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# Solution Riley

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```MECHANICS OF MATERIALS, 6th Edition RILEY, STURGES AND MORRIS
Chapter 1
1-1*
From a free-body diagram of the forearm,
the equilibrium equations give
0 :yF\u2191\u3a3 = 20 0T F W\u2212 \u2212 \u2212 =
0 :FM\u3a3 =4 ( )1.5 5.5 11.5 20 0T W\u2212 \u2212 =
............................... Ans. 3.667 153.33 lbT W= +
............................... Ans. 2.667 133.33 lbF W= +
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MECHANICS OF MATERIALS, 6th Edition RILEY, STURGES AND MORRIS
1-2*
From a free-body diagram of the ring, the equations of equilibrium
0 :xF\u2192\u3a3 = 2 1cos10 sin10 0T T° \u2212 ° =
0 :yF\u2191 \u3a3 = ( )1 2cos10 sin10 175 9.81 0T T° \u2212 °\u2212 =
are solved to get
1 25.67128T T=
.............................................................................. Ans. 1 1799 NT =
................................................................................ Ans. 2 317 NT =
....................................................... Ans. ( )3 175 9.81 1717 NT = =
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition RILEY, STURGES AND MORRIS
1-3
The equations of equilibrium
0 :xF\u2192\u3a3 = sin 30 sin 30 0A BN N° \u2212 ° =
0 :yF\u2191 \u3a3 = cos30 cos30 800 0A BN N° + ° \u2212 =
are solved to get
A BN N=
462 lbA =N ................................................................ Ans. 60°
462 lbB =N ................................................................ Ans. 60°
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition RILEY, STURGES AND MORRIS
1-4*
The equations of equilibrium
0 :xF\u2192\u3a3 = sin 45 0x BA N\u2212 =o
0 :yF\u2191 \u3a3 = cos 45 300 0y BA N+ \u2212o =
0 :AM\u3a3 =4 ( ) ( )1.5 cos 45 1.5 300 0BN \u2212 =o
are solved to get
424.264 N 424 NBN = \u2245 45............................. Ans.
300 NxA = 0 NyA =
........................................................ Ans. 300 N=A \u2192

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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
MECHANICS OF MATERIALS, 6th Edition RILEY, STURGES AND MORRIS
1-5
The equations of equilibrium
0 :xF\u2192\u3a3 = 0xA =
0 :yF\u2191 \u3a3 = 250 0yA \u2212 =
0 :AM\u3a3 =4 ( )3 250 0AM \u2212 =
are solved to get
0 lbxA = 250 lbyA =
.....................................................Ans. 250 lb=A \u2191
.................................................Ans. 750 lb ft AM = \u22c5 4
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
MECHANICS OF MATERIALS, 6th Edition RILEY, STURGES AND MORRIS
1-6
From an overall free-body diagram, the equations of equilibrium
0 :xF\u2192\u3a3 = 0xA =
0 :yF\u2191 \u3a3 = 10 15 0y FA N\u2212 \u2212 + =
0 :AM\u3a3 =4 ( ) ( )9 3 10 6 15FN \u2212 \u2212 = 0
are solved to get
0 kNxA =
11.6667 kNyA = \u2191
13.3333 kNFN = \u2191
Then, from a free-body diagram of the right hand section of the truss,
the equations of equilibrium
0 :CM\u3a3 =4 ( )3 13.3333 3 0DET\u2212 =
0 :DM\u3a3 =4 ( ) ( )3 3 15 6 13.3333 0BCT \u2212 + =
0 :yF\u2191 \u3a3 = 13.3333 15 cos 45 0CDT\u2212 \u2212 =o
are solved to get
............................................................................................................ Ans. 13.33 kN (T)DET =
.................................................................................. Ans. 11.67 kN 11.67 kN (C)BCT = \u2212 =
...................................................................................... Ans. 2.36 kN 2.36 kN (C)CDT = \u2212 =
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition RILEY, STURGES AND MORRIS
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
1-7*
0 :AM\u3a3 =4 ( ) ( )27 13 33cos15 4 35sin15 0FN \u2212 ° + ° =
14.93561 lb 14.94 lbFN = \u2245 .............Ans. 75°
0 :xF\u3a3 =Z cos30 35sin15 0P ° \u2212 ° =
10.46005 lb 10.46 lbP = \u2245 15 ................Ans. °
0 :yF\u3a3 =^ sin 30 35cos15 0R FN N P+ \u2212 °\u2212 ° =
24.1 lbRN = ..........................................Ans. 75°
MECHANICS OF MATERIALS, 6th Edition RILEY, STURGES AND MORRIS
1-8
0 :yF\u2191\u3a3 = 4060 5210 0P + \u2212 =
.......................................................................................Ans. 1150 NP =
0 :AM\u3a3 =4 ( ) ( )16 4060 37 5210 0C \u2212 \u2212 =
...............................................Ans. 257,730 N mm 258 N mC = \u22c5 \u2245 \u22c5

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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition RILEY, STURGES AND MORRIS
1-9*
From an overall free-body diagram, the equations of equilibrium
0 :xF\u2192\u3a3 = 0xA =
0 :yF\u2191 \u3a3 = 10 20 0y EA N\u2212 \u2212 + =
0 :AM\u3a3 =4 ( ) ( )45 15 10 30 20 0EN \u2212 \u2212 =
are solved to get
0 kipxA =
13.3333 kipyA = \u2191
16.6667 kipEN = \u2191
Then, from a free-body diagram of the right hand section
of the truss, the equations of equilibrium
0 :FM\u3a3 =4 ( )15 15 sin 30 0E DEN T+ ° =
=
G
0 :EM\u3a3 =4 ( ) ( )15 20 15 sin 60 0CFT\u2212 °
0 :CM\u3a3 =4 ( ) ( )22.5 7.5 20 15cos30 0E FN T\u2212 \u2212 ° =
are solved to get
............................................................................... Ans. 33.3333 kip 33.3 kip (C)CDT = \u2212 \u2245
..................................................................................```
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