Machine Design Problems

MACHINE DESIGN - An Integrated Approach, 4th Ed.

1-1-1

PROBLEM 1-1Statement: 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: 1. Cost less than $1.00 per use or application. 2. Allow disposal without human contact with mouse. 3. Be safe for other animals such as house pets. 4. Provide no threat to children or adults in normal use. 5. Be a humane method for the mouse. 6. Be environmentally friendly. 7. Have a shelf-life of at least 3 months. 8. Leave no residue. 9. Create minimum audible noise in use. 10. Create no detectable odors within 1 day of use. 11. Be biodegradable. 12. Be simple to use with minimal written instructions necessary. Concepts and sketches are left to the student. There are an infinity of possibilities.


1-2-1

PROBLEM 1-2Statement: 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: 1. Cost no more than $2 000. 2. Portable by no more than two able-bodied adults. 3. Fit through a standard doorway. 4. Provide no threat of injury to user in normal use. 5. Operate from a 110 V, 60 Hz, 20 amp circuit. 6. Be visually unthreatening. 7. Be easily positioned at bowling alley. 8. Have ball-aiming ability, controllable by user. 9. Automatically reload returned balls. 10. Require no more than 1 able-bodied adult for assistance in use. 11. Ball release requires no more than a mouth stick-switch closure. 12. Be simple to use with minimal written instructions necessary. Concepts and sketches are left to the student. There are an infinity of possibilities.


1-3-1

PROBLEM 1-3Statement: A quadriplegic needs an automated page turner to allow her to read books without assistance. 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 read standard books with minimum assistance. Task Specifications: 1. Cost no more than $1 000. 2. Useable in bed or from a seated position 3. Accept standard books from 8.5 x 11 in to 4 x 6 in in planform and up to 1.5 in thick. 4. Book may be placed, and device set up, by able-bodied person. 5. Operate from a 110 V, 60 Hz, 15 amp circuit or by battery power. 6. Be visually unthreatening and safe to use. 7. Require no more than 1 able-bodied adult for assistance in use. 8. Useable in absence of assistant once set up. 9. Not damage books. 10. Timing controlled by user. 11. Page turning requires no more than a mouth stick-switch closure. 12. Be simple to use with minimal written instructions necessary. Concepts and sketches are left to the student. There are an infinity of possibilities.


1-4-1

PROBLEM 1-4Statement: Units: Convert a mass of 1 000 lbm to (a) lbf, (b) slugs, (c) blobs, (d) kg. blob := lbf sec in Given: Solution: Mass See Mathcad file P0104. M := 1000 lb2

1. 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

2. Convert mass units by assigning different units to the units place-holder when displaying the mass value. Slugs Blobs Kilograms M = 31.081 slug M = 2.59 blob M = 453.592 kg


1-5-1

PROBLEM 1-5Statement: Given: A 250-lbm mass is accelerated at 40 in/sec2. Find the force in lb needed for this acceleration. Mass M := 250 lb Acceleration in a := 40 sec Solution: 1. See Mathcad file P0105.2

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 a = 3.333s-2

Convert acceleration to feet per second squared: F := M a F = 25.9 lbf

ft


1-6-1

PROBLEM 1-6Statement: Units: Given: Express a 100-kg mass in units of slugs, blobs, and lbm. How much does this mass weigh? blob lbf sec in2

M 100 kg

Assumptions: The mass is at sea-level and the gravitational acceleration is g 32.174 ft sec Solution: 1.2

or

g 386.089

in sec2

or

g 9.807

m sec2

See Mathcad file P0106.

Convert mass units by assigning different units to the units place-holder when displaying the mass value. The mass, in slugs, is The mass, in blobs, is The mass, in lbm, is M 6.85 slug M 0.571 blob M 220.5 lb

Note: Mathcad uses lbf for pound-force, and lb for pound-mass. 2. To determine the weight of the given mass, multiply the mass value by the acceleration due to gravity, g. The weight, in lbf, is The weight, in N, is W M g W M g W 220.5 lbf W 980.7 N


1-7-1

PROBLEM 1-7Statement: 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. See the inside front cover and Mathcad file P0107.

Solution: 1.

Rectangle, let: b 3 in Area h 4 in A b h A 12.000 in A 7742 mm Moment about x-axis Ix b h 12 h b 123 3 2

2 4 6 4

Ix 16.000 in

Ix 6.660 10 mm Iy 9.000 in4 6

Moment about y-axis

Iy

Iy 3.746 10 mm Radius of gyration about x-axis kx Ix A Iy A kx 1.155 in kx 29.329 mm ky 0.866 in ky 21.997 mm Jz 25.000 in4 7

4

Radius of gyration about y-axis

ky

Polar moment of inertia

Jz Ix Iy

Jz 1.041 10 mm 2. Solid circle, let: D 3 in Area A

4

D4

2

A 7.069 in

2 2

A 4560 mm4

Moment about x-axis

Ix

D64

Ix 3.976 in

4 6 4

Ix 1.655 10 mm4

Moment about y-axis

Iy

D64

Iy 3.976 in

4 6 4

Iy 1.655 10 mm Radius of gyration about x-axis kx Ix A kx 0.750 in kx 19.05 mm


1-7-2


ky

Iy A

ky 0.750 in ky 19.05 mm4


Jz

D32

Jz 7.952 in

4 6 4

Jz 3.310 10 mm

3.

Hollow circle, let: D 3 in Area d 1 in A

4

D d

2

2

A 6.283 in

2 2

A 4054 mm D d4 4

Moment about x-axis

Ix

64 64

Ix 3.927 in

4 6 4

Ix 1.635 10 mm Moment about y-axis Iy D d

4

4

Iy 3.927 in

4 6 4

Iy 1.635 10 mm Radius of gyration about x-axis kx Ix A Iy A kx 0.791 in kx 20.08 mm ky 0.791 in ky 20.08 mm


ky


Jz

32

D d

4

4

Jz 7.854 in

4 6 4

Jz 3.269 10 mm

4.

Solid semicircle, let: D 3 in Area R 0.5 D A R 1.5 in A 3.534 in2 2

D8

2

A 2280 mm Moment about x-axis Ix 0.1098 R4

Ix 0.556 in

4 5 4

Ix 2.314 10 mm Moment about y-axis Iy

R8

4

Iy 1.988 in

4 5 4

Iy 8.275 10 mm


1-7-3

Radius of gyration about x-axis

kx

Ix A Iy A

kx 0.397 in kx 10.073 mm ky 0.750 in ky 19.05 mm Jz 2.544 in4 6 4


ky


Jz Ix Iy

Jz 1.059 10 mm Distances to centroid a 0.4244 R b 0.5756 R a 0.637 in a 16.17 mm b 0.863 in b 21.93 mm

5.

Right triangle, let: b 2 in Area h 1 in A b h 2 A 1.000 in A 645 mm3 2

2

Moment about x-axis

Ix

b h 36 h b 36

Ix 0.056 in

4 4 4

Ix 2.312 10 mm3

Moment about y-axis

Iy

Iy 0.222 in

4 4 4

Iy 9.250 10 mm Radius of gyration about x-axis kx Ix A kx 0.236 in kx 5.987 mm ky 0.471 in ky 11.974 mm Jz 0.278 in4 5


ky

Iy A


Jz Ix Iy

Jz 1.156 10 mm

4


1-8-1

PROBLEM 1-8Statement: 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. blob lbf sec in3 2

Units: Solution: 1.

