Mechanics of materials/Problem set 3

Problem 3.1 (Pb-10.1 in sec.10.)

Problem Statement

Find the normal and shear stresses ($\sigma$, $\tau$) on the inclined facet in these triangles, with thickness t, angle $\theta$, vertical edge dy, and given normal stress ${\sigma }_{max}$ and shear stress ${\tau }_{max}$. 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: $\theta =30^{\circ }$ $\theta =45^{\circ }$

(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 $\theta =30^{\circ }$ and $\theta =45^{\circ }$

Problem 10.1

Given

${\sigma }_{max}$
${\tau }_{max}$

Solution

Step 1

Draw free body diagrams for top triangle.

Free Body Diagram

Step 2

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

$\sum F_x=0=-{\tau }_{max}{A}_0+F_{S}cos(\theta )+F_{N}sin(\theta )$

(3.1-1)

$\sum F_y=0=-{\tau }_{max}{A}_{0}tan(\theta )-F_{S}sin(\theta )+F_{N}cos(\theta )$

(3.1-2)

Equation 3.1-1 can be rearranged as so,

$F_S=\frac{{\tau }_{max}{A}_0-F_{N}sin\theta }{cos\theta }$

(3.1-3)

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

$F_N=2{\tau }_{max}{A}_{0}sin\theta$

(3.1-4)

And substituted back into equation (3.1-3)

$F_S={\tau }_{max}{A}_0(2cos\theta -\frac{1}{cos\theta })$

(3.1-5)

Step 3

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

$\sigma =\frac{F_N}{A}$

(3.1-6)

$\tau =\frac{F_S}{A}$

(3.1-7)

Where $A=\frac{A_0}{cos\theta }$

Substituting $F_N$ and $A$ back into (3.1-6),

$\sigma =\frac{2{\tau }_{max}{A}_{0}sin\theta }{A_0/cos\theta }={\tau }_{max}sin(2\theta )$

(3.1-8)

Substituting $F_S$ and $A$ back into (3.1-7),

$\tau ={\tau }_{max}{A}_0(\frac{2cos\theta }{A_0/cos\theta }-\frac{1/cos\theta }{A_0/cos\theta })={\tau }_{max}cos(2\theta )$

(3.1-9)

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

$F_S={\sigma }_{max}{A}_{0}cos\theta$

(3.1-10)

$F_N=-{\sigma }_{max}{A}_{0}cos\theta$

(3.1-11)

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

$\tau =\frac{F_S}{A}={\sigma }_{max}sin\theta cos\theta$

(3.1-12)

$\sigma =\frac{F_N}{A}=-{\sigma }_{max}co{s}^2\theta$

(3.1-13)

Part (b)

From these results, we can see that $\sigma$ and $\tau$ are only dependent on ${\tau }_{max}$ and $\theta$ for the upper triangle and ${\sigma }_{max}$ and $\theta$ for the lower triangle, not $dy$ or $t$

Part (c)

Upper Triangle

Using equation (3.1-8) and $\theta =30^{\circ }$,

$\sigma ={\tau }_{max}sin(60)=.866{\tau }_{max}$

(3.1-14)

$\tau ={\tau }_{max}cos(60)=.5{\tau }_{max}$

(3.1-15)

When $\theta =45^{\circ }$,

$\sigma ={\tau }_{max}sin(90)={\tau }_{max}$

(3.1-16)

$\tau ={\tau }_{max}cos(90)=0$

(3.1-17)

Lower Triangle

Using equation (3.1-12) and (3.1-13) when $\theta =30^{\circ }$,

$\sigma ={\sigma }_{max}co{s}^2(30)=.75{\sigma }_{max}$

(3.1-18)

$\tau ={\sigma }_{max}sin(30)cos(30)=.433{\sigma }_{max}$

(3.1-19)

When $\theta =45^{\circ }$,

$\sigma ={\sigma }_{max}co{s}^2(45)=.5{\sigma }_{max}$

(3.1-20)

$\tau ={\sigma }_{max}sin(45)cos(45)=.5{\sigma }_{max}$

(3.1-21)

Problem 3.2 (P3.2, Beer 2012)

Problem Statement

(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.

File:3.2.png

Given

Inner radius of cylinder: $c_1=30mm=0.030m$
Outer radius of cylinder: $c_2=45mm=0.045m$

Solution

Part (a): Determining torque in a hollow cylinder:

Consider a hollow cylindrical shaft having torque T, with inner radius $c_1$ and the outer radius $c_2$, causing a maximum shear stress $({\tau }_{max})$.

