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ch01-03 stress & strain & properties

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02 Solutions 46060 5/6/10 1:45 PM Page 10<br />

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*2–16. The square deforms into the position shown by the<br />

dashed lines. Determine the average normal <strong>strain</strong> along<br />

each diagonal, AB and CD. Side D¿B¿ remains horizontal.<br />

Geometry:<br />

D¿<br />

y<br />

D<br />

B¿<br />

B<br />

3 mm<br />

AB = CD = 250 2 + 50 2 = 70.7107 mm<br />

C¿D¿ =253 2 + 58 2 - 2(53)(58) cos 91.5°<br />

= 79.5860 mm<br />

B¿D¿ =50 + 53 sin 1.5° - 3 = 48.3874 mm<br />

53 mm<br />

A<br />

91.5<br />

C<br />

50 mm<br />

C¿<br />

x<br />

AB¿ =253 2 + 48.3874 2 - 2(53)(48.3874) cos 88.5°<br />

= 70.8243 mm<br />

50 mm<br />

8 mm<br />

Average Normal Strain:<br />

e AB =<br />

AB¿ -AB<br />

AB<br />

=<br />

70.8243 - 70.7107<br />

70.7107<br />

= 1.61A10 - 3 B mm>mm<br />

Ans.<br />

e CD =<br />

C¿D¿ -CD<br />

CD<br />

=<br />

79.5860 - 70.7107<br />

70.7107<br />

= 126A10 - 3 B mm>mm<br />

Ans.<br />

•2–17. The three cords are attached to the ring at B.When<br />

a force is applied to the ring it moves it to point B¿ , such<br />

that the normal <strong>strain</strong> in AB is P AB and the normal <strong>strain</strong> in<br />

CB is P CB . Provided these <strong>strain</strong>s are small, determine the<br />

normal <strong>strain</strong> in DB. Note that AB and CB remain<br />

horizontal and vertical, respectively, due to the roller guides<br />

at A and C.<br />

A<br />

A¿<br />

L<br />

B<br />

B¿<br />

Coordinates of B (L cos u, L sin u)<br />

Coordinates of B¿ (L cos u + e AB L cos u, L sin u + e CB L sin u)<br />

L DB¿ = 2(L cos u + e AB L cos u) 2 + (L sin u + e CB L sin u) 2<br />

D<br />

u<br />

C<br />

C¿<br />

L DB¿ = L2cos 2 u(1 + 2e AB + e 2 AB) + sin 2 u(1 + 2e CB + e 2 CB)<br />

e AB<br />

e CB<br />

Since and are small,<br />

L DB¿ = L21 + (2 e AB cos 2 u + 2e CB sin 2 u)<br />

Use the binomial theorem,<br />

L DB¿ = L ( 1 + 1 2 (2 e AB cos 2 u + 2e CB sin 2 u))<br />

= L ( 1 + e AB cos 2 u + e CB sin 2 u)<br />

Thus,<br />

e DB = L( 1 + e AB cos 2 u + e CB sin 2 u) - L<br />

L<br />

e DB = e AB cos 2 u + e CB sin 2 u<br />

Ans.<br />

10

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