See the page opposite the inside front cover and Mathcad file P0108. a 2 in b 3 in V a b c c 4 in

Rectangular prism, let: Volume

0.28 lbf inV 24.000 in3

V 393290 mm Mass M V g M 0.017 blob M 3.048 kg2

3

Moment about x-axis

Ix

M a b 12

2 2 2

Ix 0.019 blob in

2 2

Moment about y-axis

Iy

M a c 12

2

Ix 2130.4 kg mm Iy 0.029 blob in

2 2

Iy 3277.6 kg mm Iz M b c 122

Moment about z-axis

Iz 0.036 blob in

2 2

Iz 4097.0 kg mm kx Ix M Iy M Iz M3


kx 1.041 in kx 26.437 mm ky 1.291 in ky 32.791 mm kz 1.443 in kz 36.662 mm


ky

Radius of gyration about z-axis

kz

2.Cylinder, let:

r 2 in Volume

L 3 in

0.30 lbf inV r L V g2

V 37.699 in

3 3

V 617778 mm Mass M M 0.029 blob M 5.13 kg

MACHINE DESIGN - An Integrated Approach, 4th Ed.Moment about x-axis Ix M r 2 M 3 r L 12 M 3 r L 122

1-8-2Ix 0.059 blob in2 2

Moment about y-axis

Iy

2

2


2 2

Iy 5791.9 kg mm Iz 0.051 blob in

2

2

Moment about z-axis

Iz

2 2

Iz 5791.9 kg mm Radius of gyration about x-axis kx Ix M Iy M Iz M kx 1.414 in kx 35.921 mm ky 1.323 in ky 33.601 mm kz 1.323 in kz 33.601 mm


ky


kz

3.

Hollow cylinder, let: a 2 in Volume b 3 in L 4 in

0.28 lbf in

3 3 3

V b a L

2

2

V 62.832 in

V 1029630 mm Mass M V g M 0.046 blob M 7.98 kg

Moment about x-axis

Ix

M 2 M 12 M 12

a b

2

2

2 2

Ix 0.296 blob in4

2 2

Ix 3.3 10 kg mm 3 a 3 b L2

Moment about y-axis

Iy

Iy 0.209 blob in4

2 2

Iy 2.4 10 kg mm 3 a 3 b L2 2 2

Moment about z-axis

Iz

Iz 0.209 blob in4

2 2

Iz 2.4 10 kg mm kx 2.550 in kx 64.758 mm ky 2.141 in ky 54.378 mm


kx

Ix M Iy M


ky

MACHINE DESIGN - An Integrated Approach, 4th Ed.Radius of gyration about z-axis kz Iz M kz 2.141 in kz 54.378 mm

1-8-3

4.

Right circular cone, let: r 2 in Volume h 5 in2

0.28 lbf inV

3 3 3

r h3 V g

V 20.944 in

V 343210 mm Mass M M 0.015 blob M 2.66 kg2

Moment about x-axis

Ix

3 10

M r

Ix 0.018 blob in

2 2

Moment about y-axis

Iy M

12r2 3h280


2 2

Iy 2638.5 kg mm Iz M

Moment about z-axis

12r2 3h280

Iz 0.023 blob in

2 2

Iz 2638.5 kg mm Radius of gyration about x-axis kx Ix M Iy M Iz M kx 1.095 in kx 27.824 mm ky 1.240 in ky 31.495 mm kz 1.240 in kz 31.495 mm


ky


kz

5.

Sphere, let: r 3 in Volume V 4 3 r3

V 113.097 in

3 3

V 1853333 mm M 0.082 blob M 14.364 kg2

Mass

M

V g 2 5

Moment about x-axis

Ix

M r

Ix 0.295 blob in

2 2

Ix 33362 kg mm

MACHINE DESIGN - An Integrated Approach, 4th Ed.Moment about y-axis Iy 2 5 2 5 M r2

1-8-4Iy 0.295 blob in2 2

Iy 33362 kg mm Moment about z-axis Iz M r2

Iz 0.295 blob in

2 2

Iz 33362 kg mm Ix M Iy M Iz M kx 1.897 in kx 48.193 mm ky 1.897 in ky 48.193 mm kz 1.897 in kz 48.193 mm


kx


ky


kz


1-9-1

PROBLEM 1-9Statement: 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. See inside front cover and Mathcad file P0109. Area Moment about x-axis A ( b h ) b h Ix( b h ) Iy( b h ) b h 12 h b 122 3 3

Solution: 1. Rectangle:

Moment about y-axis

2. Solid circle:

Area

A ( D) Ix( D) Iy( D)

D4

Moment about x-axis

D64

4

Moment about y-axis

D64

4

3. Hollow circle: Area

A ( D d ) Ix( D d )

4

D d

2

2

Moment about x-axis

64

D d

4

4

Moment about y-axis

Iy( D d )

64

D d

4

4

4. Solid semicircle: Area A ( D)

D8

2

Moment about x-axis

Ix( R) 0.1098 R Iy( R)

4

Moment about y-axis 5. Right triangle: Area

R8

4

A ( b h )

b h 2 b h 36 h b 363 3

Moment about x-axis

Ix( b h ) Iy( b h )

Moment about y-axis


1-10-1

PROBLEM 1-10Statement: Convert the template in Problem 1-8 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 page opposite the inside front cover. See the page opposite the inside front cover and Mathcad file P0110.

Solution: 1

Rectangular prism: Volume Mass V ( a b c) a b c M ( a b c ) V ( a b c) g M ( a b c ) a b 12 M ( a b c ) a c 12 M ( a b c ) b c 12

Moment about x-axis

Ix( a b c ) Iy( a b c )

2 2 2

2

2

Moment about y-axis

2

Moment about z-axis 2. Cylinder: Volume Mass

Iz( a b c )

V ( r L) r L M ( r L ) V ( r L) g M ( r L ) r 22

2

Moment about x-axis

Ix( r L ) Iy( r L ) Iz( r L )

Moment about y-axis

M ( r L ) 3 r L 12 M ( r L ) 3 r L 12

2

2

2

2

Moment about z-axis

3.

Hollow cylinder: Volume Mass V ( a b L) b a L M ( a b L ) Ix( a b L ) Iy( a b L ) Iz( a b L ) V ( a b L) g M ( a b L ) 2 M ( a b L ) 12 M ( a b L ) 12 a b

2

2

Moment about x-axis

2

2

2 2

Moment about y-axis

3 a 3 b L 3 a 3 b L2 2

2

Moment about z-axis

2

MACHINE DESIGN - An Integrated Approach, 4th Ed. 4. Right circular cone:Volume V ( r h )

1-10-22

r h3 V ( r h ) g 3 10 M ( r h ) r2

Mass

M ( r h )

Moment about x-axis

Ix( r h )

Moment about y-axis

Iy( r h ) M ( r h )

12r2 3h280

Moment about z-axis 5. Sphere: Volume

Iz( r h ) M ( r h )

12r2 3h280

V ( r)

4 3

r

3

Mass

M ( r )

V ( r) g 2 5 2 5 2 5 M ( r ) r M ( r ) r M ( r ) r2

Moment about x-axis

Ix( r ) Iy( r ) Iz( r )

Moment about y-axis

2

Moment about z-axis

2


2-1-1

PROBLEM 2-1Statement: 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? See Figure P2-1 and Mathcad file P0201.

Solution: 1.

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) P2-1(b) P2-1(c)

2.

