A member is composed of three solid rods: e Rod A is 12 feet long, has a diamete

Question

A member is composed of three solid rods: e Rod A is 12 feet long, has a diameter of 2.00 inch, and an elastic modulus of 29,000 ksi e Rod B is 6 feet long, has a diameter of 1.00 inch, and an elastic modulus of 10,000 ksi . Rod C is 3 feet long, has a diameter of 0.50 inch, and an elastic modulus of 6,500 ksi All three rods have a Poisson's ratio of 0.3. Determine the increase in member length after a 20 kip weight is suspended from the end of the member. 1) 2) Determine the diameter of Rod C to the nearest 0.00001 inch, after the weight is added. 20 kips

Explanation / Answer

First individually finding increase in length of each member

Rod A

Length L=12 ft=144 inch and diameter d= 2 in. Now area of rod A is A=pi*d²/4. = 3.147*2²/4 = 3.147 in². Therefore increase in length is PL/AE where P is force applied = 20kips = 20000 lb,E is elastic modulus= 29000 ksi = 29000*1000 lb/in2

Therefore , l1= 20000*144/3.147*(29000*1000)

l1= 0.03155 inch

Now similarly change in rod B l2 is

l2=PL/AE, now area for rod B is pi*d²/4= 3.147*1²/4 = 0.787 in² and length of rod B is L =6 ft= 72 in and E = 10000 ksi= 10000*1000 psi

l2= 20000*72/0.787*(10000*1000)

l2= 0.183 in

Now for rod C l3 is PL/AE where Area A= pi*d²/4= 3.147*0.5²/4= 0.197 in² and L= 3 ft= 36 in , E = 6500 ksi= 6500*1000 psi

l3= 20000*36/0.197*(6500*1000)

l3 = .562 in

Therefore total change in member length l =l1+l2+l3 = 0.03155+0.183+0.562 = 0.7765 inch

Now change in diameter of rod C is to be findout

Initial volume and final volume of rod remains same,therefore V1=V2

(pi*d1²/4)*L1= (pi*d2²/4)*L2 where d1 and d2 are initial and final diameter and L1 and L2 are initial and final length. Therefore equation becomes,

d2= d1*?L1/L2 Now L1= 36 in and L2 =36+0.562= 36.562 in

d2= 0.5*?36/36.562

d2= 0.5* ?0.984628

d2=0.5*0.99228

d2= 0.49614 in Therefoe final diameter is 0.49614 in

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