Strength of materials |
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Page 114
Note that the shaded part of the triangular load applied between .4 and F totals
240 lb and acts at the centroid of the triangular area, i.e., at J of 6 ft from F. Thus
we obtain [M = (2M)L] Mr = 240(6) - 240(f) = 960 ft-lb Similarly, the moment at B ...
Note that the shaded part of the triangular load applied between .4 and F totals
240 lb and acts at the centroid of the triangular area, i.e., at J of 6 ft from F. Thus
we obtain [M = (2M)L] Mr = 240(6) - 240(f) = 960 ft-lb Similarly, the moment at B ...
Page 212
Af=-120x6=-720 ft-lb 120x8=960 ft-lb -720 ft-lb i -120x2= -240 ft-lb 30 lb/ft^ ^3f=-(
30x4)x2= -240 ft-lb M=4x 120^3 ? 1 =480 ft-lb ^ I V= 120 lb -720 ft-lb 30 lb/ft V/ 4'
4' Af=(30x4)x2=240ft-lb ^30 lb 8' M=- (30 x 8) x 4= - 960 ft-lb -960 ft-lb Fig. 6-21.
Af=-120x6=-720 ft-lb 120x8=960 ft-lb -720 ft-lb i -120x2= -240 ft-lb 30 lb/ft^ ^3f=-(
30x4)x2= -240 ft-lb M=4x 120^3 ? 1 =480 ft-lb ^ I V= 120 lb -720 ft-lb 30 lb/ft V/ 4'
4' Af=(30x4)x2=240ft-lb ^30 lb 8' M=- (30 x 8) x 4= - 960 ft-lb -960 ft-lb Fig. 6-21.
Page 345
400 lb 1001b Ah=25Q lb 700 lb 100 lb i4„=7251b ^1450 ft-lb 100 lb 400 lb (a)
Loading on line shaft -1150 ft-lb (b) Bending moments in vertical plane Is 500 ft-lb
1500 ft-lb -2000 ft-lb (c) Bending moments in horizontal plane 900 ft-lb 450 ft-lb At
...
400 lb 1001b Ah=25Q lb 700 lb 100 lb i4„=7251b ^1450 ft-lb 100 lb 400 lb (a)
Loading on line shaft -1150 ft-lb (b) Bending moments in vertical plane Is 500 ft-lb
1500 ft-lb -2000 ft-lb (c) Bending moments in horizontal plane 900 ft-lb 450 ft-lb At
...
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allowable stresses aluminum angle assumed axes axial load beam in Fig beam loaded beam shown bending bolt cantilever beam caused centroid CN CN column compressive stress Compute the maximum concentrated load concrete cover plate cross section deformation Determine the maximum diameter elastic curve end moments equal equivalent Euler's formula factor of safety fibers flange flexure formula free-body diagram ft long ft-lb Hence hinged Hooke's law horizontal ILLUSTRATIVE PROBLEMS lb/ft length loaded as shown main plate maximum shearing stress maximum stress midspan midspan deflection modulus Mohr's circle moments of inertia neutral axis obtain plane plastic positive product of inertia proportional limit radius ratio reaction Repeat Prob resisting restrained beam resultant segment shaft shear center shear diagram shearing force shown in Fig Solution Solve Prob span static steel strain tensile stress thickness torque torsional uniformly distributed load vertical shear weld zero
Bibliographic information
| Title | Strength of materials |
| Author | Ferdinand Leon Singer |
| Edition | 2 |
| Publisher | Harper, 1962 |
| Original from | the University of Michigan |
| Digitized | 29 Nov 2007 |
| Length | 590 pages |
| Subjects | Strength of materials |
| Export Citation | BiBTeX EndNote RefMan |