3.

4.

ya := 5 yb := 9 yc := 5

ya := 5 yb := 2 yc := 1

Using equation (2.7), the modulus of resiliency for each material is, approximately, P21a := 1 2 1 2 1 2 ya ya P21a = 12.5

P21b :=

yb yb

P21b = 9

P21c :=

yc yc

P21c = 2.5

P2-1 (a) has the largest modulus of resilience 5. 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.


2-2-1

PROBLEM 2-2Statement: Determine an approximate ratio between the yield strength and ultimate strength for each material shown in Figure P2-1. See Figure P2-1 and Mathcad file P0202.

Solution: 1.

The yield strength is the value of stress at which the stress-strain curve begins to be nonlinear. The ultimate strength is the maximum value of stress attained during the test. From the figure, we have the following values of yield strength and tensile strength: Figure P2-1(a) Figure P2-1(b) Figure P2-1(c) S ya := 5 S yb := 9 S yc := 5 S ua := 6 S ub := 10 S uc := 8

2.

The ratio of yield strength to ultimate strength for each material is: Figure P2-1(a) ratioa := S ya S ua S yb S ub S yc S uc ratioa = 0.83

Figure P2-1(b)

ratiob :=

ratiob = 0.90

Figure P2-1(c)

ratioc :=

ratioc = 0.63


2-3-1

PROBLEM 2-3Statement: 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 Young's modulus for steel E 207 GPa

Given: Solution: 1.

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 AISI 1020: AISI 1095: AISI 4142: Yield Strength Sy1020 300 MPa Sy1095 550 MPa Sy4142 1600 MPa Ultimate Strength Sut 1020 400 MPa Sut 1095 1050 MPa Sut 4142 2430 MPa Fracture Strain

f 1020 0.365 f 1095 0.11 f 4142 0.06

Note: The 0.2% offset method was used to define a yield strength for the AISI 1095 and the 4142 steels. 2. 3. 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. 1 Sy1020 UR1020 2 E 1 Sy1095 UR1095 2 E 1 Sy4142 UR4142 2 E2

UR1020 0.222

MN m m3

UR1095 0.732

MN m m3

UR4142 6.18

MN m3

m Even though the data is approximate, the AISI 4142 clearly has the largest modulus of resilience. 4. Using equation (2-8) and the data above, determine the modulus of toughness. UT1020 UT1095 UT4142 1 2 1 2 1 2 Sy1020 Sut 1020 f 1020 Sy1095 Sut 1095 f 1095 Sy4142 Sut 4142 f 4142 UT1020 128 UT1095 88 MN m m3 3

MN m m MN m m3

UT4142 121

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. 5. All three materials are steel therefore, the stiffnesses are the same.


2-4-1

PROBLEM 2-4Statement: 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 Young's modulus for aluminum E 71.7 GPa

Given: Solution: 1.


Determine, from the graph, values for yield strength, ultimate strength and strain at fracture for each material. Alum 1100: 2024-T351: 7075-T6: Yield Strength Sy1100 120 MPa Sy2024 330 MPa Sy7075 510 MPa Ultimate Strength Sut 1100 130 MPa Sut 2024 480 MPa Sut 7075 560 MPa Fracture Strain

f 1100 0.170 f 2024 0.195 f 7075 0.165

Note: The 0.2% offset method was used to define a yield strength for all of the aluminums. 2. 3. 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. 1 Sy1100 UR1100 2 E UR2024 1 Sy2024 2 E2

UR1100 0.102

MN m m3

UR2024 0.762

MN m m3

1 Sy7075 UR7075 2 E

UR7075 1.81

MN m m3

Even though the data is approximate, the 7075-T6 clearly has the largest modulus of resilience. 4. Using equation (2-8) and the data above, determine the modulus of toughness. UT1100 UT2024 UT7075 1 2 1 2 1 Sy1100 Sut 1100 f 1100 Sy2024 Sut 2024 f 2024 Sy7075 Sut 7075 f 7075 UT1100 21 UT2024 79 UT7075 88 MN m m3

MN m m3

MN m

2 3 m Even though the data is approximate, the 7075-T6 has the largest modulus of toughness. 5. All three materials are aluminum therefore, the stiffnesses are the same.


2-5-1

PROBLEM 2-5Statement: Which of the thermoplastic polymers shown in Figure 2-22 would you choose to obtain (a) Maximum strength (b) Maximum modulus of resilience (c) Maximum modulus of toughness (d) Maximum stiffness See Figure 2-22 and Mathcad file P0205.

Solution: 1.

Determine, from the graph, values for yield strength, ultimate strength, strain at fracture, and modulus of elasticity for each material. Plastic Nylon 101: HDPE: PTFE: Yield Strength SyNylon 63 MPa SyHDPE 15 MPa SyPTFE 8.3 MPa Ultimate Strength Sut Nylon 80 MPa Sut HDPE 23 MPa Sut PTFE 13 MPa Fracture Strain Mod of Elasticity ENylon 1.1 GPa EHDPE 0.7 GPa EPTFE 0.8 GPa

f Nylon 0.52 f HDPE 3.0 f PTFE 0.51

2. 3.

From the values of Sut above it is clear that the Nylon 101 has maximum strength. Using equation (2-7) and the data above, determine the modulus of resilience. 1 SyNylon URNylon 2 ENylon2

URNylon 1.82

MN m m3

1 SyHDPE URHDPE 2 EHDPE 1 SyPTFE URPTFE 2 EPTFE2

URHDPE 0.16

MN m m3

URPTFE 0.04

MN m m3

Even though the data is approximate, the Nylon 101 clearly has the largest modulus of resilience. 4. Using equation (2-8) and the data above, determine the modulus of toughness. UTNylon 1 2 1 2 1 2 SyNylon Sut Nylon f Nylon SyHDPE SutHDPE f HDPE UTNylon 37 MN m m UTHDPE 573

UTHDPE

MN m m3

UTPTFE

SyPTFE SutPTFE f PTFE

UTPTFE 5

MN m m3

Even though the data is approximate, the HDPE has the largest modulus of toughness. 5. The Nylon 101 has the steepest slope in the (approximately) elastic range and is, therefore, the stiffest of the three materials..


2-6-1

PROBLEM 2-6Statement: A metal has a strength of 414 MPa at its elastic limit and the strain at that point is 0.002. What is the modulus of elasticity? What is the strain energy at the elastic limit? Assume that the test speimen is 12.8-mm dia and has a 50-mm gage length. Can you define the type of metal based on the given data? Elastic limit: Strength S el 414 MPa Strain

Given:

el 0.002

Test specimen: Diameter d o 12.8 mm Solution: 1. See Mathcad file P0206.

Length Lo 50 mm

The modulus of elasticity is the slope of the stress-strain curve, which is a straight line, in the elastic region. Since one end of this line is at the origin, the slope (modulus of elasticity) is E S el E 207 GPa

el

2.

The strain energy per unit volume at the elastic limit is the area under the stress-strain curve up to the elastic limit. Since the curve is a straight line up to this limit, the area is one-half the base times the height, or U'el 1 2 S el el U'el 414 kN m m3

The total strain energy in the specimen is the strain energy per unit volume times the volume,

Uel U'el

d o4

2

Lo

Uel 2.7 N m

3.

Based on the modulus of elasticity and using Table C-1, the material is steel.


