MT225 Homework 6 - Solution. Problem 1: [34] 5 ft 5 ft 5 ft A B C D. 12 kips. 10ft E F G

Transcription

1 MT225 Homework 6 - Solution Problem 1: [34] 5 ft 5 ft 5 ft C D 10ft 12 kips E G To begin, we need to find the reaction forces at supports and E. To do so, we draw a free-body diagram of the entire structure isolating it from its supports. y 5 ft 5 ft 5 ft x C D 10ft 12 kips y Ex E G x Taking the sum of forces in the x and y directions and setting the sums equal to zero, x + Ex = 0 y - 12 kips = 0

2 y = 12 kips [5] Taking the sum of moments about point and setting the sum equal to zero yields -(15ft)(12 kips) + Ex (10ft) = 0. Theefore, Ex = 18 kips [4] and x = -18 kips. [4] To find the forces in members C, G, and G, we must draw a system boundary that isolates the joints and E from their supports and cuts through the members we are interested in. The new D with the reaction forces drawn in are shown below. 12 kips 5 ft 18 kips C o 10ft G y G 18 kips E x Taking the sum of forces in the x and y directions and setting the sums equal to zero yields: C + G cos( o ) + G -18 kips + 18 kips = 0 12kips - G sin( o ) = 0 Solving the y direction equation yields

3 G = kips Since G came out positive, the force is in the assumed direction. member G is in tension. Taking the moments about point and setting the sum equal to zero yields -(5ft)(12 kips) + (10ft)(18 kips) + (10ft) G = 0. G = -12 kips Since G came out negative, the direction is opposite to the one we assumed. member G must be in compression. Going back to the x direction equilibrium equation and solving for C yields C = - G - G cos( o ) = -(-12 kips) - ( kips) cos( o ) = kips Since C is positive, the force is in the direction shown so member C is in tension. Summarizing, G: kips (Tension) [7] G: 12 kips (Compression) [7] C: 6 kips (Tension) [7]

4 Problem 2: [33] 2000 lbs C 1000 lb 12 ft E D 8 ft 8 ft 8 ft The first step is to draw a free-body diagram of the whole structure to find the reactions due to the supports. C 2000 lbs 1000 lb 12 ft R x R y E D R Dy 8 ft 8 ft 8 ft Taking the sum of forces in the x and y directions and setting the sums equal to zero yields:

5 R x = 0 lbs [3] R y + R Dy = 3000 lbs Taking the moments about point and setting the sum equal to zero yields -(1000lbs)(8ft) - (2000lbs)(16ft) + R Dy (24ft) = 0 R Dy = lbs [3] and R y = lbs. [3] Re-drawing the free-body diagram with the reaction forces in place: y x 1000 lb C 2000 lbs 12 ft D lbs E lbs 8 ft 8 ft 8 ft We are looking for the loads in members C,, and E. The easiest way to find them is to use the method of sections and draw a system boundary that cuts through those three members. Drawing a system boundary that includes point leads to the free body diagram below.

6 y 1000 lb C x lbs E 6 ft o o E 8 ft The unknown angles and sides in the diagram have been found using trigonometry. Summing the forces in the x direction and setting the sum equal to zero yields C cos( o ) + cos( o ) + E = 0. Summing the forces in the y direction and setting the sum equal to zero yields C sin( o ) - sin( o ) lbs lbs = 0. inally, taking the moment about point and setting the sum equal to zero yields -(8ft)( lbs)+ E (6ft) = 0. E = lbs. Simplifying the x and y equations, C cos( o ) + cos( o ) = lbs C sin( o ) - sin( o ) = lbs Dividing the x equation through by cos( o ) yields C + = lbs Dividing the y equation through by sin( o ) yields

7 C - = lbs dding these two equations together yields 2 C = lbs. C = lbs Substituting this into the x equation yields = lbs lbs = lbs. The negative results mean that the assume direction was incorrect. Since all members were assumed to be in tension, those that have negative forces must be in compression. C = 1389lbs (C) [8] = 833.3lbs (C) [8] E = 1778lbs (T) [8]

8 Problem 3: [33] 6 kips 8 kips C D E 9 ft I H 10 kips G 4 bays at 10 ft = 40 ft irst, draw a free body diagram and find the reaction forces at and. 6 kips 8 kips C D E 9 ft R x R y I H 10 kips G R y 4 bays at 10 ft = 40 ft Taking the sum of forces in the x and y directions and setting the sums equal to zero: Rx = 0 [4] R y + R y - 24 = 0

9 Taking the moment about, -10(20) - 6(25) - 8(35) + 40R y = 0 Ry = kips [4] R y = 8.25 kips [4] Drawing a section through members CD, CH, and HI: C o CD CH 8.25 kips I HI Taking the sum of forces in the x and y directions and setting the sums equal to zero: CD + CH cos( ) + HI = kips - CH sin( ) = 0 o o Taking the sum of moments about C and setting the sum equal to zero yields: 9 HI (15) = 0 HI = kips CH = kips

10 CD = kips Since HI and CH came out positive, they are in tension. Member CD is in compression. So, CD: kips (C) [7] CH: kips (T) [7] To find member DH, we can do another cut through CD, DH, and GH: C kips DH 8.25 kips I H 10 kips GH Only the y direction sum is needed here: DH sin( ) = 0 o DH = kips. Since it came out positive, member DH is in Tension DH: kips (T) [7]