## Strength of materials |

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Page 114

The moment at F where x = 6 ft is found by applying the definition of bending

moment.* 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

at B where x = 9 ft is found to be [M = (2M)d Mb = 240(9) - 540(f) = 540

moment at C can also be computed from this basic definition, but whevener the

shear ...

The moment at F where x = 6 ft is found by applying the definition of bending

moment.* 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 momentat B where x = 9 ft is found to be [M = (2M)d Mb = 240(9) - 540(f) = 540

**ft**-**lb**Themoment at C can also be computed from this basic definition, but whevener the

shear ...

Page 212

Af=-120x6=-720

30x4)x2= -240

4' Af=(30x4)x2=240ft-lb ^30 lb 8' M=- (30 x 8) x 4= - 960

— Variations of moment diagrams by parts. Although identical results will be

obtained by using any of the above diagrams, we shall apply Eq. (a) to the third

one (Fig. G-21d). Noting that 7.4 means to take the moment of area about A, we

obtain 1 ...

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.— Variations of moment diagrams by parts. Although identical results will be

obtained by using any of the above diagrams, we shall apply Eq. (a) to the third

one (Fig. G-21d). Noting that 7.4 means to take the moment of area about A, we

obtain 1 ...

Page 345

400 lb 1001b Ah=25Q lb 700 lb 100 lb i4„=7251b ^1450

Loading on line shaft -1150

1500

C: and at D: (d) Torque distribution in shaft Fig. 9-25. 7% = VM2 + = V(1890)2 + (

900)2 = 2090

2000)2+ (450)2 = 2050

largest of ...

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**AtC: and at D: (d) Torque distribution in shaft Fig. 9-25. 7% = VM2 + = V(1890)2 + (

900)2 = 2090

**ft**-**lb**M. = i(M + T.) - £(1890 + 2090) = 1990**ft**-**lb**T, = VM2 + n = V (2000)2+ (450)2 = 2050

**ft**-**lb**M. = l(M + T.) = $(2000 + 2050) = 2025**ft**-**lb**Thelargest of ...

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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