ENGR 8
Test 3 100 pts
Name:___SOLUTIONS______________
1)
Sum Fx = 0
Ax + 500*3/5 = 0, Ax = -300 (left)
Sum Fy = 0
Ay + By – 500*4/5 = 0, Ay = 400 - By
Sum Ma = 0
-500*4/5*(5) + By*10 - 600 = 0
By = (600 + 2000)/10, By = 260 (up)
Ay = 400 – 260 = 140 (up)
2) There is an Ax, Az, Max, Maz and Bz reaction
Sum Fx = 0 , Ax = 0
Sum Fz = 0
Az + Bz – 80 =0, Az = 80 - Bz
Sum Max = 0
-80*6 + Bz*6 + Max =0, Max = 480 – 6*Bz
Sum May =0
-80*1.5 + Bz*3 = 0, Bz = 40 (up)
Az = 80 – Bz , Az = 40 (up)
Therefore Max = 480 – 6*Bz , Max = 240 lb-ft
Sum Maz = 0
Nothing is providing moment in Z, so Maz = 0
3) Find the force in each member using the Method of Joints.
There is a roller at A. State whether T or C.
3) Use method of joints, start at D (1 force, 2 unkn)
Assume Fdb, Fda are in tension
DAB is a 1.5, 2,2.5 triangle (a 3-4-5 divided by 2)
Joint D
Sum Fy = 0
Fdb*3/5 – 300 = 0, Fdb = 1500/3 , Fdb= 500 (T)
Sum Fx = 0
Fdb*4/5 + Fda = 0, Fda = -500*4/5, Fda= -400 (C)
Joint A
Sum Fy=0
Fab = 0 (no other forces in y at A)
4) Use the Method of Sections to find the force in
members FE, BE and BC and state whether T or C
4) First, find reactions at B and C
Sum Fx = 0
Bx = 0
Sum Fy = 0
Bx + Cx = 600
Sum Mc = 0
-600*3 + 300*6 – By*3 = 0
By =0
Cy = 900
Now look at left section:
Assume Ffe, Fbe, Fbc in tension
Sum Mb = 0
-Ffe*3 + 300*3 = 0, Ffe = 300 (T)
Sum Me = 0
Fbc*3 + 300*6 = 0, Fbc = -600(C)
Sum Fy = 0
Fbe*sin(45) – 300 = 0, Fbe = 300/sin(45) = 424 (T)
5) Fba acts along the line AB (i.e. a 45 degree angle)
First find Fab, then Cx, Cy
Sum Mc = 0
-Fab*sin(45)*3 + Fab*cos(45)*1 + 800 + 400*2=0
Fab(cos(45)-3sin(45)) + 1600=0
Fab = 1600/(2 sqrt(2)/2) = 1600/1.414,
Fab = 1131 (C)
Sum Fx = 0
Fab*cos(45) + Cx = 0, Cx = -800 (left)
Sum Fy = 0
Fab*sin(45) – 400 + Cy = 0, Cy= -400 (down)