2-7-1

PROBLEM 2-7Statement: A metal has a strength of 41.2 kpsi (284 MPa) at its elastic limit and the strain at that point is 0.004. What is the modulus of elasticity? What is the strain energy at the elastic limit? Assume that the test speimen is 0.505-in dia and has a 2-in gage length. Can you define the type of metal based on the given data? Elastic limit: Strength S el 41.2 ksi Strain

Given:

el 0.004

S el 284 MPa

Test specimen: Diameter d o 0.505 in Solution: 1. See Mathcad file P0207.

Length Lo 2.00 in

The modulus of elasticity is the slope of the stress-strain curve, which is a straight line, in the elastic region. Since one end of this line is at the origin, the slope (modulus of elasticity) is E S el E 10.3 10 psi6

el

E 71 GPa

2.

The strain energy per unit volume at the elastic limit is the area under the stress-strain curve up to the elastic limit. Since the curve is a straight line up to this limit, the area is one-half the base times the height, or 1 2 lbf in in2 3

U'el

S el el

U'el 82.4

U'el 568

kN m m3

The total strain energy in the specimen is the strain energy per unit volume times the volume, Uel U'el 3.

d o4

Lo

Uel 33.0 in lbf

Based on the modulus of elasticity and using Table C-1, the material is aluminum.


2-8-1

PROBLEM 2-8Statement: A metal has a strength of 134 MPa at its elastic limit and the strain at that point is 0.006. What is the modulus of elasticity? What is the strain energy at the elastic limit? Assume that the test speimen is 12.8-mm dia and has a 50-mm gage length. Can you define the type of metal based on the given data? Elastic limit: Strength S el 134 MPa Strain

Given:

el 0.003

Test specimen: Diameter d o 12.8 mm Solution: 1. See Mathcad file P0208.

Length Lo 50 mm

The modulus of elasticity is the slope of the stress-strain curve, which is a straight line, in the elastic region. Since one end of this line is at the origin, the slope (modulus of elasticity) is E S el E 45 GPa

el

2.

The strain energy per unit volume at the elastic limit is the area under the stress-strain curve up to the elastic limit. Since the curve is a straight line up to this limit, the area is one-half the base times the height, or U'el 1 2 S el el U'el 201 kN m m3


Uel U'el

d o4

2

Lo

Uel 1.3 N m

3.

Based on the modulus of elasticity and using Table C-1, the material is magnesium.


2-9-1

PROBLEM 2-9Statement: A metal has a strength of 100 kpsi (689 MPa) at its elastic limit and the strain at that point is 0.006. What is the modulus of elasticity? What is the strain energy at the elastic limit? Assume that the test speimen is 0.505-in dia and has a 2-in gage length. Can you define the type of metal based on the given data? Elastic limit: Strength S el 100 ksi S el 689 MPa Test specimen: Diameter d o 0.505 in Solution: 1. See Mathcad file P0209. Length Lo 2.00 in Strain

Given:

el 0.006

The modulus of elasticity is the slope of the stress-strain curve, which is a straight line, in the elastic region. Since one end of this line is at the origin, the slope (modulus of elasticity) is E S el E 16.7 10 psi6

el

E 115 GPa

2.

The strain energy per unit volume at the elastic limit is the area under the stress-strain curve up to the elastic limit. Since the curve is a straight line up to this limit, the area is one-half the base times the height, or U'el 1 2 S el el U'el 300 lbf in in3

U'el 2 10

3 kN m

m

3


Uel U'el

d o4

2

Lo

Uel 120.18 in lbf

3.

Based on the modulus of elasticity and using Table C-1, the material is titanium.


2-10-1

PROBLEM 2-10Statement: A material has a yield strength of 689 MPa at an offset of 0.6% strain. What is its modulus of resilience? MJ 10 joule Yield strength Yield strain Solution: 1. S y 689 MPa6

Units: Given:

y 0.006


The modulus of resilience (strain energy per unit volume) is given by Equation (2.7) and is approximately 1 2 MJ m3

UR

S y y

UR 2.067

UR 2.1 MPa


2-11-1

PROBLEM 2-11Statement: A material has a yield strength of 60 ksi (414 MPa) at an offset of 0.2% strain. What is its modulus of resilience? MJ 10 joule Yield strength Yield strain Solution: 1. S y 60 ksi S y 414 MPa6

Units: Given:

y 0.002


The modulus of resilience (strain energy per unit volume) is given by Equation (2.7) and is approximately 1 2 in lbf in3

UR

S y y

UR 60

UR 0.414

MJ m3

UR 0.414 MPa


2-12-1

PROBLEM 2-12Statement: A steel has a yield strength of 414 MPa, an ultimate tensile strength of 689 MPa, and an elongation at fracture of 15%. What is its approximate modulus of toughness? What is the approximate modulus of resilience? S y 414 MPa See Mathcad file P0212. S ut 689 MPa

Given: Solution: 1.

f 0.15

Determine the modulus of toughness using Equation (2.8).

UT

Sy S ut f 2

UT 82.7

MN m m3

UT 82.7 MPa

2.

Determine the modulus of resilience using Equation (2.7) and Young's modulus for steel: E 207 GPa 1 Sy UR 2 E2

UR 414

kN m m3

UR 0.41 MPa


2-13-1

PROBLEM 2-13Statement: The Brinell hardness of a steel specimen was measured to be 250 HB. What is the material's approximate tensile strength? What is the hardness on the Vickers scale? The Rockwell scale? Brinell hardness of specimen See Mathcad file P0213. HB 250

Given: Solution:

1. Determine the approximate tensile strength of the material from equations (2.10), not Table 2-3. S ut 0.5 HB ksi S ut 125 ksi S ut 862 MPa

2. From Table 2-3 (using linear interpolation) the hardness on the Vickers scale is HV HB 241 277 241 ( 292 253 ) 253 HV 263

3. From Table 2-3 (using linear interpolation) the hardness on the Rockwell C scale is HRC HB 241 277 241 ( 28.8 22.8) 22.8 HRC 24.3


2-14-1

PROBLEM 2-14Statement: The Brinell hardness of a steel specimen was measured to be 340 HB. What is the material's approximate tensile strength? What is the hardness on the Vickers scale? The Rockwell scale? Brinell hardness of specimen See Mathcad file P0214. HB 340

Given: Solution:

1. Determine the approximate tensile strength of the material from equations (2.10), not Table 2-3. S ut 0.5 HB ksi S ut 170 ksi S ut 1172 MPa

2. From Table 2-3 (using linear interpolation) the hardness on the Vickers scale is HV HB 311 341 311 ( 360 328 ) 328 HV 359

3. From Table 2-3 (using linear interpolation) the hardness on the Rockwell C scale is HRC HB 311 341 311 ( 36.6 33.1) 33.1 HRC 36.5


2-15-1

PROBLEM 2-15Statement: Solution: What are the principal alloy elements of an AISI 4340 steel? How much carbon does it have? Is it hardenable? By what techniques? See Mathcad file P0215.

1. Determine the principal alloying elements from Table 2-5 for 43xx steel.. 1.82% Nickel 0.50 or 0.80% Chromium 0.25% Molybdenum 2. From "Steel Numbering Systems" in Section 2.6, the carbon content is From the last two digits, the carbon content is 0.40%. 3. Is it hardenable? Yes, all of the alloying elements increase the hardenability. By what techniques? It can be through hardened by heating, quenching and tempering; and it can also be case hardened (See Section 2.4).


