20.1104 Intro. to Engng Analysis (Studio/Laptop Version)
Test #4, December 6, 1996
Solution
For all problems, 95% of credit is given for complete FBD, statement of theory,
definition of required vectors, substitute of vectors into theory, and identification of
scalar equations to be solved.
1. (30 points) The bar AB has a built in support at A. The tension in the cable BC is 10
kN.
(a) Draw the free-body diagram of the bar.
(b) Determine the reactions at A.
(a) Draw the FBD of the bar AB:
(b) To determine the reaction forces at A
first, define all forces:
A = (AXi + AYj + Azk) kN
M = (MXi + MYj + Mzk) kN-m
T = 10 kN
-2i - 6j + 3k
= (-2.857i-8.571j+4.286k) kN
{(-2)2+(-6)2+(3)2)}1/2
rT/A = (5i + 6j + k) m
Theory: Static equilibrium (F = 0, M = 0)
Summing the forces and setting them equal to zero:
FX = 0  AX - 2.857 kN = 0
AX = 2.857 kN
FY = 0  AY - 8.571 kN = 0
AY = 8.571 kN
FZ = 0  AZ + 4.286 kN = 0
AZ = -4.286 kN
Summing the moments about point A and setting them equal to zero:
MA = MXi + Myj + MZk + (5i + 6j + k)x(-2.857i-8.571j+4.286k) kN-m
Solve the cross product on Maple to get:
MA = MXi + Myj + MZk + (34.287i-24.287j-25.713k) kN-m
MX = 0  MX + 34.287 kN-m = 0
MX = 34.287 kN-m
MY = 0  MY - 24.287 kN-m = 0
MY = 24.287 kN-m
MZ = 0  MZ - 25.713 kN-m = 0
MZ = 25.713 kN-m
2. (35 points) The bridge truss supports five forces (F = 75 kip). The dimension L = 25 ft.
(a) Determine the reactions at the supports A and H.
(b) Determine the force in members JK, CK and DK.
(a) First, draw the FBD of the entire truss to solve for reaction forces:
Define all forces:
A = (AXi + AYj) kip
F = -75j kip
H = CYj kip
Theory: Static equilibrium (F = 0, M = 0)
Summing the forces and setting them equal to zero:
FX = 0  AX = 0
FY = 0  CY + AY - 5(75) kip = 0
(1)
Summing the moments about point A and setting them equal to zero:
MA = -(25)(75)k kip-ft - (50)(75)k kip-ft - (75)(75)k kip-ft - (100)(75)k kip-ft
- (125)(75)k kip-ft + (150)(CY)k ft
CY = 187.5 kip
Substituting into equation (1) to get
AY = 187.5 kip
Next, cut through members JK, CK and DK, leaving the left hand section with which to
solve for static equilibrium:
Define all forces:
A = (187.5) kip
F = -75j kip
CD = CDi kip
JK = JKi kip
CK = (CKcos45i - CKsin45j) kip
Theory: Static equilibrium (F = 0, M = 0)
Summing the forces and setting them equal to zero:
FY = 0  187.5 kip - 2(75) kip - CKsin45 = 0
CK = 53.03 kip (T)
Summing the moments about point C and setting them equal to zero:
MC = (25)(75)k kip-ft + (25)(JK)k ft - (50)(187.5)k kip-ft
so JK = 300 kip (T)
Finally, drawing joint D:
Define all forces:
CD = -CDi kip
DE = DEi kip
DK = (-DKj) kip
Theory: Static equilibrium (F = 0, M = 0)
FY = 0  DK = 0
We can see that DK is a zero force member.
3. (35 points) For the frame shown, determine the reactions at the built-in support A and
the force exerted on the structure by the smooth floor at C.
First, draw the FBD of the entire frame to solve for reaction forces:
Define all vectors:
A = (AXi + AYj) lb
MA = (MAk) lb-ft
F = -200j lb
C = CYj lb
Theory: Static equilibrium (F = 0, M = 0)
Summing the forces and setting them equal to zero:
FX = 0  AX = 0
FY = 0  CY + AY - 200 lb = 0
(1)
Summing the moments about point A and setting them equal to zero:
MA = MAk - (9)(200)k lb-ft + (12)(CY)k ft
MA + 12CY ft = 1800 lb-ft
(2)
Next, draw the FBD of member BC to solve for the remaining unknowns:
Define all forces:
B = (BXi + BYj) lb
F = -200j lb
C = CYj lb
Theory: Static equilibrium (F = 0, M = 0)
Summing the forces and setting them equal to zero:
FX = 0  BX = 0
FY = 0  BY + CY - 200 lb = 0
(3)
Summing the moments about point B and setting them equal to zero:
MB = -(3)(200)k lb-ft + (6)(CY)k ft
so CY = 100 lb
Substituting into equation (3) to get
BY = 100 lb
Substituting into equation (1) to get
AY = 100 lb
Substituting into equation (2) to get
MA = 600 lb-ft