2004 Final
McGill University, Faculty of Engineering, Department Mechanical Engineering
MECHANICS 1 – MECH 210-001 FINAL EXAMINATION Monday December 13th 2004, 14:00-17:00 (3 hours)
Examiner: Prof. Pascal Hubert Associate Examiner: Prof. Larry Lessard • • •
• • •
Closed book, Faculty approved calculator The exam has 7 questions (100 points total) o Question 1 Short answers (10 points) o Question 2 3-D equilibrium (15 points) o Question 3 Trusses and beams (15 points) o Question 4 Frames and beams (15 points) o Question 5 Principle of virtual work (15 points) o Question 6 Potential energy and stability (15 points) o Question 7 Application (15 points) Draw free body diagrams and reference axis. State your assumptions. Equations sheet on last page
Page 1 of 8
Fall 2004
McGill University, Faculty of Engineering, Department Mechanical Engineering
Question 1 (10 points)
Short answers
a) (5 points) Can you assume symmetry to compute the forces in the members? Justify in one line. A is a pin joint and O is a roller.
500 lb 500 lb 500 lb 500 lb 500 lb 500 lb 250 lb 250 lb 250 lb
b) (5 points) Draw the free body diagram of the gate ABCD. D is a roller and A is a pin joint. The water has a constant density. Do not solve, just show the forces and their direction.
WATER
Page 2 of 8
Fall 2004
McGill University, Faculty of Engineering, Department Mechanical Engineering
Question 2 (15 points)
Fall 2004
3-D equilibrium
Under the action of the 40 Nm torque (couple) applied to the vertical shaft, the restraining cable AC limits the rotation of the arm OA and attached shaft to an angle of 60° measured from the y-axis. • • • •
The collar D fastened to the shaft prevents downward motion of the shaft in the bearing. The bearing prevents rotation around xaxis and y-axis. Bending moment expressed as a vector is normal to the shaft axis (z-axis). Shear force is also normal to the shaft axis (z-axis).
D
Calculate in the shaft at section B: • the bending moment M • the compression P • the shear force V.
Question 3 (15 points)
Trusses and beams
A floor truss is loaded as shown. a) Determine the force in members CF, EF and EG. State whether the force is in tension or compression. Support A is a roller and K is a pin joint. b) If you replace the floor truss by a beam of constant cross section, draw the shear force and bending moment diagram in the beam and find the absolute maximum shear force and bending moment and their location.
Page 3 of 8
C
McGill University, Faculty of Engineering, Department Mechanical Engineering
Question 4 (15 points)
Fall 2004
Frames and beams
Neglect friction and the weights of the members. A is a pin joint. B
a) Compute the magnitudes of the pin reactions at A and C for the folding table shown.
Detail
b) Draw the shear force and bending moment diagram in the table top AB and find the absolute maximum shear force and bending moment and their location. Question 5 (15 points)
Principle of virtual work
The slender rod AB is attached to a collar A and rests on a small wheel C. Neglect the radius of the wheel, the effect of friction and the weight of the rod. a) Draw the diagram of active forces and moments. b) Write the expression for the virtual work of the system. c) Derive an expression for the magnitude of force Q required to maintain equilibrium of the rod. Question 6 (15 points)
Potential energy and stability
Rigid rod AB is attached to a hinge at A and two springs, each of constant k. h = 30 in, k = 4 lb/in and W = 40 lb. Each spring can act in either tension or compression. The spring unstretched position is shown. Assume that the springs remain horizontal. a) Write the expression of the potential energy of the system. b) Determine the smallest value of d for which the equilibrium of the rod is stable in the position shown (Angle between the rod AB and the vertical is 0°).