2-16-1


1. Determine the principal alloying elements from Table 2-5 for 10xx steel. Carbon only, no alloying elements 2. From "Steel Numbering Systems" in Section 2.6, the carbon content is From the last two digits, the carbon content is 0.95%. 3. Is it hardenable? Yes, as a high-carbon steel, it has sufficient carbon content for hardening. By what techniques? It can be through hardened by heating, quenching and tempering; and it can also be case hardened (See Section 2.4).


2-17-1


1. Determine the principal alloying elements from Table 2-5 for 61xx steel.. 0.15% Vanadium 0.60 to 0.95% Chromium 2. From "Steel Numbering Systems" in Section 2.6, the carbon content is From the last two digits, the carbon content is 0.80%. 3. Is it hardenable? Yes, all of the alloying elements increase the hardenability. By what techniques? It can be through hardened by heating, quenching and tempering; and it can also be case hardened (See Section 2.4).


2-18-1

PROBLEM 2-18Statement: Solution: Which of the steels in Problems 2-15, 2-16, and 2-17 is the stiffest? See Mathcad file P0218.

1. None. All steel alloys have the same Young's modulus, which determines stiffness.


2-19-1

PROBLEM 2-19Statement: Calculate the specific strength and specific stiffness of the following materials and pick one for use in an aircraft wing spar. Material Steel Code st 0 al 1 ti 2 Ultimate Strength Sut 80 ksist

Given:

Young's Modulus E 30 10 psist 6

Weight Density

0.28st

lbf in3

Aluminum

Sut

al

60 ksi

E

al

10.4 10 psi6

6

0.10al

lbf in3

Titanium

Sut 90 ksiti

E 16.5 10 psiti

0.16ti

lbf in3

Index Solution: 1.

i 0 1 2


Specific strength is the ultimate tensile strength divided by the weight density and specific stiffness is the modulus of elasticity divided by the weight density. The text does not give a symbol to these quantities. Sut Specific strengthi 1

E Specific stiffness Steel Aluminum Titanium

i 1

i

in

ini

286103 600103 563103

107106 104106 103106

2.

Based on the results above, all three materials have the same specific stiffness but the aluminum has the largest specific strength. Aluminum for the aircraft wing spar is recommended.


2-20-1

PROBLEM 2-20Statement: Solution: If maximum impact resistance were desired in a part, which material properties would you look for? See Mathcad file P0220.

1. Ductility and a large modulus of toughness (see "Impact Resistance" in Section 2.1).


2-21-1

PROBLEM 2-21Statement:

_____

Refer to the tables of material data in Appendix A and determine the strength-to-weight ratios of the following material alloys based on their tensile yield strengths: heat-treated 2024 aluminum, SAE 1040 cold-rolled steel, Ti-75A titanium, type 302 cold-rolled stainless steel. Material1

Given:

Yield Strength1

Specific Weight

Mat "2024 Aluminum, HT" Sy 290 MPa Mat "1040 CR Steel"2

0.10 lbf in1

3

27.14 1

kN m3

Sy 490 MPa2

0.28 lbf in2

3

76.01 2

kN m3

Mat "Ti-75A Titanium"3

Sy 517 MPa3

0.16 lbf in3

3

43.43 3

kN m3

Mat "Type 302 CR SS"4

Sy 1138 MPa4

0.28 lbf in4

3

76.01 4

kN m3

i 1 2 4 Solution: 1. See Mathcad file P0221.

Calculate the strength-to-weight ratio for each material as described in Section 2.1. Sy SWR i

i

SWR4

i

i

10 m

"2024 Aluminum, HT" "1040 CR Steel" Mat i "Ti-75A Titanium" "Type 302 CR SS"

1.068 0.645 1.190 1.497


2-22-1


_____

Refer to the tables of material data in Appendix A and determine the strength-to-weight ratios of the following material alloys based on their ultimate tensile strengths: heat-treated 2024 aluminum SAE 1040 cold-rolled steel, unfilled acetal plastic, Ti-75A titanium, type 302 cold-rolled stainless steel. Material1 2 3 4 5

Given:

Tensile Strength1 2 3 4 5

Specific Weight

Mat "2024 Aluminum, HT" Sut 441 MPa Mat "1040 CR Steel" Mat "Acetal, unfilled" Mat "Ti-75A Titanium" Mat "Type 302 CR SS" i 1 2 5 Solution: 1. See Mathcad file P0222. Sut 586 MPa Sut 60.7 MPa Sut 586 MPa Sut 1310 MPa

0.10 lbf in1 2 3 4 5

3 3 3

27.14 kN m1 2 3 4 5

3 3 3 3 3

0.28 lbf in

76.01 kN m 13.84 kN m 43.43 kN m 76.01 kN m

0.051 lbf in 0.16 lbf in 0.28 lbf in

3 3

Calculate the strength-to-weight ratio for each material as described in Section 2.1.

Sut SWR i

i

SWR4

i

i

10 m

"2024 Aluminum, HT" "1040 CR Steel" Mat "Acetal, unfilled" i "Ti-75A Titanium" "Type 302 CR SS"

1.625 0.771 0.438 1.349 1.724


2-23-1


_____

Refer to the tables of material data in Appendix A and calculate the specific stiffness of aluminum, titanium, gray cast iron, ductile iron, bronze, carbon steel, and stainless steel. Rank them in increasing order of this property and discuss the engineering significance of these data. Mg 10 kg Material Mat "Aluminum"1 2 3 4 5 6 7 3

Units: Given:

Modulus of Elasticity E 71.7 GPa1 2 3 4 5 6 7 1 2 3 4 5 6 7

Density

2.8 Mg m 4.4 Mg m 7.2 Mg m 6.9 Mg m 8.6 Mg m 7.8 Mg m 7.8 Mg m

3 3 3 3 3 3 3

Mat "Titanium" Mat "Gray cast iron" Mat "Ductile iron" Mat "Bronze" Mat "Carbon steel" Mat "Stainless steel" i 1 2 7 Solution: 1. See Mathcad file P0223.

E 113.8 GPa E 103.4 GPa E 168.9 GPa E 110.3 GPa E 206.8 GPa E 189.6 GPa

Calculate the specific stiffness for each material as described in Section 2.1. E E' i

i i

E'

i 6

s

2 2

"Aluminum" "Titanium" "Gray cast iron" Mat "Ductile iron" i "Bronze" "Carbon steel" "Stainless steel" E'2 5 s 6

10

m

25.6 25.9 14.4 24.5 12.8 26.5 24.3

2.

Rank them in increasing order of specific stiffness. Mat "Bronze"5

10 E' Mat "Gray cast iron"3

m

2

12.8

2 3 s 6

10 E' Mat "Stainless steel"7

m

2

14.4

2 7 s 6

10

m

2

24.3


2-23-2

E' Mat "Ductile iron"4

2 4 s 6

10 E' Mat "Aluminum"1

m

2

24.5

2 1 s 6

10 E' Mat "Titanium"2

m

2

25.6

2 2 s 6

10 E' Mat "Carbon steel"6

m

2

25.9

2 6 s 6

10 3.

m

2

26.5

Bending and axial deflection are inversely proportional to the modulus of elasticity. For the same shape and dimensions, the material with the highest specific stiffness will give the smallest deflection. Or, put another way, for a given deflection, using the material with the highest specific stiffness will result in the least weight.


