University of Florida/Egm3520/s13.team1.r3.1

Team 1 : Report 3

Problem R3.1

PB-10.1, Sec. 10

Find the normal and shear stresses $\left(\sigma ,\tau \right)$ 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 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

Problem R3.2

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

Given(s):

$\displaystyle \tau _{max_{h}}=45MPa$

$\displaystyle c_{1}$ (outer radius) $\displaystyle =45mm$

$c_{2}$ (inner radius) $\displaystyle =30mm$

$\displaystyle c_{3}$ : radius of solid cylinder

$\displaystyle A_{s}$ : cross-sectional area of solid cylinder

$\displaystyle A_{h}$ : 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

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

Solve the definite integral for the centroidal polar moment of inertia

${\begin{aligned}J=\int c^{2}dA,&&A=\pi c^{2},&&dA=2\pi c\end{aligned}}$ $dc$

$J_{h}=2\pi \int _{c_{2}}^{c_{1}}c^{3}dc={\frac {\pi }{2}}(c_{1}^{4}-c_{2}^{4})=5.1689*10^{-6}m^{4}$

The torque experienced by the hollow cylindrical steel shaft is

$\displaystyle T={\frac {J_{h}\tau _{max_{h}}}{c_{1}}}=5.169KN*m$

(b) Given that both the hollow and solid cylinders have the same cross-sectional area $\displaystyle A_{h}=A_{s}$, we can find the radius of the solid cylinder

$\displaystyle A_{h}=\pi (c_{1}^{2}-c_{2}^{2})$

$\displaystyle A_{s}=\pi c_{3}^{2}$

$\displaystyle c_{3}={\sqrt {c_{1}^{2}-c_{2}^{2}}}=0.0335m$

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

$J_{s}=2\pi \int _{0}^{c_{3}}c^{3}dc={\frac {\pi }{2}}c_{3}^{4}=1.988*10^{-6}m^{4}$

Using $\displaystyle J_{s}$ and $\displaystyle c_{3}$ solve for $\displaystyle \tau _{max_{s}}$

$\displaystyle \tau _{max_{s}}={\frac {Tc_{3}}{J_{s}}}=87.21MPa$

Problem R3.3

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

Given:

Internal Diameter= 1.2 in

Outer Diameter= 1.6 in

$\displaystyle \tau _{max}=7.5ksi$

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

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

Where

T= applied torque

$\displaystyle c_{1}$ (Inner Radius) $\displaystyle =0.6in$

$\displaystyle c_{2}$ (outer radius) :$\displaystyle =0.8in$

$J={\frac {\pi }{2}}(c_{2}^{4}-c_{1}^{4})$ as given by equation 3.11 from the textbook

Therefore,

$\displaystyle \tau _{max}={\frac {T*C_{max}}{J}}={\frac {T*C_{max}}{\pi 2\left(c_{max}^{4}-c_{min}^{4}\right)}}$

Problem R3.4

Problem Number 3.7, p.155

The solid spindle AB has a diameter d_s = 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

For solid spindle AB

$c=0.5d_{s}=0.5*1.5=0.75in$

$J={\frac {\pi }{2}}c^{4}={\frac {\pi }{4}}(0.75)^{4}=0.49701in^{4}$

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

$T_{AB}={\frac {J*T_{d}}{C}}={\frac {(0.49701in^{4})(12ksi)}{0.75in}}=7.952kip-in$

For sleeve CD

$c_{2}=(0.5)d_{2}=(0.5)(3)=1.5in$

$c_{1}=c_{2}-t\implies 1.5in-0.25in=1.25in$

$J={\frac {\pi }{2}}(c_{2}^{4}-c_{1}^{4})={\frac {\pi }{2}}(1.5^{4}-1.25^{4})=4.1172in^{4}$

$T_{CD}\implies {\frac {JT_{d}}{C_{d}}}\implies {\frac {(4.1172in^{4})(7ksi)}{1.5in}}=19.213kip-in$

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

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

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:

$\displaystyle T_{A}=T_{AB}=300N*m$

and from disk B

$\displaystyle T_{AB}+T_{B}=T_{BC}=700N*m$

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

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

Where T = torque, c = the radius of the cross section, J = centroidal polar moment of inertia (${\frac {1}{2}}\pi c^{4}$ for a solid shaft)

By analyzing shaft AB we find:

$\displaystyle \tau _{(max)AB}={\frac {T_{AB}c_{AB}}{J_{AB}}}={\frac {2T_{AB}}{\pi c_{AB}^{3}}}=56.6MPa$

By analyzing shaft BC we find:

$\displaystyle \tau _{(max)BC}={\frac {T_{BC}c_{BC}}{J_{BC}}}={\frac {2T_{BC}}{\pi c_{BC}^{3}}}=36.6MPa$

Problem R3.6

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.

