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ressy
Answers (3)
How many columns are needed to provide a safety factor of two against rupture.
Given: t=3.3sec. Vo=0m/s g=9.81m/s^2 V=340m/s^2 Solution: H=Vot+0.5gt^2 H=(0.5)(9.81)(3.3^2) H=53.42m H=vt H=340(5) H=1700m H=53.42+1700 [H=1753.42m] :)
Given: t=5sec. Vo=0m/s g=9.81m/s^2 V=320m/s^2 Solution: H=Vot+0.5gt^2 H=(0.5)(9.81)(5^2) H=122.625m H=vt H=320(5) H=1600m H=122.625+1600 [H=1722.625m] :)
