Baixe Solucionário - Projeto de Máquinas 4 Ed. e outras Exercícios em PDF para Engenharia Mecânica, somente na Docsity!
PROBLEM 1-
Statement: It is often said, "Build a better mousetrap and the world will beat a path to your door." Consider this problem and write a goal statement and a set of at least 12 task specifications that you would apply to its solution. Then suggest 3 possible concepts to achieve the goal. Make annotated, freehand sketches of the concepts.
Solution:
Goal Statement: Create a mouse-free environment. Task Specifications:
- Cost less than $1.00 per use or application.
- Allow disposal without human contact with mouse.
- Be safe for other animals such as house pets.
- Provide no threat to children or adults in normal use.
- Be a humane method for the mouse.
- Be environmentally friendly.
- Have a shelf-life of at least 3 months.
- Leave no residue.
- Create minimum audible noise in use.
- Create no detectable odors within 1 day of use.
- Be biodegradable.
- Be simple to use with minimal written instructions necessary. Concepts and sketches are left to the student. There are an infinity of possibilities.
PROBLEM 1-
Statement: A bowling machine is desired to allow quadriplegic youths, who can only move a joystick, to engage in the sport of bowling at a conventional bowling alley. Consider the factors involved, write a goal statement, and develop a set of at least 12 task specifications that constrain this problem. Then suggest 3 possible concepts to achieve the goal. Make annotated, freehand sketches of the concepts.
Solution:
Goal Statement: Create a means to allow a quadriplegic to bowl. Task Specifications:
- Cost no more than $2 000.
- Portable by no more than two able-bodied adults.
- Fit through a standard doorway.
- Provide no threat of injury to user in normal use.
- Operate from a 110 V, 60 Hz, 20 amp circuit.
- Be visually unthreatening.
- Be easily positioned at bowling alley.
- Have ball-aiming ability, controllable by user.
- Automatically reload returned balls.
- Require no more than 1 able-bodied adult for assistance in use.
- Ball release requires no more than a mouth stick-switch closure.
- Be simple to use with minimal written instructions necessary. Concepts and sketches are left to the student. There are an infinity of possibilities.
PROBLEM 1-
Statement: Convert a mass of 1 000 lbm to (a) lbf, (b) slugs, (c) blobs, (d) kg.
Units: blob lbf sec
^2
in
Given: Mass M := 1000 lb
Solution: See Mathcad file P0104.
- To determine the weight of the given mass, multiply the mass value by the acceleration due to gravity, g.
W := M g W = 1000 lbf
- Convert mass units by assigning different units to the units place-holder when displaying the mass value.
Slugs (^) M =31.081 slug Blobs (^) M =2.59 blob Kilograms (^) M =453.592 kg
PROBLEM 1-
Statement: A 250-lbm mass is accelerated at 40 in/sec2. Find the force in lb needed for this acceleration.
Given: Mass M := 250 lb Acceleration a 40 in sec^2
Solution: See Mathcad file P0105.
- To determine the force required, multiply the mass value, in slugs, by the acceleration in feet per second squared.
Convert mass to slugs: (^) M =7.770 slug Convert acceleration to feet per second squared: (^) a =3.333 s -^2 ft F := M a F =25.9 lbf
PROBLEM 1-
Statement: Prepare an interactive computer program (using, for example, Excell, Mathcad, or TKSolver) from which the cross-sectional properties for the shapes shown in the inside front cover can be calculated. Arrange the program to deal with both ips and SI unit systems and convert the results between those systems.
Solution: See the inside front cover and Mathcad file P0107.
