Problem 1: FBD:

Equilibrium equation: 𝐹" − 𝐹$ − 𝐹% − 2𝑃 = 0 (1)

𝑒$ =𝐹$𝐿$𝐸𝐴$

+ 𝛼𝛥𝑇𝐿$ =2𝐹$𝐿

𝐸 14 𝜋𝑑%+ 2𝛼𝛥𝑇𝐿 =

8𝐹$𝐿𝐸𝜋𝑑% + 2𝛼𝛥𝑇𝐿

𝑒% =𝐹%𝐿%𝐸𝐴%

=𝐹%𝐿

𝐸 14 𝜋(9𝑑% − 4𝑑%)

=4𝐹%𝐿5𝐸𝜋𝑑%

𝑒" =𝐹"𝐿"𝐸𝐴"

=𝐹"𝐿

𝐸 14 𝜋(4𝑑% − 𝑑%)

=4𝐹"𝐿3𝐸𝜋𝑑%

Compatibility equations:

𝑒$ = 𝑒%

8𝐹$𝐿𝐸𝜋𝑑% + 2𝛼𝛥𝑇𝐿 =

4𝐹%𝐿5𝐸𝜋𝑑%

𝐹$ =110𝐹% −

14𝛼𝛥𝑇𝐸𝜋𝑑

%

𝑒% + 𝑒" = 0

4𝐹%𝐿5𝐸𝜋𝑑% +

4𝐹"𝐿3𝐸𝜋𝑑% = 0

𝐹" = −35𝐹%

solve equation (1):

𝐹$ = −417𝛼𝛥𝑇𝐸𝜋𝑑

% −217𝑃

𝐹% =534𝛼𝛥𝑇𝐸𝜋𝑑

% −2017𝑃

𝐹" = −334𝛼𝛥𝑇𝐸𝜋𝑑

% +1217𝑃

Axial stresses:

𝜎$ =𝐹$𝐴$

=− 417𝛼𝛥𝑇𝐸𝜋𝑑

% − 217𝑃

14𝜋𝑑

%=−16𝛼𝛥𝑇𝐸𝜋𝑑% − 8𝑃

17𝜋𝑑%

𝜎% =𝐹%𝐴%

=534𝛼𝛥𝑇𝐸𝜋𝑑

% − 2017𝑃54𝜋𝑑

%=2𝛼𝛥𝑇𝐸𝜋𝑑% − 16𝑃

17𝜋𝑑%

𝜎" =𝐹"𝐴"

=− 334𝛼𝛥𝑇𝐸𝜋𝑑

% + 1217𝑃34𝜋𝑑

%=−2𝛼𝛥𝑇𝐸𝜋𝑑% + 16𝑃

17𝜋𝑑%

Displacement of the connector C:

𝑢A = 0 + 𝑒% =4𝐹%𝐿5𝐸𝜋𝑑% =

217𝛼𝛥𝑇𝐿 −

16𝑃𝐿17𝐸𝜋𝑑%

ME 323 – Mechanics of Materials Examination #1 February 12th, 2020

Name (Print) ______________________________________________

(Last) (First) PROBLEM # 3 (25 points) Shafts (1) and (2) in Fig. 3(a) are connected by a rigid connector at B, and are fixed to the rigid walls at the ends A and C. Both shafts have length L and shear modulus G. Shaft (1) has a solid cross section of diameter 2d, while shaft (2) has a hollow cross section of outer diameter 2d and inner diameter d as shown in Figs. 3(b) and 3(c). An external torque TB is applied at the connect B.

(a) Determine if the assembly is statically determinate or indeterminate. (b) Determine the internal torque carried by each shaft. (c) Consider the points M and N on the outer radius of shaft (1) and (2), respectively. The cross-

section views are shown in Figs. 3(b) and 3(c). Determine the state of stress at M and N. (d) Draw the stress elements to represent the state of stress at M and N with clear labels of the

stress components. Express your results in terms of TB, d, L, G, and π.

L LA

(1)

2d

xTB

(1) (2)

Fig. 3(a)

Fig. 3(b) Fig. 3(c)

CB

y

z

yM (2)

d 2d

z

y

N

ME 323 – Mechanics of Materials Examination #1 February 12th, 2020

Name (Print) ______________________________________________

(Last) (First) PROBLEM # 3 (cont.)

FBI 7 7 0 2 72 31Er HE statically

Assumeallmembersunder positivetorque FBI

tiffsEI

Equilibrium IT Tz Ta TBAs Id

Torquetwist behavior 4954 YIp.EE z

zTEtEpzw Ipz tTl3dYzdB It5zd

Compatibilityconditions

g 4BOla z e B0 5 04 0

ok ok

ME 323 – Mechanics of Materials Examination #1 February 12th, 2020

Name (Print) ______________________________________________

(Last) (First) PROBLEM # 3 (cont.)

Solvefor Ta and Tz

Tz I 4k sdYzz

I I a s

IEEfa

Tryto

III FII.uis.su

I e soExz 40 tht

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