2-24-1

PROBLEM 2-24Statement: Call your local steel and aluminum distributors (consult the Yellow Pages) and obtain current costs per pound for round stock of consistent size in low-carbon (SAE 1020) steel, SAE 4340 steel, 2024-T4 aluminum, and 6061-T6 aluminum. Calculate a strength/dollar ratio and a stiffness/dollar ratio for each alloy. Which would be your first choice on a cost-efficiency basis for an axial-tension-loaded round rod (a) If maximum strength were needed? (b) If maximum stiffness were needed?

Solution:

Left to the student as data will vary with time and location.


2-25-1

PROBLEM 2-25Statement: Call your local plastic stock-shapes distributors (consult the Yellow Pages) and obtain current costs per pound for round rod or tubing of consistent size in plexiglass, acetal, nylon 6/6, and PVC. Calculate a strength/dollar ratio and a stiffness/dollar ratio for each alloy. Which would be your first choice on a cost-efficiency basis for an axial-tension-loaded round rod or tube of particular diameters. (a) If maximum strength were needed? (b) If maximum stiffness were needed?

Solution:

Left to the student as data will vary with time and location.


2-26-1

PROBLEM 2-26Statement: A part has been designed and its dimensions cannot be changed. To minimize its deflections under the same loading in all directions irrespective of stress levels, which material woulod you choose among the following: aluminum, titanium, steel, or stainless steel? See Mathcad file P0226.

Solution: 1.

Choose the material with the highest modulus of elasticity because deflection is inversely proportional to modulus of elasticity. Thus, choose steel unless there is a corrosive atmosphere, in which case, choose stainless steel.


2-27-1

PROBLEM 2-27Statement: Assuming that the mechanical properties data given in Appendix Table A-9 for some carbon steels represents mean values, what is the value of the tensile yield strength for 1050 steel quenched and tempered at 400F if a reliability of 99.9% is required? Mean yield strength See Mathcad file P0227. S y 117 ksi S y 807 MPa

Given: Solution: 1.

From Table 2-2 the reliability factor for 99.9% is Re 0.753. Applying this to the mean tensile strength gives S y99.9 S y Re S y99.9 88.1 ksi S y99.9 607 MPa


2-28-1

PROBLEM 2-28Statement: Assuming that the mechanical properties data given in Appendix Table A-9 for some carbon steels represents mean values, what is the value of the ultimate tensile strength for 4340 steel quenched and tempered at 800F if a reliability of 99.99% is required? Mean ultimate tensile strength See Mathcad file P0228. S ut 213 ksi S ut 1469 MPa

Given: Solution: 1.

From Table 2-2 the reliability factor for 99.99% is Re 0.702. Applying this to the mean ultimate tensile strength gives S ut99.99 S ut Re S ut99.99 150 ksi S ut99.99 1031 MPa


2-29-1

PROBLEM 2-29Statement: Assuming that the mechanical properties data given in Appendix Table A-9 for some carbon steels represents mean values, what is the value of the ultimate tensile strength for 4130 steel quenched and tempered at 400F if a reliability of 90% is required? Mean ultimate tensile strength See Mathcad file P0229. S ut 236 ksi S ut 1627 MPa

Given: Solution: 1.

From Table 2-2 the reliability factor for 90% is Re 0.897. Applying this to the mean ultimate tensile strength gives S ut99.99 S ut Re S ut99.99 212 ksi S ut99.99 1460 MPa


2-30-1

PROBLEM 2-30Statement: Assuming that the mechanical properties data given in Appendix Table A-9 for some carbon steels represents mean values, what is the value of the tensile yield strength for 4140 steel quenched and tempered at 800F if a reliability of 99.999% is required? Mean yield strength See Mathcad file P0230. S y 165 ksi S y 1138 MPa

Given: Solution:

1. From Table 2-2 the reliability factor for 99.999% is Re 0.659. Applying this to the mean tensile strength gives S y99.9 S y Re S y99.9 109 ksi S y99.9 750 MPa


2-31-1

PROBLEM 2-31Statement: A steel part is to be plated to give it better corrosion resistance. Two materials are being considered: cadmium and nickel. Considering only the problem of galvanic action, which would you chose? Why? See Mathcad file P0231.

Solution: 1.

From Table 2-4 we see that cadmium is closer to steel than nickel. Therefore, from the standpoint of reduced galvanic action, cadmium is the better choice. Also, since cadmium is less noble than steel it will be the material that is consumed by the galvanic action. If nickel were used the steel would be consumed by galvanic action.


2-32-1

PROBLEM 2-32Statement: A steel part with many holes and sharp corners is to be plated with nickel. Two processes are being considered: electroplating and electroless plating. Which process would you chose? Why? See Mathcad file P0232.

Solution: 1.

Electroless plating is the better choice since it will give a uniform coating thickness in the sharp corners and in the holes. It also provides a relatively hard surface of about 43 HRC.


2-33-1

PROBLEM 2-33Statement: What is the common treatment used on aluminum to prevent oxidation? What other metals can also be treated with this method? What options are available with this method? See Mathcad file P0233.

Solution: 1.

Aluminum is commonly treated by anodizing, which creates a thin layer of aluminum oxide on the surface. Titanium, magnesium, and zinc can also be anodized. Common options include tinting to give various colors to the surface and the use of "hard anodizing" to create a thicker, harder surface.


2-34-1

PROBLEM 2-34Statement: Steel is often plated with a less nobel metal that acts as a sacrificial anode that will corrode instead of the steel. What metal is commonly used for this purpose (when the finished product will not be exposed to saltwater), what is the coating process called, and what are the common processes used to obtain the finished product? See Mathcad file P0234.

Solution: 1.

The most commonly used metal is zinc. The process is called "galvanizing" and it is accomplished by electroplating or hot dipping.


2-35-1

PROBLEM 2-35Statement: A low-carbon steel part is to be heat-treated to increase its strength. If an ultimate tensile strength of approximately 550 MPa is required, what mean Brinell hardness should the part have after treatment? What is the equivalent hardness on the Rockwell scale? Approximate tensile strength See Mathcad file P0235. S ut 550 MPa

Given: Solution: 1.

Use equation (2.10), solving for the Brinell hardness, HB. S ut = 3.45 HB HB S ut 3.45 MPa HB 159

2.

From Table 2-3, the equivalent hardness on the Rocwell scale is 83.9HRB.


2-36-1

PROBLEM 2-36Statement: A low-carbon steel part has been tested for hardness using the Brinell method and is found to have a hardness of 220 HB. What are the approximate lower and upper limits of the ultimate tensile strength of this part in MPa? Hardness HB 220

Given: Solution: 1.


Use equation (2.10), solving for ultimate tensile strength. Minimum: S utmin ( 3.45 HB 0.2 HB) MPa S utmax ( 3.45 HB 0.2 HB) MPa S utmin 715 MPa S utmax 803 MPa

Maximum:


2-37-1

PROBLEM 2-37Statement: Figure 2-24 shows "guide lines" for minimum weight design when failure is the criterion. The guide line, or index, for minimizing the weight of a beam in bending is f2/3/, where f is the yield strength of a material and is its mass density. For a given cross-section shape the weight of a beam with given loading will be minimized when this index is maximized. The following materials are being considered for a beam application: 5052 aluminum, cold rolled; CA-170 beryllium copper, hard plus aged; and 4130 steel, Q & T @ 1200F. The use of which of these three materials will result in the least-weight beam? Units: Given: Mg kg3

5052 Aluminum

S ya 255 MPa

a 2.8 Mg m b 8.3 Mg m s 7.8 Mg m

3 3

CA-170 beryllium copper S yb 1172 MPa 4130 steel Solution: 1. 2. See Mathcad file P0237. S ys 703 MPa

3

The values for the mass density are taken from Appendix Table A-1 and the values of yield strength come from from Tables A-2, A-4, and A-9 for aluminum, beryllium copper, and steel, respectively. Calculate the index value for each material. Index S y Sy0.667

Mg m MPa

3

0.667

Aluminum Beryllium copper Steel

Ia Index S ya a Ib Index S yb b Is Index S ys s

Ia 14.4 Ib 13.4 Is 10.2

The 5052 aluminum has the highest value of the index and would be the best choice to minimize weight.