From the torsion equation,

${\tau }_{max}=\frac{T{c}_2}{J}$

(3.2-1)

Calculate polar moment of inertia fro the hollow cylindrical shaft.
Substitute 0.045mm $c_2$ and 0.030mm for $c_1$

$J=\frac{1}{2}\pi (0.045^4-0.030^4)$

(3.2-2)

$J=5.168*10^{-6}{m}^4$

(3.2-3)

Calculate the torque for the hollow cylindrical shaft

$T=\frac{{\tau }_{max}J}{c_2}$

(3.2-4)

Substitute in values.

$T=\frac{(45*10^6)(5.168*10^{-}6)}{0.045}$

(3.2-4)

Therefore, the torque is

$T=5.168kN$

(3.2-5)

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

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

${\tau }_{max}=\frac{Tc}{J}$

(3.2-6)

Calculate polar moment of inertia for the solid cylindrical shaft.

$J=\frac{1}{2}\pi (c{)}^4$

(3.2-7)

$J=\frac{1}{2}\pi (0.045{)}^4$

(3.2-8)

$J=6.4412*10^{-}6m^4$

(3.2-9)

Insert all values into equations 3.2-6

${\tau }_{max}=\frac{(5.168*10^3)(0.045)}{6.4412*10^{-}6}$

(3.2-10)

Therefore, maximum shearing stress required is

${\tau }_{max}=36.1050MPa$

(3.2-11)

Problem 3.3 ( P3.4, Beer 2012)

Problem Statement

Knowing that the internal diameter of the hollow shaft shown is $d=0.9in$, determine the maximum shearing stress
caused by a torque of magnitude $T=9kip*in$

Torque acting on shaft

Given

Internal Diameter

$D_1=.9in$

(3.4-1)

External Diameter

$D_2=1.6in$

(3.4-2)

Torque

$T=9kip*in$

(3.4-3)

Inner Radius

$C_1=0.45$

(3.4-4)

Outer Radius

$C_2=0.8in$

(3.4-5)

Solution

Step One:

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

$(\tau )=\frac{TC}{J}$

(3.4-6)

In this case $J$ can be represented as $C_1$ and $C_2$to find the polar moment of inertia:

$(\tau )=\frac{TC}{\frac{\pi }{2}[c_2^4-C_1^4]}$

(3.4-7)

Substituting values into the equation:

$(\tau )=\frac{9*0.8}{\frac{\pi }{2}[0.8^4-.45^4]}$

(3.4-8)

$(\tau )=12.44ksi$

(3.4-9)

Problem 3.4 ( P3.7, Beer 2012)

Problem Statement

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?

File:P3.7.svg

Figure for 3.4

Solution

From the torsion equation,

${\tau }_{AB}=\frac{T_{AB}c}{J}$

(3.4-1)

Rearranging Equation 3.4-1 to solve for the the torque, $T_{AB}$, gives,

$T_{AB}=\frac{{\tau }_{AB}J}{c}$

(3.4-2)

Now $T_{AB}$ can be calculated with the given parameters in the problem statement.

Step Two: Calculating c and J

The allowable shearing stress of solid spindle AB is ${\tau }_a=12ksi$

The diameter of the solid spindle AB is $d_s=1.5in$

Free Body Diagram of Solid Spindle AB

File:FBD3.7.png File:FBDC.png

Radius c is half the diameter $d_s$

$c=\frac{1}{2}{d}_s=\frac{1}{2}1.5in=0.75in$

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

$J=\frac{1}{2}\pi c^4=\frac{1}{2}\pi (0.75in.{)}^4=0.4970in{.}^4$

(3.4-4)

The maximum sheer stress is equal to the torque at the radius in this case $c_2$ over the polar moment of inertia

${\tau }_{max}=\frac{Tc}{J}$

Solving for $T_{AB}$

$T_{AB}=\frac{J{\tau }_a}{c}=\frac{(0.49701i{n}^4)(12ksi)}{0.75in}=7.952kip*in$

The allowable shearing stress of the sleeve CD is is ${\tau }_c=7ksi$

The diameter of the sleeve CD is $d_2=3in$

The thickness of the sleeve CD is $t=\frac{1}{4}in=0.25in$

Free Body Diagram of Sleeve CD

File:Fbd3.7 2.png

Radius $c_2$ is equal to half of diameter of sleeve $d_2$

$c_2=\frac{1}{2}{d}_2=\frac{1}{2}(3in)=1.5in$

Radius $c_1$ is equal to the radius of the sleeve minus the thickness of the sleeve $t$

