User:Jshminer46/Mom-s13-team4-R3

=Report 3=

Problem 3.1
sec. 10 Pb- 10.1

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: $$ \theta = 30^\circ $$ $$ \theta = 45^ \circ $$

test test

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 steel shaft shown. (b) Determine the maximm shearing stress caused by the same torque T in a solid cylindrical shaft of the same cross-sectional area.



GIVENS:
$$\tau = 45 MPa$$ (THE MAXIMUM SHEARING STRESS)

$$C_2=45mm$$ (THE EXTERNAL RADIUS)

$$C_1=30mm$$ (THE INTERNAL RADIUS)

a) DETERMINE THE TORQUE T

WE KNOW THAT $$\tau = 45 MPa$$

AND WE ALSO KNOW THAT $$\tau = \frac{T\times C_2}{J}$$

$$\tau = \frac{T\times C_2}{J}$$

$$\Rightarrow T=\frac{\tau \times J}{C_2}$$

WE ALSO KNOW THAT $$J=\frac{\pi }{2}\times C^4$$

HOWEVER IN THIS CASE WE HAVE A HOLLOW CYLINDRICAL STEEL SHAFT SO $$C^4=C^4_2-C^4_1$$

THEREFORE $$J=\frac{\pi }{2}\times (C^4_2-C^4_1)$$

WHICH MEANS THAT $$T=\frac{\tau \times \pi \times (C^4_2-C^4_1)}{2\times C_2}$$

$$\Rightarrow T=\frac{45MPa\times \pi \times (0.045^4-0.03^4)}{2\times 0.045}$$

SO $$T=5.2kN$$

b) DETERMINE THE MAXIMUM SHEARING STRESS CAUSED BY THE TORQUE IN PART (a)

WE FIGURED OUT IN PART (a) THAT THE TORQUE T WAS EQUAL 5.2kN

AND WE KNOW THAT $$\tau = \frac{T\times C_2}{J}$$

IN THIS CASE WE HAVE A SOLID CYLINDRICAL SHAFT SO $$J=\frac{\pi }{2}\times C^4$$

THEREFORE $$\tau = \frac{T}{\frac{\pi }{2}\times C^3_2}$$

$$\Rightarrow \tau = \frac{45kN}{\frac{\pi }{2}\times 0.045^3}$$

SO $$\tau =3.63MPa$$

On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

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.



Solution
GIVENS:

$$d_1=0.9in$$ (THE INTERNAL DIAMETER)

THEREFORE $$c_1=0.45in$$ (THE INTERNAL RADIUS)

AND $$d_2=1.6in$$ (THE EXTERNAL DIAMETER)

THEREFORE $$c_2=0.8in$$ (THE EXTERNAL RADIUS)

$$T=9kip.in$$ ( THE TORQUE)

LET'S FIND THE MAXIMUM SHEARING STRESS

WE KNOW THAT THE ELASTIC TORSION IS:

$$\tau_{max} =\frac{T\times C}{J}$$

$$\Rightarrow \tau_{max} =\frac{2T\times C^4_2}{\pi \times\left ( C^4_2-C^4_1 \right )}$$

$$\Rightarrow \tau_{max} =\frac{2\times 9Kip.in\times 0.8}{\pi \times \left ( 0.8^4-0.45^4 \right )}$$

$$\Rightarrow \tau_{max} =12.44kip$$

On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

Problem 3.4
P3.7, Beer 2012

Problem Statement
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.



Solution
The allowable shearing stress of solid spindle AB is $$\tau_{a}=12 ksi$$

The diameter of the solid spindle AB is $$d_{s}=1.5in$$

Free Body Diagram of Solid Spindle AB



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 inertia of a circle of radius c is

$$J=\frac{1}{2}\pi c^{4}=\frac{\pi }{2}(0.75in)^{4}=0.49701in^{4}$$

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

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

Solving for $$T_{AB}$$

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

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

The diameter of the sleeve CD is $$d_{2}=3 in$$

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

Free Body Diagram of Sleeve CD



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

$$c_2=\frac{1}{2}d_2=\frac{1}{2}(3in)=1.5 in$$

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

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

The polar moment of inertia is

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

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

$$\tau _{max}=\frac{Tc_2}{J}$$

Solving for $$T_{CD}$$

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

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

$$T=7.952kip*in$$

On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

Problem 3.5
P3.9, Beer 2012

Problem Statement
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.