2-38-1

PROBLEM 2-38Statement: Figure 2-24 shows "guide lines" for minimum weight design when failure is the criterion. The guide line, or index, for minimizing the weight of a member in tension is f/, where f is the yield strength of a material and is its mass density. The weight of a member with given loading will be minimized when this index is maximized. For the three materials given in Problem 2-37, which will result in the least weight tension member? Units: Given: Mg kg3

5052 Aluminum

S ya 255 MPa

a 2.8 Mg m b 8.3 Mg m s 7.8 Mg m

3 3

CA-170 beryllium copper S yb 1172 MPa 4130 steel Solution: 1. 2. See Mathcad file P0238. S ys 703 MPa

3

The values for the mass density are taken from Appendix Table A-1 and the values of yield strength come from from Tables A-2, A-4, and A-9 for aluminum, beryllium copper, and steel, respectively. Calculate the index value for each material. Index S y S y Mg m 3 MPa Ia 91.1 Ib 141.2 Is 90.1


Ia Index S ya a Ib Index S yb b Is Index S ys s

The beryllium copper has the highest value of the index and would be the best choice to minimize weight.


2-39-1

PROBLEM 2-39Statement: Figure 2-23 shows "guide lines" for minimum weight design when stiffness is the criterion. The guide line, or index, for minimizing the weight of a beam in bending is E1/2/, where E is the modulus of elasticity of a material and is its mass density. For a given cross-section shape the weight of a beam with given stiffness will be minimized when this index is maximized. The following materials are being considered for a beam application: 5052 aluminum, cold rolled; CA-170 beryllium copper, hard plus aged; and 4130 steel, Q & T @ 1200F. The use of which of these three materials will result in the least-weight beam? Units: Given: Mg kg3

5052 Aluminum

Ea 71.7 GPa

a 2.8 Mg m b 8.3 Mg m s 7.8 Mg m

3 3

CA-170 beryllium copper Eb 127.6 GPa 4130 steel Solution: 1. 2. See Mathcad file P0239. Es 206.8 GPa

3

The values for the mass density and modulus are taken from Appendix Table A-1. Calculate the index value for each material. E0.5

Index( E )

Mg m GPa

3

0.5


Ia Index Ea a Ib Index Eb b Is Index Es s

Ia 3.0 Ib 1.4 Is 1.8

The 5052 aluminum has the highest value of the index and would be the best choice to minimize weight.


2-40-1

PROBLEM 2-40Statement: Figure 2-24 shows "guide lines" for minimum weight design when stiffness is the criterion. The guide line, or index, for minimizing the weight of a member in tension is E/, where E is the modulus of elasticity of a material and is its mass density. The weight of a member with given stiffness will be minimized when this index is maximized. For the three materials given in Problem 2-39, which will result in the least weight tension member? Mg kg3

Units: Given:

5052 Aluminum

Ea 71.7 GPa

a 2.8 Mg m b 8.3 Mg m s 7.8 Mg m

3 3

CA-170 beryllium copper Eb 127.6 GPa 4130 steel Solution: 1. 2. See Mathcad file P0240. Es 206.8 GPa

3

The values for the mass density and modulus are taken from Appendix Table A-1. Calculate the index value for each material. E Mg m GPa3

Index( E )


Ia Index Ea a Ib Index Eb b Is Index Es s

Ia 25.6 Ib 15.4 Is 26.5

The steel has the highest value of the index and would be the best choice to minimize weight.


3-1-1

PROBLEM 3-1Statement: Which load class from Table 3-1 best suits these systems? (a) Bicycle frame (b) Flag pole (c) Boat oar (d) Diving board (e) Pipe wrench (f) Golf club. See Mathcad file P0301.

Solution:

1. Determine whether the system has stationary or moving elements, and whether the there are constant or time-varying loads. (a) Bicycle frame Class 4 (Moving element, time-varying loads) (b) Flag pole (c) Boat oar Class 2 (Stationary element, time-varying loads) Class 2 (Low acceleration element, time-varying loads)

(d) Diving board Class 2 (Stationary element, time-varying loads) (e) Pipe wrench Class 2 (Low acceleration elements, time-varying loads) (f) Golf club Class 4 (Moving element, time-varying loads)


3-2a-1

PROBLEM 3-2aStatement: Draw free-body diagrams for the system of Problem 3-1a (bicycle frame). Assumptions: 1. A two-dimensional model is adequate. 2. The lower front-fork bearing at C takes all of the thrust load from the front forks. 3. There are no significant forces on the handle bars. Solution: 1. See Mathcad file P0302a.

A typical bicycle frame is shown in Figure 3-2a. There are five points on the frame where external forces and moments are present. The rider's seat is mounted through a tube at A. This is a rigid connection, capable of transmitting two force components and a moment. The handle bars and front-wheel forks are supported by the frame through two bearings, located at B and C. These bearings are capable of transmitting radial and axial loads. The pedal-arm assembly is supported by bearings at D. These bearings are capable of transmitting radial loads. The rear wheel-sprocket assembly is supported by bearings mounted on an axle fixed to the frame at E.Ra Rb Fbr Rc Fay Rd Re Fey C

Ma Fax

A

B Fct Fcr

E

Fex Fdx D

Fdy

FIGURE 3-2aFree Body Diagram for Problem 3-2a

2. The loads at B and C can be determined by analyzing a FBD of the front wheel-front forks assembly. The loads at D can be determined by analyzing a FBD of the pedal-arm and front sprocket (see Problem 3-3), and the loads at E can be determined by analyzing a FBD of the rear wheel-sprocket assembly. 3. With the loads at B, C, D, and E known, we can apply equations 3.3b to the FBD of the frame and solve for Fax , Fay , and Ma.

Fx : Fy : Mz:

Fax Fbr cos( ) + Fcr cos( ) Fct sin( ) Fdx + Fex = 0 Fay Fbr sin( ) + Fcr sin( ) + Fct cos( ) Fdy + Fey = 0

(1) (2) (3)

Ma + ( Rbx Fby Rby Fbx ) + ( Rcx Fcy Rcy Fcx) ... = 0 + ( R F R F ) + ( R F R F ) ex ey ey ex dx dy dy dx


3-2e-1

PROBLEM 3-2eStatement: Draw free-body diagrams for the system of Problem 3-1e (pipe wrench). Assumptions: A two-dimensional model is adequate. Solution: See Mathcad file P0302e.

1. A typical pipe wrench with a pipe clamped in its jaw is shown in Figure 3-2e(a). When a force Fhand is applied on the wrench, the piping system provides an equal and opposite force and a resisting torque, Tpipe.

