Strength of Materials |
From inside the book
Results 1-3 of 92
Page 212
30 lb / ft M = - 120 x 6 = - 720 ft - lb - = A1c - - - Bop - tbc 1V = 120 lb - 474 ( a )
120 x8 = 960 ' ft - lb M = 120x8 = 960 ft - lb TV = 120 lb M = - 720 ft - lb M = - 720 ft
- lb - 720 ft - lb - 120 x2 = - 240 ft - lb 11 30 lb / ft 4 M = - 120 x2 = - 240 ft - lb 30 lb
...
30 lb / ft M = - 120 x 6 = - 720 ft - lb - = A1c - - - Bop - tbc 1V = 120 lb - 474 ( a )
120 x8 = 960 ' ft - lb M = 120x8 = 960 ft - lb TV = 120 lb M = - 720 ft - lb M = - 720 ft
- lb - 720 ft - lb - 120 x2 = - 240 ft - lb 11 30 lb / ft 4 M = - 120 x2 = - 240 ft - lb 30 lb
...
Page 284
distributed load of w lb / ft over the middle span . 1 + 2B Ans . M2 = - 4 ( 1 + a ) ( a
+ B ) – 1 wL21 + 2a M3 = - 4 4 ( 1 + a ) ( a + B ) – 1 823 . A continuous beam
simply supported over three 10 - ft spans carries a concentrated load of 400 lb at
the ...
distributed load of w lb / ft over the middle span . 1 + 2B Ans . M2 = - 4 ( 1 + a ) ( a
+ B ) – 1 wL21 + 2a M3 = - 4 4 ( 1 + a ) ( a + B ) – 1 823 . A continuous beam
simply supported over three 10 - ft spans carries a concentrated load of 400 lb at
the ...
Page 345
1 400 400 lb 100 lb AS Da = 750 16 100 16 1 Do = 575 lb I V 11 . 5 100 lb 400 lb
An = 250 lb Dulle ( a ) Loading on line shaft 100 lb 700 lb A , = 725 lb 1450 ft - lb
1150 ft - lb ( b ) Bending moments in vertical plane 500 ft - lb 1500 ft - lb x 2000 ft
...
1 400 400 lb 100 lb AS Da = 750 16 100 16 1 Do = 575 lb I V 11 . 5 100 lb 400 lb
An = 250 lb Dulle ( a ) Loading on line shaft 100 lb 700 lb A , = 725 lb 1450 ft - lb
1150 ft - lb ( b ) Bending moments in vertical plane 500 ft - lb 1500 ft - lb x 2000 ft
...
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Contents
List of Symbols and Abbreviations | xvi |
SIMPLE STRAIN | 26 |
TORSION | 60 |
Copyright | |
18 other sections not shown
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Common terms and phrases
acting actual allowable angle applied assumed axes axial axis beam beam shown bending bending moment cantilever carries caused centroid circle column compressive compressive stress compute concentrated concrete consider constant couple cross section deflection deformation Determine developed diameter direction distance distributed load effect elastic curve element equal equation equivalent expressed flange flexural stress force formula ft-lb given gives Hence horizontal ILLUSTRATIVE inertia joint lb/ft length limit load material maximum maximum shearing method midspan moments negative neutral axis normal obtain occurs plane plate positive principal Prob PROBLEMS produce radius reaction reduces reference reinforced relation resisting respect resultant rivet segment shaft shearing stress shown in Fig shows slope Solution Solve span steel strain strength supported Table tangent tensile thickness torsional uniformly varies vertical weight weld yield zero
Bibliographic information
| Title | Strength of Materials Harper International Student Repr Harper international edition International student reprints |
| Author | Ferdinand Leon Singer |
| Edition | 2 |
| Publisher | Harper, 1962 |
| Original from | the University of Michigan |
| Digitized | 29 Nov 2007 |
| Length | 590 pages |
| Export Citation | BiBTeX EndNote RefMan |