Problem 1: FBD:
Equilibrium equation: πΉ" β πΉ$ β πΉ% β 2π = 0 (1)
π$ =πΉ$πΏ$πΈπ΄$
+ πΌπ₯ππΏ$ =2πΉ$πΏ
πΈ 14 ππ%+ 2πΌπ₯ππΏ =
8πΉ$πΏπΈππ% + 2πΌπ₯ππΏ
π% =πΉ%πΏ%πΈπ΄%
=πΉ%πΏ
πΈ 14 π(9π% β 4π%)
=4πΉ%πΏ5πΈππ%
π" =πΉ"πΏ"πΈπ΄"
=πΉ"πΏ
πΈ 14 π(4π% β π%)
=4πΉ"πΏ3πΈππ%
Compatibility equations:
π$ = π%
8πΉ$πΏπΈππ% + 2πΌπ₯ππΏ =
4πΉ%πΏ5πΈππ%
πΉ$ =110πΉ% β
14πΌπ₯ππΈππ
%
π% + π" = 0
4πΉ%πΏ5πΈππ% +
4πΉ"πΏ3πΈππ% = 0
πΉ" = β35πΉ%
solve equation (1):
πΉ$ = β417πΌπ₯ππΈππ
% β217π
πΉ% =534πΌπ₯ππΈππ
% β2017π
πΉ" = β334πΌπ₯ππΈππ
% +1217π
Axial stresses:
π$ =πΉ$π΄$
=β 417πΌπ₯ππΈππ
% β 217π
14ππ
%=β16πΌπ₯ππΈππ% β 8π
17ππ%
π% =πΉ%π΄%
=534πΌπ₯ππΈππ
% β 2017π54ππ
%=2πΌπ₯ππΈππ% β 16π
17ππ%
π" =πΉ"π΄"
=β 334πΌπ₯ππΈππ
% + 1217π34ππ
%=β2πΌπ₯ππΈππ% + 16π
17ππ%
Displacement of the connector C:
π’A = 0 + π% =4πΉ%πΏ5πΈππ% =
217πΌπ₯ππΏ β
16ππΏ17πΈππ%
ME 323 β 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 Ο.
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 β 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 β 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
New Section 2 Page 1
New Section 2 Page 2
New Section 2 Page 3
New Section 2 Page 2
New Section 2 Page 4