Page 4 of 8
McGill University, Faculty of Engineering, Department Mechanical Engineering
Question 7 (15 points)
Fall 2004
Application
You are working for a wind turbine manufacturing firm and just obtained a major contract from Hydro Quebec. You need to provide specific information to your sub-contractors. The 1.5 MW three blades turbine shown has a 60 m rotor diameter. The rotor-nacelle-tower weight is 80 000 kg and acts at G. In the configuration shown and in a 20m/s wind, the aerodynamic forces generated from the blades act at points 1, 2 and 3. The blade aerodynamic forces consist of an in-plane (plane of rotation: y-z plane) and out-of-place (perpendicular to plane of rotation: xaxis) components. Table 1 gives the coordinate of these points and Table 2 the magnitude and orientation of the forces calculated by your aerodynamic engineer. Table 2
Table 1 Coordinate (x,y,z) (m) (0,0,0) (0,0,60) (-2,0,40) (-2,17.3,70) (-2,-17.3,70)
Point A Point G Point 1 Point 2 Point 3
Magnitude (kN) Blade 1 Blade 2 Blade 3
29.1 18.4 18.4
Orientation vunit v v vector i , j , k (0.961,0.275,0) (0.976,-0.108,0.188) (0.976,-0.108,-0.188)
(
)
Note: the force vector represented on the figure are not to scale
3 z
z
2
G
3 2
2 3
G
1
1
A
A y
Page 5 of 8
G
1
A x
McGill University, Faculty of Engineering, Department Mechanical Engineering
Fall 2004
a) What is the equivalent force at point G? b) Compute the reactions at the fixed tower base (Point A) Decompose the reaction in x,y,z components. c) What is the magnitude of the dominant bending moment in the tower, where is acting and what force component causes this moment? (use information found in a) and b) to justify your answer). d) Based on the information found in c), draw the most important deformation mode of the tower. e) The blade cross section is shown below. The blade is composed of an airfoil skin (1) and two spar caps (2) and (3). The airfoil (1) moment of inertia Ix (about the x-axis) is 2.18x108 mm4. The spar caps have a rectangular cross section with a length of 840 mm and a thickness of 30 mm. Calculate the moment of inertia of the blade section about the x-axis.
200 mm 3 420 mm 420 mm
x
1 2 840 mm
Page 6 of 8
McGill University, Faculty of Engineering, Department Mechanical Engineering
Trigonometry sin θ = a c cosθ = b c tan θ = a b
Fall 2004
c a
c = a2 + b2 sin 2 θ + cos 2 = 1
θ b
sin θ 2 = 1 2 (1 − cosθ ) cosθ 2 = 1 2 (1 + cosθ ) sin 2θ = 2 sin θ cosθ
cos 2θ = cos 2 θ − sin 2 θ
sin (a ± b ) = sin a cos b ± cos a sin b cos(a ± b ) = cos a cos b m sin a sin b a sin A = b sin B c 2 = a 2 + b 2 − 2ab cos C
B c
a
A
c 2 = a 2 + b 2 + 2ab cos D
C b
Dot product v v P ⋅ Q = PQ cos θ v v P ⋅ Q = Px Q x + Py Q y + Pz Q z Cross v v v product v v P × Q = (Py Qz − Pz Q y )i − (Pz Qx − Px Qz ) j + (Px Q y − Py Qx )k Derivatives and integrals dx n x n +1 = nx n −1 ∫ x n dx = dx n +1 ⎛u⎞ d ⎜ ⎟ v du − u dv d (uv ) dv du ⎝ v ⎠ = dx dx =u +v 2 dx dx dx dx v d sin x d cos x = cos x = − sin x dx dx d tan x 1 = dx cos 2 x
Page 7 of 8
D
McGill University, Faculty of Engineering, Department Mechanical Engineering
Position vector v v v v rAB = ( x B − x A )i + ( y B − y A ) j + ( z B − z A )k v v v Unit vector u AB = rAB rAB Force definition v v v v F = F cos θ x i + cos θ y j + cos θ z k v v v v F = Fx i + Fy j + Fz k v v F = Fu AB v v v Moment M = r × F Equilibrium ∑ Fx = 0 ∑ Fy = 0
(
)
∑M
o
=0
Internal forces V
x
M
x
Vo
xo
Mo
xo
∫ dV = − ∫ wdx ∫ dM = ∫ Vdx
Virtual work δU = Fδs δU = Mδθ Potential energy V g = mgh Ve = (1 2)ks 2 Section properties I = I + Ad 2
Units and constants 1 lb = 4.448 N 1 kip = 1000 lbs 1 ft = 0.3048 m 1 ft = 12 in Gravitational acceleration g = 9.81 m/s2
Page 8 of 8
Fall 2004
MECHANICS 1 – MECH 210-001 FINAL EXAMINATION Monday December 13th 2004, 14:00-17:00 (3 hours)
Examiner: Prof. Pascal Hubert Associate Examiner: Prof. Larry Lessard • • •
• • •
Closed book, Faculty approved calculator The exam has 7 questions (100 points total) o Question 1 Short answers (10 points) o Question 2 3-D equilibrium (15 points) o Question 3 Trusses and beams (15 points) o Question 4 Frames and beams (15 points) o Question 5 Principle of virtual work (15 points) o Question 6 Potential energy and stability (15 points) o Question 7 Application (15 points) Draw free body diagrams and reference axis. State your assumptions. Equations sheet on last page
Page 1 of 8
Fall 2004
McGill University, Faculty of Engineering, Department Mechanical Engineering
Question 1 (10 points)
Short answers
a) (5 points) Can you assume symmetry to compute the forces in the members? Justify in one line. A is a pin joint and O is a roller.
