## Strength of materials |

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Page 215

Because the elastic curve is very flat, dc is practically equivalent to dx; so from

Eqs. (c) and (b) we

formula in Art. 5-2, we

Because the elastic curve is very flat, dc is practically equivalent to dx; so from

Eqs. (c) and (b) we

**obtain**1 d0 d0 1 dzy raw °' 5'3; '”'> In deriving the flexureformula in Art. 5-2, we

**obtained**on page 157 the relation l M — = — 5-1 p E1 V ...Page 519

from which we

these several theories of failure, experimental work shows best agreement with

the Mises yield theory when applied to ductile materials. For such materials the ...

from which we

**obtain**' 2°yp2 = (U1 T °2)2 + ('72 T °3)2 'l' ('73 _ °1)2 Summary Ofthese several theories of failure, experimental work shows best agreement with

the Mises yield theory when applied to ductile materials. For such materials the ...

Page 627

This conclusion could also have been

the polar moment of inertia J, is the sum of the moments of inertia with respect to

rectangular axes passing through the polar axis. Hence, since J, is a constant, we

...

This conclusion could also have been

**obtained**from Art. A-2, which shows thatthe polar moment of inertia J, is the sum of the moments of inertia with respect to

rectangular axes passing through the polar axis. Hence, since J, is a constant, we

...

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allowable stresses aluminum angle applied load area-moment method assumed axial load beam in Fig beam loaded beam shown bolt bronze cantilever beam caused centroid column compressive stress concentrated load continuous beam cross section deﬂection deformation Determine the maximum diameter elastic curve element end moments equal equilibrium equivalent exploratory section factor of safety fibers ﬂexural flexural stress Flgure free-body diagram Hence horizontal ILLUSTRATIVE PROBLEMS kN-m kN/m length loaded as shown maximum shearing stress maximum stress midspan midspan deﬂection Mohr’s circle neutral axis normal stress obtain plane positive principal stresses proportional limit radius reaction resisting restrained beam resultant rivet segment shaft shear center shear diagram shearing force shearing strain shown in Fig simply supported beam slope Solution span static equilibrium statically indeterminate steel strain tangent drawn tensile stress three-moment equation torque torsional uniformly distributed load value of E18 vertical shear weld