$c_1=c_2-t=1.5in-0.25in=1.25in$

The polar moment of inertia is

$J=\frac{1}{2}\pi (c_2^4-c_1^4)=\frac{\pi }{2}((1.5in{)}^4-(1.25in{)}^4)=4.1172i{n}^4$

The maximum sheer stress is equal to the torque at the radius in this case $c_2$ over the polar moment of inertia

${\tau }_{max}=\frac{T{c}_2}{J}$

Solving for $T_{CD}$

$T_{CD}=\frac{J{\tau }_c}{c_2}=\frac{(4.1172i{n}^4)(7ksi)}{1.5in}=19.213kip*in$

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

Step Three: Substitute given and calculated values and solve for T_AB

Substituting ${\tau }_{AB}=12ksi$, $J=0.4970in{.}^4$, and $c=0.75in.$ into Equation 3-4.1 gives,

$T_{AB}=\frac{12ksi\times 0.4970in{.}^4}{0.75in.}=7.95\times kipsin.$

Problem 3.5 ( P3.9, Beer 2012)

Problem Statement

File:P3.9.svg

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

Diameter of the shaft AB, $d_{AB}=30mm$
Diameter of the shaft BC, $d_{BC}=46mm$
Acting torque at A, $T_A=300N-m$
Acting torque at B, $T_B=400N-m$

Solution

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

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

$c_{AB}=\frac{30}{2}=15mm$

(3.9-1)

$c_{BC}=\frac{46}{2}=23mm$

(3.9-2)

For shaft AB acting torque is $T_{AB}=300N-m$
Therefore, the maximum sheer stress for shaft AB can be given by

$({\tau }_{max}{)}_{AB}=\frac{T_{AB}{c}_{AB}}{J_{AB}}$

(3.9-3)

Substituting the cross sectional area for J

$({\tau }_{max}{)}_{AB}=\frac{T_{AB}*c}{\frac{\pi }{2}{c}^4}$

(3.9-4)

Simplifying the equation, we get

$({\tau }_{max}{)}_{AB}=\frac{2T_{AB}}{\pi {c^3}_{AB}}$

(3.9-5)

Now, insert the values

$({\tau }_{max}{)}_{AB}=\frac{2*300}{\pi *15^3*10^{-9}}$

(3.9-6)

The torque in shaft AB is,

$({\tau }_{max}{)}_{AB}=56.58MPa$

(3.9-7)

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

For shaft BC acting torque

$T_{BC}=T_A+T_B$

(3.9-8)

$T_{BC}=300+400$

(3.9-9)

$T_{BC}=700N-m$

(3.9-9)

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

$({\tau }_{max}{)}_{BC}=\frac{T_{BC}{L}_{CB}}{J_{BC}}$

(3.9-10)

The equations then simplifies similarly to equation 3.9-5

$({\tau }_{max}{)}_{BC}=\frac{2T_{BC}}{\pi {c^3}_{BC}}$

(3.9-11)

Now, insert the values

$({\tau }_{max}{)}_{BC}=\frac{2*700}{\pi *23^3*10^{-9}}$

(3.9-11)

Torque in shaft BC is,

$({\tau }_{max}{)}_{BC}=36.62MPa$

(3.9-7)

Problem 3.6 (P3.17, Beer2012)

Problem Statement

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

${\tau }_{max,brass}=50MPa$
${\tau }_{max,alum}=25MPa$

Solution

Step One:

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

${\tau }_{max}=\frac{Tc}{J}$

(3.6-1)

Where $J$ for a cylindrical rod can be expressed as

$J=\frac{1}{2}\pi c^4$

(3.6-2)

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

${\tau }_{max}=\frac{Tc}{\frac{1}{2}\pi c^4}$

(3.6-3)

${\tau }_{max}=\frac{2T}{\pi c^3}$

(3.6-4)

Solving for $c$ gives us

$c^3=\frac{2T}{{\tau }_{max}\pi }$

(3.6-5)

$c=\sqrt[3]{\frac{2T}{{\tau }_{max}\pi }}$

(3.6-6)

Since the diameter is twice the bar's radius,

$d=2*c$

(3.6-7)

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

Step Two:

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

$c_{brass}=\sqrt[3]{\frac{2(1250N\cdot m)}{(50MPa)\pi }}$

(3.6-8)

$d_{brass}=2*c_{brass}$

(3.6-9)

$d_{brass}=.0503m$

(3.6-10)

Now for the aluminum rod,

$c_{alum}=\sqrt[3]{\frac{2(1250N\cdot m)}{(25MPa)\pi }}$

(3.6-11)

$d_{alum}=2*c_{alum}$

(3.6-12)

$d_{alum}=.0634m$

(3.6-13)

Problem 3.7 (P3.8, Beer2012)

Problem Statement

File:P3.7.svg

Figure for 3.7

Given

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

File:Fbd3.7 2.png

Free Body Diagram of Sleeve CD. Credit: diaz.