500 lb 500 lb 500 lb 500 lb 500 lb 500 lb 250 lb 250 lb 250 lb
b) (5 points) Draw the free body diagram of the gate ABCD. D is a roller and A is a pin joint. The water has a constant density. Do not solve, just show the forces and their direction.
WATER
Page 2 of 8
Fall 2004
McGill University, Faculty of Engineering, Department Mechanical Engineering
Question 2 (15 points)
Fall 2004
3-D equilibrium
Under the action of the 40 Nm torque (couple) applied to the vertical shaft, the restraining cable AC limits the rotation of the arm OA and attached shaft to an angle of 60° measured from the y-axis. • • • •
The collar D fastened to the shaft prevents downward motion of the shaft in the bearing. The bearing prevents rotation around xaxis and y-axis. Bending moment expressed as a vector is normal to the shaft axis (z-axis). Shear force is also normal to the shaft axis (z-axis).
D
Calculate in the shaft at section B: • the bending moment M • the compression P • the shear force V.
Question 3 (15 points)
Trusses and beams
A floor truss is loaded as shown. a) Determine the force in members CF, EF and EG. State whether the force is in tension or compression. Support A is a roller and K is a pin joint. b) If you replace the floor truss by a beam of constant cross section, draw the shear force and bending moment diagram in the beam and find the absolute maximum shear force and bending moment and their location.
Page 3 of 8
C
McGill University, Faculty of Engineering, Department Mechanical Engineering
Question 4 (15 points)
Fall 2004
Frames and beams
Neglect friction and the weights of the members. A is a pin joint. B
a) Compute the magnitudes of the pin reactions at A and C for the folding table shown.
Detail
b) Draw the shear force and bending moment diagram in the table top AB and find the absolute maximum shear force and bending moment and their location. Question 5 (15 points)
Principle of virtual work
The slender rod AB is attached to a collar A and rests on a small wheel C. Neglect the radius of the wheel, the effect of friction and the weight of the rod. a) Draw the diagram of active forces and moments. b) Write the expression for the virtual work of the system. c) Derive an expression for the magnitude of force Q required to maintain equilibrium of the rod. Question 6 (15 points)
Potential energy and stability
Rigid rod AB is attached to a hinge at A and two springs, each of constant k. h = 30 in, k = 4 lb/in and W = 40 lb. Each spring can act in either tension or compression. The spring unstretched position is shown. Assume that the springs remain horizontal. a) Write the expression of the potential energy of the system. b) Determine the smallest value of d for which the equilibrium of the rod is stable in the position shown (Angle between the rod AB and the vertical is 0°).