Solution
Torque on pulley A is

$$T_A=300N*m$$

Diameter of shaft AB is

$$d_{AB}=0.03m$$

Free Body Diagram of Shaft AB



The radius of shaft AB is half the diameter of shaft AB

$$r_{AB}=\frac{1}{2}d_{AB}=\frac{1}{2}*0.03m=0.015m$$

Torque at A $$T_A$$ is equal to torque of shaft AB $$T_{AB}$$

The sheer stress of shaft AB is equal to

$$\tau _{AB}=\frac{T_{AB}*r_{AB}}{J_{AB}}=\frac{T_{AB}r_{AB}}{\frac{\pi }{2}(r_{AB})^{4}}=\frac{2T_{AB}}{\pi (r_{AB})^{3}}=\frac{2*300N*m}{\pi (0.015m)^{3}}=56.588*10^{6}\frac{N}{m^{2}}=56.588MPa$$

Free Body Diagram of shaft BC



The radius of shaft BC is half the diameter of shaft BC

$$r_{BC}=\frac{1}{2}d_{BC}=\frac{1}{2}*0.046m=0.023m$$

Torque at BC $$T_{BC}$$ is equal to torque of B $$T_{B}$$ plus torque of A $$T_{A}$$

The sheer stress of shaft BC is equal to

$$\tau _{BC}=\frac{(T_{B}+T_A)*r_{BC}}{J_{BC}}=\frac{(T_{B}+T_A)r_{BC}}{\frac{\pi }{2}(r_{BC})^{4}}=\frac{2(T_{B}+T_A)}{\pi (r_{BC})^{3}}=\frac{2(400N*m+300N*m)}{\pi (0.023m)^{3}}=36.626*10^{6}\frac{N}{m^{2}}=36.626MPa$$

On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

Problem 3.6
P3.17, Beer 2012, p.156

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



Solution
GIVENS:

$$\tau _{maxAB}=50MPa$$ (THE SHEAR STRESS ON ROD AB)

$$\tau _{maxBC}=25MPa$$ (THE SHEAR STRESS ON ROD BC)

$$T=1250Nm$$ (THE TORQUE APPLIED ON THE ASSEMBLY)

a) DETERMINE THE REQUIRED DIAMETER OF THE ROD AB WITH AN APPLIED TORQUE T

WE KNOW THAT

$$\tau _{maxAB}=\frac{2T\times C_{AB}}{\pi \times C^4_{AB}}$$

$$\Rightarrow \tau _{maxAB}=\frac{2T}{\pi \times C^3_{AB}}$$

$$\Rightarrow C^3_{AB}=\frac{2T}{\pi \times \tau _{maxAB}}$$

$$\Rightarrow C_{AB}=(\frac{2T}{\pi \times \tau _{maxAB}})^(\frac{1}{3})$$

$$\Rightarrow d_{AB}=2(\frac{2T}{\pi \times \tau _{maxAB}})^(\frac{1}{3})$$

$$\Rightarrow d_{AB}=2(\frac{2\times 1250Nm}{\pi\times 5\times 10^7}})^(\frac{1}{3})$$

$$\Rightarrow d_{AB}=0.050m=50mm$$

b) DETERMINE THE REQUIRED DIAMETER OF ROD BC

WE KNOW THAT

$$\tau _{maxBC}=\frac{2T\times C_{BC}}{\pi \times C^4_{BC}}$$

$$\Rightarrow \tau _{maxBC}=\frac{2T}{\pi \times C^3_{BC}}$$

$$\Rightarrow C^3_{BC}=\frac{2T}{\pi \times \tau _{maxBC}}$$

$$\Rightarrow C_{BC}=(\frac{2T}{\pi \times \tau _{maxBC}})^(\frac{1}{3})$$

$$\Rightarrow d_{BC}=2(\frac{2T}{\pi \times \tau _{maxBC}})^(\frac{1}{3})$$

$$d_{BC}=2(\frac{2\times 1250Nm}{\pi\times 25\times 10^6}})^(\frac{1}{3})$$

$$d_{BC}=0.063m=63mm$$

Problem 3.7
P3.8, Beer 2012

Problem Statement
The solid spindle AB is made of a steel with an allowable shearing stress of 12 ksi, and sleeve CD is made of a 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 $$d_s$$ of spindle AB.