- Rectangle, let: b 3 in h 4 in Area (^) A b h A 12.000 in^2 A 7742 mm^2
Moment about x -axis (^) Ix b h
^3
12 Ix 16.000 in ^4 Ix 6.660 106 mm^4
Moment about y -axis (^) Iy h b
^3
12 Iy 9.000 in ^4 Iy 3.746 106 mm^4
Radius of gyration about x -axis (^) k (^) x Ix (^) A k (^) x 1.155 in k (^) x 29.329 mm
Radius of gyration about y -axis (^) k (^) y Iy (^) A k (^) y 0.866 in k (^) y 21.997 mm
Polar moment of inertia (^) J (^) z Ix Iy J (^) z 25.000 in^4 J (^) z 1.041 107 mm^4
- Solid circle, let: D 3 in Area (^) A^ π^ D
^2
4 A 7.069 in ^2 A 4560 mm^2
Moment about x -axis (^) Ix^ π^ D
^4
64 Ix 3.976 in ^4 Ix 1.655 106 mm^4
Moment about y -axis (^) Iy^ π^ D
^4
64 Iy 3.976 in ^4 Iy 1.655 106 mm^4
Radius of gyration about x -axis (^) k (^) x Ix (^) A k (^) x 0.750 in k (^) x 19.05 mm
Radius of gyration about y -axis (^) k (^) y Iy (^) A k k (^) y (^) y 0.75019.05 inmm
Polar moment of inertia (^) J (^) z^ π^ D
^4
32 J (^) z 7.952 in ^4 J (^) z 3.310 106 mm^4
- Hollow circle, let:
D 3 in d 1 in Area (^) A^ π 4 D
^2 d^2 A 6.283 in^2
A 4054 mm^2
Moment about x -axis (^) Ix^ π 64 D
^4 d^4 Ix 3.927 in^4
Ix 1.635 106 mm^4
Moment about y -axis (^) Iy^ π 64 D
^4 d^4 Iy 3.927 in^4
Iy 1.635 106 mm^4
Radius of gyration about x -axis (^) k (^) x Ix (^) A k (^) x 0.791 in k (^) x 20.08 mm
Radius of gyration about y -axis (^) k (^) y Iy (^) A k (^) y 0.791 in k (^) y 20.08 mm
Polar moment of inertia (^) J (^) z^ π 32 D
^4 d^4 J z 7.854 in^4
J (^) z 3.269 106 mm^4
- Solid semicircle, let:
D 3 in R 0.5 D R 1.5 in
Area (^) A^ π^ D
^2
8 A 3.534 in ^2 A 2280 mm^2
Moment about x -axis (^) Ix 0.1098 R^4 Ix 0.556 in^4 Ix 2.314 105 mm^4 Moment about y -axis (^) Iy^ π^ R
^4
8 Iy 1.988 in ^4 Iy 8.275 105 mm^4
PROBLEM 1-
Statement: Prepare an interactive computer program (using, for example, Excell, Mathcad, or TKSolver) from which the mass properties for the solids shown in the page opposite the inside front cover can be calculated. Arrange the program to deal with both ips and SI unit systems and convert the results between those systems.
Units: blob lbf sec
^2
in
Solution: See the page opposite the inside front cover and Mathcad file P0108.
- Rectangular prism, let: a 2 in b 3 in c 4 in γ 0.28 lbf in ^3
Volume (^) V a b c V 24.000 in^3 V 393290 mm^3 Mass (^) M V ^ γ (^) g M 0.017 blob M 3.048 kg
Moment about x -axis (^) Ix M^ a
^2 b^2
12 Ix 0.019 blob in ^2 Ix 2130.4 kg mm ^2
Moment about y -axis (^) Iy M^ a
^2 c^2
12 Iy 0.029 blob in ^2 Iy 3277.6 kg mm ^2
Moment about z -axis (^) Iz M^ b
^2 c^2
12 Iz 0.036 blob in ^2 Iz 4097.0 kg mm ^2
Radius of gyration about x -axis (^) k (^) x Ix (^) M k (^) x 1.041 in k (^) x 26.437 mm
Radius of gyration about y -axis (^) k (^) y Iy (^) M k (^) y 1.291 in k (^) y 32.791 mm
Radius of gyration about z -axis (^) k (^) z Iz (^) M k (^) z 1.443 in k (^) z 36.662 mm
2.Cylinder, let: r 2 in L 3 in γ 0.30 lbf in ^3
Volume (^) V π r 2 L V 37.699 in^3 V 617778 mm^3 Mass (^) M V ^ γ (^) g M 0.029 blob M 5.13 kg
Moment about x -axis (^) Ix M r
^2
2 Ix 0.059 blob in ^2 Ix 6619.4 kg mm ^2 Moment about y -axis (^) Iy M^^3 r
^ 2 L^2
12 Iy 0.051 blob in ^2 Iy 5791.9 kg mm ^2 Moment about z -axis (^) Iz M^^3 r
^ 2 L^2
12 Iz 0.051 blob in ^2 Iz 5791.9 kg mm ^2
Radius of gyration about x -axis (^) k (^) x Ix (^) M k (^) x 1.414 in k (^) x 35.921 mm
Radius of gyration about y -axis (^) k (^) y Iy (^) M k (^) y 1.323 in k (^) y 33.601 mm
Radius of gyration about z -axis (^) k (^) z Iz (^) M k (^) z 1.323 in k (^) z 33.601 mm
- Hollow cylinder, let:
a 2 in b 3 in L 4 in γ 0.28 lbf in ^3
Volume V π b^2 a^2 L V 62.832 in^3
V 1029630 mm^3
Mass (^) M V ^ γ (^) g M 0.046 blob M 7.98 kg
Moment about x -axis (^) Ix M 2 a
^2 b^2 Ix 0.296 blob in ^2
Ix 3.3 104 kg mm ^2
Moment about y -axis (^) Iy M 12 3 a
^ 2 3 b^2 L^2 Iy 0.209 blob in ^2
Iy 2.4 104 kg mm ^2
Moment about z -axis (^) Iz M 12 3 a
^ 2 3 b^2 L^2 Iz 0.209 blob in ^2
Iz 2.4 104 kg mm ^2
Radius of gyration about x -axis (^) k (^) x Ix (^) M k (^) x 2.550 in k (^) x 64.758 mm
Radius of gyration about y -axis (^) k (^) y Iy (^) M k (^) y 2.141 in k (^) y 54.378 mm
Moment about y -axis (^) Iy^2 5 ^ Mr ^2 Iy 0.295 blob in ^2 Iy 33362 kg mm ^2
Moment about z -axis (^) Iz^2 5 ^ Mr ^2 Iz 0.295 blob in ^2 Iz 33362 kg mm ^2
Radius of gyration about x -axis (^) k (^) x Ix (^) M k (^) x 1.897 in k (^) x 48.193 mm
Radius of gyration about y -axis (^) k (^) y Iy (^) M k (^) y 1.897 in k (^) y 48.193 mm
Radius of gyration about z -axis (^) k (^) z Iz (^) M k (^) z 1.897 in k (^) z 48.193 mm
PROBLEM 1-
Statement: Convert the template in Problem 1-7 to have and use a set of functions or subroutines that can be called from within any program in that language to solve for the cross-sectional properties of the shapes shown on the inside front cover.
