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Homework 4 Solutions

# Homework 4 Solutions - Mechanics of Deformables Spring 2007...

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Unformatted text preview: Mechanics of Deformables Spring 2007 Homework 4 (Due Tuesday 02/130007) 4.1 The ship is pushed through the water using an A-36 steel propeller shaft that is 8 m long, measured from the propeller to the thrust bearing D at the engine. If it has an outer diameter of 400 mm and a wall thickness of 50 mm, determine the amount of axial contraction of the shaft when the propeller exerts a force on the shaft of 5 kN. The bearings at B and C are journal bearings. 17- #30 kN H -——> 5' KN ‘ R S __ FL .5.0 (log) __ ﬂ 2: f a E Tr (0.4i—1'r0-3L) (202))“ +%: (l0) #6)”. --- r 3.63800 7. - 3- 6+ (to #3)“ Z -— V1 Means WOJQS "1th Nels D - 4.2 The A—36 steel column is used to support the symmetric loads from the two ﬂoors of a building. Determine the vertical displacement of its top A if P1 = 40 kip, P2 = 62 kip, and the column has a cross—sectional area of 23.4 - 2 1n . K S=ZPL 13 E 1' ago-o ((2)020 93-+c29-o)“°3) U“? \L it») w? .1— (—- 20‘?) (l1)(l2.] Q‘I-c - ~93 ‘H 003)) 62 ,1, 62 WP \h‘ 1’ k‘P “230*,de S {UM Moms 9 W‘s,qu dowoi’ds C. 4.3 The post is made of Douglas ﬁr and has a diameter of 60 m. If it is subjected to the load of 20 kN and the soil provides a frictional resistance that is uniformly distributed along its sides of w = 4 kN/m, determine the force F at its bottom needed for equilibrium. Also, what is the displacement of the top of the post A with respect to its bottom B? Neglect the weight of the post. T’sr’ce. TFF.5 = , \$30“; F+6'00--20=O " F”- lQo kid 3‘“ \$20 “4 BN5 4(1) = 8-0 KN . j . ‘ti D‘s IMO-Went E L __ r: gens: j 1-1;) 9s \i/ 0 mpg Fiﬁ) 2—H : _l.. [(43,20'Jc1} HE 0 Linn '” _‘ (ag‘rﬂzogﬂo PIE 3 7 #‘8ét1 ~ * 32 W m .w m 4.4 The assembly consists of two A—36 steel suspender rods AC and BD attached to the 100-lb uniform rigid beam AB. Determine the position x for the 300—lb loading so that the beam remains in a horizontal position both before and after the load is applied. Each rod has a diameter of 0.5 in. =r-———— Bum—4 +23!“ 2 0‘. Flat”) - 30th - 100(15) = 0 + 2M. :2 0; -F:4c(30) + MSO—x} + 10005} I: 0 ﬁc- - 350 — 10:: {2) 54 I 8’ From Eq. (1) and (2). (350 - 10:)(15) = (10: + 50x25) .415 u: A 3 (5" 5250-111: = 250.: + 1250 molt x- 10m Ans ...
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Homework 4 Solutions - Mechanics of Deformables Spring 2007...

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