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
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 63.435 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(63.435 o ) + G -18 kips + 18 kips = 0 12kips - G sin(63.435 o ) = 0 Solving the y direction equation yields
G = 13.416 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(63.435 o ) = -(-12 kips) - (13.416 kips) cos(63.435 o ) = 6.000 kips Since C is positive, the force is in the direction shown so member C is in tension. Summarizing, G: 13.416 kips (Tension) [7] G: 12 kips (Compression) [7] C: 6 kips (Tension) [7]
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:
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 = 1666.67 lbs [3] and R y = 1333.33 lbs. [3] Re-drawing the free-body diagram with the reaction forces in place: y x 1000 lb C 2000 lbs 12 ft D 1333.33 lbs E 1666.67 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.
y 1000 lb C x 1333.33 lbs E 6 ft 36.870 o 36.870 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(36.870 o ) + cos(36.870 o ) + E = 0. Summing the forces in the y direction and setting the sum equal to zero yields C sin(36.870 o ) - sin(36.870 o ) -1000 lbs + 1333.33 lbs = 0. inally, taking the moment about point and setting the sum equal to zero yields -(8ft)(1333.33 lbs)+ E (6ft) = 0. E = 1777.77 lbs. Simplifying the x and y equations, C cos(36.870 o ) + cos(36.870 o ) = -1777.77 lbs C sin(36.870 o ) - sin(36.870 o ) = -333.333 lbs Dividing the x equation through by cos(36.870 o ) yields C + = -2222.22lbs Dividing the y equation through by sin(36.870 o ) yields
C - = -555.556 lbs dding these two equations together yields 2 C = -2777.78 lbs. C = -1388.89 lbs Substituting this into the x equation yields = -2222.22lbs + 1388.89 lbs = -833.33 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]
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
Taking the moment about, -10(20) - 6(25) - 8(35) + 40R y = 0 Ry = 15.75 kips [4] R y = 8.25 kips [4] Drawing a section through members CD, CH, and HI: C 60.945 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(60.945 ) + HI = 0 8.25 kips - CH sin(60.945 ) = 0 o o Taking the sum of moments about C and setting the sum equal to zero yields: 9 HI - 8.25(15) = 0 HI = 13.75 kips CH = 9.438 kips
CD = -18.333 kips Since HI and CH came out positive, they are in tension. Member CD is in compression. So, CD: 18.333 kips (C) [7] CH: 9.438 kips (T) [7] To find member DH, we can do another cut through CD, DH, and GH: C 18.333 kips DH 8.25 kips I H 10 kips GH Only the y direction sum is needed here: 8.25-10 + DH sin(60.945 ) = 0 o DH = 2.002 kips. Since it came out positive, member DH is in Tension DH: 2.002 kips (T) [7]