and (b) rod BC.

Given[edit]


Solution[edit]

Step One:[edit]

Using the elastic torsion formula, which relates the shearing stress to the torque and properties of the rod, solve for , the radius of the rod.

(3.6-1)

Where for a cylindrical rod can be expressed as

(3.6-2)

Replacing equation (3.6-2) with from equation (3.6-1), we recieve

(3.6-3)

(3.6-4)

Solving for gives us

(3.6-5)

(3.6-6)

Since the diameter is twice the bar's radius,

(3.6-7)

This can now be used to solve for the diameter of each bar separately

Step Two:[edit]

Starting with the brass rod, AB, and using equation (3.6-6)

(3.6-8)

(3.6-9)

 

(3.6-10)

Now for the aluminum rod,

(3.6-11)

(3.6-12)

 

(3.6-13)

Problem 3.7 (P3.8, Beer2012)[edit]

Problem Statement[edit]

Figure for 3.7

Given[edit]

The solid steel spindle AB has an allowable shearing stress of 12 ksi and the brass sleeve CD has 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 ds of spindle AB.

Solution[edit]

Free Body Diagram of Sleeve CD. Credit: diaz.

Step One:[edit]

The torque exerted on the shaft is calculated by

(3.7-1)

where is the maximum shearing stress allowable, is the polar moment of inertia of the cross section of a cylinder with respect to its center, and is the radius of the shaft.

To determine the largest torque that can be applied at A without exceeding the allowable shearing stress of sleeve CD implies that the value for in Eq. (3.7-1) will be the 7 ksi corresponding to the maximum shearing stress allowable for sleeve CD. One can see that since is what sets the limit for the largest torque that can be applied at A which is equivalent to the largest torque exerted by the shaft on the sleeve, we then calculate the variables from Eq. (3.7-1) with respect to the sleeve CD.

Step Two:[edit]

First, calculate the outer diameter of the sleeve CD, which is identified as and given by,


(3.7-2)



Then take to be the inner diameter for CD, which is calculated by


(3.7-3)


with t, being the thickness of the sleeve.


The subsequent step is to calculate, the polar moment of inertia of sleeve CD, which for a hollow circular shaft with an inner and outer radius is calculated by

(3.7-4)


Substituting these values into Eq (3.3-1) yields the largest torque that can be applied at A

(3.7-1)


 

Step Three:[edit]

To calculate the diameter of the solid steel spindle AB simple rearrangements of Eqs. (3.7-1) and (3.7-2) will yield a solution.

(3.7-1)

Substituting the values for = 19.213 kip•in, = 12 ksi, yields


Once again, rearranging Eq (3.7-2)will ultimately give the solution wanted, which is the diameter of spindle AB.

(3.7-2)

 

Problem 3.8 (P3.10, Beer2012)[edit]

Problem Statement[edit]

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.

Figure for 3.8

The torques shown are exerted on pulleys A and B. Determine the diameter of BC for which the maximum allowable shearing stress in the system will not increase.

Given[edit]

Diameter of the shaft AB,
Diameter of the shaft BC,
Acting torque at A,
Acting torque at B,

Solution[edit]

Step One: Determine maximum shearing stress in shaft AB:[edit]

To solve the problem, the radius for shaft AB and BC must be found. To calculate, divide the given diameter's by 2.

(3.8-1)

For shaft AB acting torque is
Therefore, the maximum sheer stress for shaft AB can be given by

(3.8-2)

Substituting the cross sectional area for J

(3.8-3)

Simplifying the equation, we get

(3.8-4)

Now, insert the values

(3.8-5)

The torque in shaft AB is,

(3.8-6)

Step Two: Determine maximum shearing stress in shaft BC[edit]

For shaft BC acting torque

(3.8-7)

(3.8-)

(3.8-9)

Therefore, the maximum sheer stress for shaft BC can be given by

(3.8-10)

The equations then simplifies similarly to equation 3.9-5

(3.8-11)

Now, insert the values

(3.8-11)

(3.8-12)

Because the shear stress is highest in part AB (56.68MPa is greater than 36.62 MPa), this will be the limiting value used to calculate the minimum allowable diameter for member BC in Step Three.

Step Three: Solve for the minimum allowable diameter of BC[edit]

The radius of CB can be determined by manipulating the maximum torque equation as shown,

(3.8-13)

This mean that the minimum diameter is

 

References[edit]

Beer, F. P., Johnston, E. R., Jr., DeWolf, J. T., & Mazurek, D. F. (2012). Mechanics of materials (6th ed.). New York, NY: McGraw Hill.

  1. Jshminer46 (31 December 2015). "User:Jshminer46/Mom-s13-team4-R3, In: Wikiversity". San Francisco, California USA: Wikimedia Foundation, Inc. Retrieved 2017-10-09.