Ch. 4.3.4 Problems(SI unit)

4. 1, 2

(1)

[eqn: d``=` {P} over {k _{1} +k _{2} +k _{3}} ,````k`=` {EA} over {L}]

(2)

[eqn: F _{1} = {Pk _{1}} over {k _{1} +k _{2} +k _{3}} ,`F _{2} = {Pk _{2}} over {k _{1} +k _{2} +k _{3}} ,`F _{3} =- {Pk _{3}} over {k _{1} +k _{2} +k _{3}}]

4. 3

[eqn: k _{a} = {(EA) _{a}} over {L} ,```````k _{s} = {(EA) _{s}} over {L}]

[eqn: F _{a} ,`F _{s}] = total forces

[eqn: F _{a0} ,`F _{s0}] = additional forces

[eqn: F _{a} `=`F _{a0} ``+`P,`````````F _{s} `=`F _{s0} ``-`P]

[eqn: F``-`F _{a} `-`F _{s} `=`0]  :  equilibrium equation                  1)

[eqn: e _{a0} =F _{a0} `/`k _{a} ,````e _{s0} =F _{s0} `/`k _{s}]  :  constitutive equation       2)

[eqn: e _{a0} =e _{s0}]  :  compatibility equation                       3)

from 1), 2), 3)

[eqn: F _{a0} = {Fk _{a}} over {k _{a} +k _{s}} ,````````F _{s0} = {Fk _{s}} over {k _{a} +k _{s}}]

[eqn: F _{a} =P+ {Fk _{a}} over {k _{a} +k _{s}} = {73F} over {185} +P,````````F _{s} = {Fk _{s}} over {k _{a} +k _{s}} -P= {112F} over {185} -P]

(1) P = 0,

[eqn: F`=` {185} over {73} F _{a,max`=`} {185} over {73} 600000`=`1520546`N# # F`=` {185} over {112} F _{s,max`=`} {185} over {112} 480000`=`792857N]

[eqn: F`=``792857N]

(2)

[eqn: F _{a} `=`600000`=` {73F} over {185} +P# # F _{s} =480000`=` {112F} over {185} -P]

[eqn: P`=`173837`N# # F`=`1080000`N]

(3)

[eqn: {1080000`-`792857} over {792857} 100`=`36.2162`%]

4. 3

(1)

3 equations (1 force equilibrium + 2 moment equilibrium)

4 unkowns ([eqn: F _{A} ,`F _{B} ,`F _{C} ,`F _{D}])

redundancy = 4 - 3 = 1

(2)

force equilibrium

[eqn: F _{A} +F _{B} +F _{C} +F _{D} +P=0]          1)

moment equilibrium from center of the plate

[eqn: F _{C} {l} over {sqrt {2}} -F _{D} {l} over {sqrt {2}} =0]               2)

[eqn: F _{A} {l} over {sqrt {2}} +Pd-F _{B} {l} over {sqrt {2}} =0]          3)

compatability

[eqn: TRIANGLE _{A} + TRIANGLE _{B} = TRIANGLE _{C} + TRIANGLE _{D}]              4)

from 4)

[eqn: F _{A} +F _{B} -F _{C} -F _{D} =0]             5)

from 1), 5)

[eqn: F _{A} +F _{B} + {P} over {2} =0]                  6)

from 3), 6)

[eqn: F _{A} =- {1} over {4} (P`+`2 sqrt {2} {d} over {l} )# F _{B} =- {1} over {4} (P-`2 sqrt {2} {d} over {l} )]

from 2), 5)

[eqn: F _{C} =F _{D} =- {1} over {4} P]

4. 5

(1) 1

(2) let [eqn: eta = {sqrt {2} d} over {l}] where P must be applied.