Solution
Part a

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

The diameter of the sleeve CD is $$d_{2}=3 in$$

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

Free Body Diagram of Sleeve CD



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

$$c_2=\frac{1}{2}d_2=\frac{1}{2}(3in)=1.5 in$$

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

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

The polar moment of inertia is

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

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

$$\tau _{max}=\frac{Tc_2}{J}$$

Solving for $$T_{CD}$$

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

To maintain equilibrium the largest torque T that can be applied at A is

$$T=19.213kip*in$$

Part b

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

Free Body Diagram of Solid Spindle AB



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

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

Solving for c

$$c=\sqrt[3]{\frac{2T}{\pi \tau }}=\sqrt[3]{\frac{2*19.213kip*in}{\pi 12ksi}}=1.0064in$$

The diameter is twice radius c

$$d_s=2c=2*1.0064in=2.01in$$

On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

Problem 3.8
P3.10, Beer 2012

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.



Solution
Free Body Diagram of Shaft BC



GIVENS: $$T_{A}=300Nm$$ (THE TORQUE ON DISK A)

$$T_{B}=400Nm$$ (THE TORQUE ON DISK B)

$$d_{AB}=30mm$$ (THE DIAMETER OF THE ROD AB)

$$d_{BC}=46mm$$ (THE DIAMETER OF THE ROD BC)

DETERMINE THE SMALLEST DIAMETER OF BC FOR WHICH THE SHEARING STRESS OF THE ASSEMBLY WILL NOT INCREASE

IF WE PERFORM A CUT BETWEEN A AND B

(FREE BODY DIAGRAM HERE)

$$\sum M_y=0 \Rightarrow T_{AB}-T_{A}=0$$

$$\Rightarrow T_{AB}=T_{A}=300Nm$$

(FREE BODY DIAGRAM HERE)

$$\sum M_{y}=0 \Rightarrow T_{BC}-T_{A}-T_{B}=0$$

$$\Rightarrow T_{BC}=T_{A}+T_{B}=300+400=700Nm$$

LET'S FIND THE MAXIMUM STRESS

$$\bullet ON AB$$

$$\tau =\frac{2 \times T_{AB}\times C_{AB}}{\pi \times C^{4}_{AB}}$$

$$\Rightarrow \tau =\frac{2\times T_{AB}}{\pi \times C^3_{AB}}$$

$$\Rightarrow \tau =\frac{2\times 300Nm}{\pi \times 0.015^3m^{3}}$$

$$\Rightarrow \tau =56.59MPa$$

$$\bullet ON BC$$

$$\tau =\frac{2\times T_{BC}}{\pi \times C^{3}_{BC}}$$

$$\tau =36.63MPa$$

THEREFORE THE LARGEST SHEARING STRESS IS $$ \tau _{ABmax}=56.59MPa$$

TO FIND THE MAXIMUM DIAMETER FOR THE ROD BC WE HAVE TO USE THE LARGEST SHEARING STRESS APPLIED TO THE ASSEMBLY

WE KNOW THAT

$$\tau_{AB}=\frac{2\times T_{BC}}{\pi \times C^{3}_{BC}}$$

$$\Rightarrow C^{3}_{BC}=\frac{2\times T_{BC}}{\pi \times \tau }$$

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

$$\Rightarrow d_{BC}=2\times (\frac{2\times T_{BC}}{\pi \times \tau })^(\frac{1}{3})$$

$$\Rightarrow d_{BC}=(\frac{2\times 700Nm}{\pi \times 56.59\times 10^6\frac{N}{m^2} })^(\frac{1}{3})$$

SO $$d_{BC}=39.8mm $$

On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.