Solution: See inside front cover and Mathcad file P0109.
- Rectangle: Area (^) A b h ( ) b h
Moment about x -axis (^) Ix ( b h ) b h
^3
Moment about y -axis (^) Iy ( b h ) h b
^3
- Solid circle: Area (^) A D ( )^ π^ D
^2
Moment about x -axis (^) Ix ( D )^ π^ D
^4
Moment about y -axis (^) Iy ( D )^ π^ D
^4
- Hollow circle: Area (^) A D d ( )^ π 4 D
^2 d^2
Moment about x -axis (^) Ix ( D d )^ π 64 D
^4 d^4
Moment about y -axis (^) Iy ( D d )^ π 64 D
^4 d^4
- Solid semicircle: Area (^) A D ( )^ π^ D
^2
Moment about x -axis (^) Ix ( R ) 0.1098 R^4
Moment about y -axis (^) Iy ( R )^ π^ R
^4
- Right triangle: Area (^) A b h ( ) b h 2
Moment about x -axis (^) Ix ( b h ) b h
^3
Moment about y -axis (^) Iy ( b h ) h b
^3
- Right circular cone:
Volume (^) V r h ( )^ π^ r 2 h 3
Mass (^) M r h ( γ) V r h (^ )^ γ g
Moment about x -axis (^) Ix ( r h γ) 3 10 ^ M r h (^ ^ γ) r
^2
Moment about y -axis (^) Iy ( r h γ) M r h ( γ) 12 r
2 3 h^2
Moment about z -axis (^) Iz ( r h γ) M r h ( γ) 12 r
2 3 h^2
- Sphere:
Volume (^) V r ( ) 4 3 ^ π r
^3
Mass (^) M r ( γ) V r ( )^ γ g
Moment about x -axis (^) Ix ( r γ) 2 5 ^ M r (^ ^ γ) r
^2
Moment about y -axis (^) Iy ( r γ) 2 5 ^ M r (^ ^ γ) r
^2
Moment about z -axis (^) Iz ( r γ) 2 5 ^ M r (^ ^ γ) r
^2
PROBLEM 2-
Statement: Figure P2-1 shows stress-strain curves for three failed tensile-test specimens. All are plotted on the same scale. (a) Characterize each material as brittle or ductile. (b) Which is the stiffest? (c) Which has the highest ultimate strength? (d) Which has the largest modulus of resilience? (e) Which has the largest modulus of toughness?
Solution: See Figure P2-1 and Mathcad file P0201.
- The material in Figure P2-1(a) has a moderate amount of strain beyond the yield point, P2-1(b) has very little, and P2-1(c) has considerably more than either of the other two. Based on this observation, the material in Figure P2-1(a) is mildly ductile , that in P2-1(b)is brittle , and that in P2-1(c) is ductile.
- The stiffest material is the one with the grearesr slope in the elastic range. Determine this by dividing the rise by the run of the straight-line portion of each curve. The material in Figure P2-1(c) has a slope of 5 stress units per strain unit, which is the greatest of the three. Therefore, P2-1(c) is the stiffest.
- Ultimate strength corresponds to the highest stress that is achieved by a material under test. The material in Figure P2-1(b) has a maximum stress of 10 units, which is considerably more than either of the other two. Therefore, P2-1(b) has the highest ultimate strength.