[eqn: A _{C} =A _{D}] and [eqn: eta =1- {2A _{B}} over {A _{A} +A _{B} +2A _{D}} - {2A _{D}} over {A _{A} +A _{B} +2A _{D}}]

4. 6

[eqn: K _{b} = {k _{b} L} over {EA _{s}}]

(1)

[eqn: {F _{y}} over {F} = {1+K _{b}} over {2+3.25K _{b}}# # {F _{s}} over {F} = {1.25K _{b}} over {2+3.25K _{b}}]

(2)

[eqn: k _{b} =0]

(3)

0.384615

(4)

4

4. 7

[eqn: sigma _{T} =- TRIANGLE T`E alpha _{s} =(120)(73 TIMES 10 ^{9} )(16.5 TIMES 10 ^{-6} )=-1.4454 TIMES 10 ^{8}]

4. 8

30731.7

4. 9

0.2

4. 10

Imperfection can be considered as a thermal expansion of the bar B.

Thus, [eqn: TRIANGLE T= {mu } over {alpha L}]

[eqn: k _{A} = {EA _{A}} over {EA _{B}}]

(1) from page 160

[eqn: F _{B} =- {2EA _{A} cos ^{3} theta } over {1+2k _{A} cos ^{3} theta } alpha TRIANGLE T=- {2 mu EA _{A} cos ^{3} theta } over {L(1+2k _{A} cos ^{3} theta )}# # F _{A} =F _{C} = {EA _{A} cos ^{2} theta } over {1+2k _{A} cos ^{3} theta } alpha TRIANGLE T= { mu EA _{A} cos ^{2} theta } over {L(1+2k _{A} cos ^{3} theta )}]

(2) from page 161

[eqn: d= {1+2(k _{A} -1)cos ^{3} theta } over {1+2k _{A} cos ^{3} theta } alpha TRIANGLE TL= {1+2(k _{A} -1)cos ^{3} theta } over {1+2k _{A} cos ^{3} theta } mu]

7. 1

(1)

[eqn: tau _{a} =7.18764 TIMES 10 ^{7}# # tau _{s} =9.75466 TIMES 10 ^{7}]

(2)

0.128351 rad

(3)

7791.15

(4)

1537.73

7. 2

(1)

[eqn: H _{AB} =G {pi } over {2} R _{o} ^{4}# # H _{BC} =G {pi } over {2} (R _{o} ^{4} -R _{i} ^{4} )# # phi = {Q _{1} L} over {GH _{AB}} + {Q _{1} L} over {GH _{BC}}# # phi = {2(-LQ _{1} R _{i} ^{4} +2LQ _{1} R _{o} ^{4} )} over {G pi R _{0} ^{4} (-R _{i} ^{4} +R _{o} ^{4} )}]

(2)

[eqn: H _{11} = {phi } over {Q _{1}}# # H _{11} = {2(-LR _{i} ^{4} +2LR _{o} ^{4} )} over {G pi R _{0} ^{4} (-R _{i} ^{4} +R _{o} ^{4} )}]

(3)

[eqn: tau _{AB,max} =G {Q _{1} R _{o}} over {H _{AB}}# # tau _{BC,max} =G {Q _{1} R _{o}} over {H _{BC}}# # {tau _{BC}} over {tau _{AB}} = {H _{AB}} over {H _{BC}} = {R _{o} ^{4}} over {-R _{i} ^{4} +R _{o} ^{4}}]

7. 3

(1)

[eqn: Q _{AB} =- {Q _{1} R _{0} ^{4}} over {R _{i} ^{4} -2R _{0} ^{4}} ,# # Q _{BC} = {Q _{1} (-R _{i} ^{4} +R _{0} ^{4} )} over {R _{i} ^{4} -2R _{0} ^{4}}]

(2)

[eqn: phi =- {2LQ _{1}} over {G pi (R _{i} ^{4} -2R _{o} ^{4} )}]

(3)

[eqn: H _{11} =- {2L} over {G pi (R _{i} ^{4} -2R _{o} ^{4} )}]

(4)

[eqn: tau _{max} =- {2Q _{1} R _{o}} over {pi (R _{i} ^{4} -2R _{o} ^{4} )}]

7. 4

(1)

204.349

(2)

0.782974