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

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

Known Values:

$\tau _{ABmax}=50MPa$

$\tau _{BCmax}=25MPa$

$T_{A}=1250Nm$

As there are no torsional forces being applied between points A and C, it can also be seen that $T_{A}=T_{AB}=T_{BC}$

Formulas: $\tau _{max}={\frac {TC}{J}}$, $J={\frac {\pi }{2}}C^{4}\implies C=({\frac {2T}{\pi \tau _{max}}})^{\frac {1}{3}}$

(a) Maximum Allowable Shear Stress in the Rod:

$\tau _{ABmax}=50MPa=50000000{\frac {N}{m^{2}}}$

Torque Applied to the Rod:

$T_{AB}=T_{A}=1250Nm$

Radius of the Rod AB:

$C_{AB}=({\frac {2T_{AB}}{\pi \tau _{ABmax}}})^{\frac {1}{3}}$

Substitute Known Values into the Radius Equation:

$C_{AB}=({\frac {2(1250Nm)}{\pi (50000000{\frac {N}{m^{2}}})}})^{\frac {1}{3}}$

Simplify and Solve for Radius:

$C_{AB}=({\frac {1m^{3}}{20000\pi }})^{\frac {1}{3}}=~.025m$

Find Diameter of AB:

$D_{AB}=2C_{AB}\implies D_{AB}=~.050m$

(b) Maximum Allowable Shear Stress in the Rod:

$\tau _{BCmax}=25MPa=25000000{\frac {N}{m^{2}}}$

Torque Applied to the Rod:

$T_{BC}=T_{A}=1250Nm$

Radius of the Rod BC:

$C_{BC}=({\frac {2T_{BC}}{\pi \tau _{BCmax}}})^{\frac {1}{3}}$

Substitute Known Values into the Radius Equation:

$C_{BC}=({\frac {2(1250Nm)}{\pi (25000000{\frac {N}{m^{2}}})}})^{\frac {1}{3}}$

Simplify and Solve for Radius:

$C_{BC}=({\frac {1m^{3}}{10000\pi }})^{\frac {1}{3}}=~.032m$

Find Diameter of BC:

$D_{BC}=2C_{BC}\implies D_{BC}=~.063m$

R3.7 Solution

A

for the sleeve CD

$\displaystyle c_{2}={\frac {1}{2}}*d_{2}=.5*3=1.5in$

$\displaystyle c_{1}=c_{2}-t$

$\displaystyle =>1.5in-0.25in=1.25in$

$\displaystyle c_{2}={\frac {d_{2}}{2}}=.5*3=1.5in$

$\displaystyle J={\frac {\pi }{2}}((c_{2})^{4}-(c_{1})^{4})$

$\displaystyle J={\frac {\pi }{2}}((1.5)^{4}-(1.25)^{4})$

$\displaystyle J=4.1172in^{4}$

$\displaystyle T_{C}D={\frac {JT_{d}}{C_{2}}}$

$\displaystyle =>{\frac {(4.1172in^{4})(7ksi)}{1.5in}}=19.213kip-in$

B

at the solid spindle AB

$\displaystyle \tau ={\frac {TC}{J}}={\frac {2T}{\pi c^{3}}}$

$\displaystyle =>{\frac {2T}{(\pi \tau )^{1/3}}}=({\frac {2*19.213kip-in}{12ksi*\pi }})^{1/3}=1.0064in$

A

for the sleeve CD

$\displaystyle =>1.5in-0.25in=1.25in$

B

at the solid spindle AB

$\displaystyle \tau ={\frac {TC}{J}}={\frac {2T}{(\pi c^{3})}}$

$=>{\frac {2T}{(\pi \tau )^{1/3}}}={\frac {2*19.213kip-in}{(12ksi*\pi )^{1/3}}}=1.0064in$

$\displaystyle ds=2c$

$\displaystyle =>2*1.0064in=2.01in$

$\displaystyle ds=2.01in\sigma$

Problem R3.8

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

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:

$\displaystyle T_{A}=T_{AB}=300N*m$

and from disk B

$\displaystyle T_{AB}+T_{B}=T_{BC}=700N*m$

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

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

Where T = torque, c = the radius of the cross section, J = centroidal polar moment of inertia (${\frac {1}{2}}\pi c^{4}$ for a solid shaft)

By analyzing shaft AB we find:

$\displaystyle \tau _{(max)AB}={\frac {T_{AB}c_{AB}}{J_{AB}}}={\frac {2T_{AB}}{\pi c_{AB}^{3}}}=56.6MPa$

By analyzing shaft BC we find:

$\displaystyle \tau _{(max)BC}={\frac {T_{BC}c_{BC}}{J_{BC}}}={\frac {2T_{BC}}{\pi c_{BC}^{3}}}=36.6MPa$

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 2c_unknown = d_unknown

$\displaystyle \tau _{max}={\frac {2T_{BC}}{\pi c_{unknown}^{3}}}$

$\displaystyle c_{unknown}^{3}={\frac {2T_{BC}}{\pi \tau _{max}}}$

$\displaystyle c_{unknown}=19.8mm$

$\displaystyle d_{unknown}=39.6mm$

Therefore the smallest possible diameter for shaft BC is 39.6mm