## Strength of materials |

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Page 197

or d6 = -j ds (a) In most practical cases the

error is made in assuming the length ds to equal its projection dx. With this

assumption, we obtain dd = jjdx (b) It is evident that tangents drawn to the

or d6 = -j ds (a) In most practical cases the

**elastic curve**is so flat that no seriouserror is made in assuming the length ds to equal its projection dx. With this

assumption, we obtain dd = jjdx (b) It is evident that tangents drawn to the

**elastic****curve**...Page 208

P-630, compute the value of (Area)^B-lA- From this result determine whether the

tangent drawn to the

to Eq. (6-5) and Fig. 6-10. Ana. (AreaWZx = 2700 lb-ft3; slope is up to right 631.

P-630, compute the value of (Area)^B-lA- From this result determine whether the

tangent drawn to the

**elastic curve**at B slopes up or down to the right. Hint: Referto Eq. (6-5) and Fig. 6-10. Ana. (AreaWZx = 2700 lb-ft3; slope is up to right 631.

Page 221

The positive value of tAic means that A on the

reference tangent at C. Hence the reference tangent at C slopes down to the left,

as shown. From the similar triangles ACE and CDF, we obtain DF = — tA/c = _4_

4400 ...

The positive value of tAic means that A on the

**elastic curve**lies above thereference tangent at C. Hence the reference tangent at C slopes down to the left,

as shown. From the similar triangles ACE and CDF, we obtain DF = — tA/c = _4_

4400 ...

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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