Page 4 of 8
McGill University, Faculty of Engineering, Department Mechanical Engineering
Question 7 (15 points)
Fall 2004
Application
You are working for a wind turbine manufacturing firm and just obtained a major contract from Hydro Quebec. You need to provide specific information to your sub-contractors. The 1.5 MW three blades turbine shown has a 60 m rotor diameter. The rotor-nacelle-tower weight is 80 000 kg and acts at G. In the configuration shown and in a 20m/s wind, the aerodynamic forces generated from the blades act at points 1, 2 and 3. The blade aerodynamic forces consist of an in-plane (plane of rotation: y-z plane) and out-of-place (perpendicular to plane of rotation: xaxis) components. Table 1 gives the coordinate of these points and Table 2 the magnitude and orientation of the forces calculated by your aerodynamic engineer. Table 2
Table 1 Coordinate (x,y,z) (m) (0,0,0) (0,0,60) (-2,0,40) (-2,17.3,70) (-2,-17.3,70)
Point A Point G Point 1 Point 2 Point 3
Magnitude (kN) Blade 1 Blade 2 Blade 3
29.1 18.4 18.4
Orientation vunit v v vector i , j , k (0.961,0.275,0) (0.976,-0.108,0.188) (0.976,-0.108,-0.188)
(
)
Note: the force vector represented on the figure are not to scale
3 z
z
2
G
3 2
2 3
G
1
1
A
A y
Page 5 of 8
G
1
A x
McGill University, Faculty of Engineering, Department Mechanical Engineering
Fall 2004
a) What is the equivalent force at point G? b) Compute the reactions at the fixed tower base (Point A) Decompose the reaction in x,y,z components. c) What is the magnitude of the dominant bending moment in the tower, where is acting and what force component causes this moment? (use information found in a) and b) to justify your answer). d) Based on the information found in c), draw the most important deformation mode of the tower. e) The blade cross section is shown below. The blade is composed of an airfoil skin (1) and two spar caps (2) and (3). The airfoil (1) moment of inertia Ix (about the x-axis) is 2.18x108 mm4. The spar caps have a rectangular cross section with a length of 840 mm and a thickness of 30 mm. Calculate the moment of inertia of the blade section about the x-axis.
200 mm 3 420 mm 420 mm
x
1 2 840 mm
Page 6 of 8
McGill University, Faculty of Engineering, Department Mechanical Engineering
Trigonometry sin θ = a c cosθ = b c tan θ = a b
Fall 2004
c a
c = a2 + b2 sin 2 θ + cos 2 = 1
θ b
sin θ 2 = 1 2 (1 − cosθ ) cosθ 2 = 1 2 (1 + cosθ ) sin 2θ = 2 sin θ cosθ
cos 2θ = cos 2 θ − sin 2 θ
sin (a ± b ) = sin a cos b ± cos a sin b cos(a ± b ) = cos a cos b m sin a sin b a sin A = b sin B c 2 = a 2 + b 2 − 2ab cos C
B c
a
A
c 2 = a 2 + b 2 + 2ab cos D
C b
Dot product v v P ⋅ Q = PQ cos θ v v P ⋅ Q = Px Q x + Py Q y + Pz Q z Cross v v v product v v P × Q = (Py Qz − Pz Q y )i − (Pz Qx − Px Qz ) j + (Px Q y − Py Qx )k Derivatives and integrals dx n x n +1 = nx n −1 ∫ x n dx = dx n +1 ⎛u⎞ d ⎜ ⎟ v du − u dv d (uv ) dv du ⎝ v ⎠ = dx dx =u +v 2 dx dx dx dx v d sin x d cos x = cos x = − sin x dx dx d tan x 1 = dx cos 2 x
Page 7 of 8
D
McGill University, Faculty of Engineering, Department Mechanical Engineering
Position vector v v v v rAB = ( x B − x A )i + ( y B − y A ) j + ( z B − z A )k v v v Unit vector u AB = rAB rAB Force definition v v v v F = F cos θ x i + cos θ y j + cos θ z k v v v v F = Fx i + Fy j + Fz k v v F = Fu AB v v v Moment M = r × F Equilibrium ∑ Fx = 0 ∑ Fy = 0
(
)
∑M
o
=0
Internal forces V
x
M
x
Vo
xo
Mo
xo
∫ dV = − ∫ wdx ∫ dM = ∫ Vdx
Virtual work δU = Fδs δU = Mδθ Potential energy V g = mgh Ve = (1 2)ks 2 Section properties I = I + Ad 2
Units and constants 1 lb = 4.448 N 1 kip = 1000 lbs 1 ft = 0.3048 m 1 ft = 12 in Gravitational acceleration g = 9.81 m/s2
Page 8 of 8
Fall 2004