Step One:

The torque exerted on the shaft is calculated by

$T=\frac{{\tau }_{max}J}{c}$

(3.7-1)

where $t_{max}$ is the maximum shearing stress allowable, $J$ is the polar moment of inertia of the cross section of a cylinder with respect to its center, and $c$ 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 ${\tau }_{max}$ 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 ${\tau }_{CD,max}$ 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:

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

$c_o=\frac{d_s}{2}$

(3.7-2)

Template:Center top$=\frac{3.0in}{2}=1.5in$Template:Center bottom

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

$c_i=c_o-t$

(3.7-3)

Template:Center top$=1.05in-0.25in=1.25in$Template:Center bottom

with t, being the thickness of the sleeve.

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

$J=\frac{\pi }{2}[c_o^4-c_i^4]$

(3.7-4)

Template:Center top$=\frac{\pi }{2}[1.5^4-1.25^4]=4.1172i{n}^4$Template:Center bottom

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

$T=\frac{7ksi\times 4.1172i{n}^4}{1.5in}$

(3.7-1)

$T=19.213kip*in$ Step Three: 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. $T=\frac{{\tau }_{max}J}{c}$ (3.7-1) $c=\frac{{\tau }_{max}J}{T}$ $c^3=\frac{2T}{\pi {\tau }_{max}}$ Substituting the values for $T$= 19.213 kip•in, ${\tau }_{max}$= 12 ksi, yields Template:Center top $c=1.0064in$Template:Center bottom Once again, rearranging Eq (3.7-2)will ultimately give the solution wanted, which is the diameter of spindle AB. $c=\frac{d_s}{2}$ (3.7-2) $d_s=2c$   $d_s=(2)(1.0064)=2.01in$ Problem 3.8 (P3.10, Beer2012) Problem Statement 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. File:P3.10.svg 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 Diameter of the shaft AB, $d_{AB}=30mm$ Diameter of the shaft BC, $d_{BC}=?$ Acting torque at A, $T_A=300N-m$ Acting torque at B, $T_B=400N-m$ Solution Step One: Determine maximum shearing stress in shaft AB: To solve the problem, the radius for shaft AB and BC must be found. To calculate, divide the given diameter's by 2. $c_{AB}=\frac{30}{2}=15mm$ (3.8-1)

For shaft AB acting torque is $T_{AB}=300N-m$
Therefore, the maximum sheer stress for shaft AB can be given by

$({\tau }_{max}{)}_{AB}=\frac{T_{AB}{c}_{AB}}{J_{AB}}$

(3.8-2)

Substituting the cross sectional area for J

$({\tau }_{max}{)}_{AB}=\frac{T_{AB}*c}{\frac{\pi }{2}{c}^4}$

(3.8-3)

Simplifying the equation, we get

$({\tau }_{max}{)}_{AB}=\frac{2T_{AB}}{\pi {c^3}_{AB}}$

(3.8-4)

Now, insert the values

$({\tau }_{max}{)}_{AB}=\frac{2*300}{\pi *15^3*10^{-9}}$

(3.8-5)

The torque in shaft AB is,

$({\tau }_{max}{)}_{AB}=56.58MPa$

(3.8-6)

Step Two: Determine maximum shearing stress in shaft BC

For shaft BC acting torque

$T_{BC}=T_A+T_B$

(3.8-7)

$T_{BC}=300+400$

(3.8-)

$T_{BC}=700N-m$

(3.8-9)

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

$({\tau }_{max}{)}_{BC}=\frac{T_{BC}{L}_{CB}}{J_{BC}}$

(3.8-10)

The equations then simplifies similarly to equation 3.9-5

$({\tau }_{max}{)}_{BC}=\frac{2T_{BC}}{\pi {c^3}_{BC}}$

(3.8-11)

Now, insert the values

$({\tau }_{max}{)}_{BC}=\frac{2*700}{\pi *23^3*10^{-9}}$

(3.8-11)

$({\tau }_{max}{)}_{BC}=36.62MPa$

(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

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

$c_{BC}^4=\frac{2T_{BC}}{\pi {\tau }_{max}}=\frac{2\times 700N\times m}{\pi \times 56.68\times 10^{6}Pa}=0.0198m$

(3.8-13)

This mean that the minimum diameter is

$d_{bc}=0.0396m$

References

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