- The modulus of resilience is the area under the elastic portion of the stress-starin curve. From observation of the three graphs, the stress and strain values at the yield points are: P2-1(a) σ ya := 5 ε ya := 5 P2-1(b) σ yb := 9 ε yb := 2 P2-1(c) σ yc := 5 ε yc := 1 Using equation (2.7), the modulus of resiliency for each material is, approximately,
P21a := 12 ⋅ σ ya ⋅ ε ya P21a =12.
P21b := 12 ⋅ σ yb ⋅ ε yb P21b = 9
P21c := 12 ⋅ σ yc ⋅ ε yc P21c =2.
P2-1 (a) has the largest modulus of resilience
- The modulus of toughness is the area under the stress-starin curve up to the point of fracture. By inspection, P2-1 (c) has the largest area under the stress-strain curve therefore, it has the largest modulus of toughness.
PROBLEM 2-
Statement: Which of the steel alloys shown in Figure 2-19 would you choose to obtain (a) Maximum strength (b) Maximum modulus of resilience (c) Maximum modulus of toughness (d) Maximum stiffness
Given: Young's modulus for steel E 207 GPa Solution: See Figure 2-19 and Mathcad file P0203.
- Determine from the graph: values for yield strength, ultimate strength and strain at fracture for each material.
Steel Yield Strength Ultimate Strength Fracture Strain AISI 1020: (^) Sy 1020 300 MPa Sut 1020 400 MPa ε f 1020 0. AISI 1095: (^) Sy 1095 550 MPa Sut 1095 1050 MPa ε f 1095 0. AISI 4142: (^) Sy 4142 1600 MPa Sut 4142 2430 MPa ε f 4142 0. Note: The 0.2% offset method was used to define a yield strength for the AISI 1095 and the 4142 steels.
- From the values of Sut above it is clear that the AISI 4142 has maximum strength.
- Using equation (2-7) and the data above, determine the modulus of resilience.
UR1020^12 Sy 1020^2 E UR1020 0. MN m m^3
UR1095^12
Sy 1095^2 E UR1095 0. MN m m^3
UR4142^12
Sy 4142^2 E UR4142 6. MN m m^3
Even though the data is approximate, the AISI 4142 clearly has the largest modulus of resilience.
- Using equation (2-8) and the data above, determine the modulus of toughness.
UT1020 12 Sy (^) 1020 Sut 1020 ε f 1020 UT1020 128 MN m m^3
UT1095 12 Sy (^) 1095 Sut 1095 ε f 1095 UT1095 88 MN m m^3
UT4142 12 Sy (^) 4142 Sut 4142 ε f 4142 UT4142 121 MN m m^3
Since the data is approximate, there is no significant difference between the 1020 and 4142 steels. Because of the wide difference in shape and character of the curves, one should also determine the area under the curves by graphical means. When this is done, the area under the curve is about 62 square units for 1020 and 66 for 4142. Thus, they seem to have about equal toughness, which is about 50% greater than that for the 1095 steel.
- All three materials are steel therefore, the stiffnesses are the same.
PROBLEM 2-
Statement: Which of the aluminum alloys shown in Figure 2-21 would you choose to obtain (a) Maximum strength (b) Maximum modulus of resilience (c) Maximum modulus of toughness (d) Maximum stiffness
Given: Young's modulus for aluminum E 71.7 GPa Solution: See Figure 2-21 and Mathcad file P0204.
- Determine, from the graph, values for yield strength, ultimate strength and strain at fracture for each material.
Alum Yield Strength Ultimate Strength Fracture Strain 1100: (^) Sy 1100 120 MPa Sut 1100 130 MPa ε f 1100 0. 2024-T351: (^) Sy 2024 330 MPa Sut 2024 480 MPa ε f 2024 0. 7075-T6: (^) Sy 7075 510 MPa Sut 7075 560 MPa ε f 7075 0. Note: The 0.2% offset method was used to define a yield strength for all of the aluminums.
- From the values of Sut above it is clear that the 7075-T6 has maximum strength.
- Using equation (2-7) and the data above, determine the modulus of resilience.
UR1100^12 Sy 1100^2 E UR1100 0. MN m m^3
UR2024^12
Sy 2024^2 E UR2024 0. MN m m^3
UR7075^12
Sy 7075^2 E UR7075 1. MN m m^3
Even though the data is approximate, the 7075-T6 clearly has the largest modulus of resilience.
- Using equation (2-8) and the data above, determine the modulus of toughness.
UT1100 12 Sy (^) 1100 Sut 1100 ε f 1100 UT1100 21 MN m m^3
UT2024 12 Sy (^) 2024 Sut 2024 ε f 2024 UT2024 79 MN m m^3
UT7075 12 Sy (^) 7075 Sut 7075 ε f 7075 UT7075 88 MN m m^3
Even though the data is approximate, the 7075-T6 has the largest modulus of toughness.
- All three materials are aluminum therefore, the stiffnesses are the same.