Fhand

Tpipe Fhand a

(a) FBD of pipe wrench and pipe

Fbt FbnFax

A Fay d b

(b) FBD of pipe wrench only FIGURE 3-2eFree Body Diagrams for Problem 3-2e

2. The pipe reacts with the wrench at the points of contact A and B. The forces here will be directed along the common normals and tangents. The jaws are slightly tapered and, as a result, the action of Fhand tends to drive the wrench further into the taper, increasing the normal forces. This, in turn, allows for increasing tangential forces. It is the tangential forces that produce the turning torque. 3. Applying equations 3.3b to the FBD of the pipe wrench,

Fx : Fy : M A:4.

Fax + Fbn cos( ) Fbt sin( ) = 0 Fay + Fbn sin( ) + Fbt cos( ) Fhand = 0 d ( Fbt cos( ) + Fbn sin( ) ) ( d + a ) Fhand = 0

(1) (2) (3)

These equations can be solved for the vertical forces if we assume is small so that sin() = 0 and cos () = 1.


3-3-1

PROBLEM 3-3Statement: Draw a free-body diagram of the pedal-arm assembly from a bicycle with the pedal arms in the horizontal position and dimensions as shown in Figure P3-1. (Consider the two arms, pedals and pivot as one piece). Assuming a rider-applied force of 1500 N at the pedal, determine the torque applied to the chain sprocket and the maximum bending moment and torque in the pedal arm. a 170 mm b 60 mm Frider 1.5 kN

Given:

Assumptions: The pedal-arm assembly is supprted by bearings at A and at B. Solution: See Figure 3-3 and Mathcad file P0303.

1. The free-body diagram (FBD) of the pedal-arm assembly (including the sprocket) is shown in Figure 3-3a. The rider-applied force is Frider and the force applied by the chain (not shown) is Fchain. The radial bearing reactions are Fax, Faz, Fbx, and Fbz. Thus, there are five unknowns Fchain, Fax, Faz, Fbx, and Fbz. In general, we can write six equilibrium equations for a three-dimensional force system, but in this system there are no forces in the y-direction so five equations are available to solve for the unknowns.z Fchain Faz a Frider b Arm Fax Fbx Pedal x y A B Arm (sectioned) Fbz Sprocket

(a) FBD of complete pedal-arm assemblyz

a Frider b Mc Arm

Tc

Fc Pedal x

y

(b) FBD of pedal and arm with section through the origin FIGURE 3-3Dimensions and Free Body Diagram of the pedal-arm assembly for Problem 3-3

2.

The torque available to turn the sprocket is found by summing moments about the sprocket axis. From Figure 3-3a, it is


3-3-2

Ty-axis:

a Frider r Fchain = a Frider Tsprocket = 0 Tsprocket a Frider Tsprocket 255 N m

where r is the sprocket pitch radius. 3. In order to determine the bending moment and twisting torque in the pedal arm, we will cut the arm with a section plane that goes through the origin and is parallel to the y-z plane, removing everything beyond that plane and replacing it with the internal forces and moments in the pedal arm at the section. The resulting FBD is shown in Figure 3-3b. The internal force at section C is Fc the internal bending moment is Mc, and the internal twisting moment (torque) is Tc. We can write three equilibrium equations to solve for these three unknowns: Shear force in pedal arm at section C

Fz : My-axis: Mx-axis:

Fc Frider = 0

Fc Frider

Fc 1.5 kN

Bending moment in pedal arm at section C a Frider Mc = 0 Mc a Frider Mc 255 N m

Twisting moment in pedal arm at section C b Frider Tc = 0 Tc b Frider Tc 90 N m

MACHINE DESIGN - An Integrated Approach,4th Ed.

3-4-1

PROBLEM 3-4Statement: The trailer hitch from Figure 1-1 has loads applied as shown in Figure P3-2. The tongue weight of 100 kg acts downward and the pull force of 4905 N acts horizontally. Using the dimensions of the ball bracket in Figure 1-5 (p. 15), draw a free-body diagram of the ball bracket and find the tensile and shear loads applied to the two bolts that attach the bracket to the channel in Figure 1-1. a 40 mm Mtongue 100 kg b 31 mm Fpull 4.905 kN c 70 mm t 19 mm d 20 mm

Given:

Assumptions: 1. The nuts are just snug-tight (no pre-load), which is the worst case. 2. All reactions will be concentrated loads rather than distributed loads or pressures. Solution: 1. See Figure 3-4 and Mathcad file P0304.

The weight on the tongue is Wtongue Mtongue g Wtongue 0.981 kN

2. The FBD of the hitch and bracket assembly is shown in Figure 3-4. The known external forces that act on the ball are Fpull and Wtongue . The reactions on the bracket are at points C and D. The bolts at C provide tensile (Fc2x) and shear (Fc2y) forces, and the bracket resists rotation about point D where the reaction force Fd2 is applied by the channel to which the bracket is bolted. 3. Solving for the reactions by summing the horizontal and vertical forces and the moments about D:

Fx : Fy : MD :

Fpull Fc2x Fd2 = 0Fc2y Wtongue = 0 Fc2x d Fpull ( a t b d ) Wtongue c = 0

(1) (2) (3)

W tongue 70 = c

1

F pull

1

40 = a 2 A

B

19 = t 31 = b Fc2x

2

B

C 20 = d D

C D

Fd2 F c2y

FIGURE 3-4Dimensions and Free Body Diagram for Problem 3-4

4.

Solving equation (3) for Fc2x

MACHINE DESIGN - An Integrated Approach,4th Ed.Fpull ( a t b d ) Wtongue c d

3-4-2

Fc2x 5.

Fc2x 30.41 kN

(4)

Substituting into (1) and solving for Fd2 Fd2 Fc2x Fpull Fd2 25.505 kN (5)

6.

Solving (2) for Fc2y Fc2y Wtongue Fc2y 0.981 kN (6)

7.

The loads applied to the two bolts that attach the bracket to the channel are: Axial force on two bolts Shear force on two bolts Fc2x 30.4 kN Fc2y 0.98 kN

We assume that each bolt would carry one half of these loads.


3-5-1

PROBLEM 3-5Statement: Given: For the trailer hitch of Problem 3-4, determine the horizontal force that will result on the ball from accelerating a 2000-kg trailer to 60 m/sec in 20 sec. Mass of trailer Final velocity Time to reach velocity Mtrailer 2000 kg vf 60 m sec

20 sec

Assumptions: 1. Acceleration is constant. 2. The rolling resistance of the tires and the wheel bearings is negligible. Solution: 1. See Mathcad file P0305.

From elementary kinematics, the acceleration required is a vf a 3.00 m sec2

(1)

2.

Using Newton's second law to find the force required to accelerate the trailer, Fhitch Mtrailer a Fhitch 6.00 kN (2)


3-6-1

PROBLEM 3-6Statement: For the trailer hitch of Problem 3-4, determine the horizontal force that will result on the ball from an impact between the ball and the tongue of the 2000-kg trailer if the hitch deflects 2.8 mm dynamically on impact. The tractor weighs 1000 kg. The velocity at impact is 0.3 m/sec. Mass of trailer Dynamic deflection Mass of tractor Impact velocity Mtrailer 2000 kg

Given:

i 2.8 mmMtractor 1000 kg vi 0.3 m sec

Assumptions: 1. The tractor is the "struck member" because the hitch is on the tractor and it is the hitch that deflects. 2. Equations (3.9) and (3.10) can be used to model the impact. Solution: 1. See Mathcad file P0306.

The weight of the trailer (the "striking member") is Wtrailer Mtrailer g Wtrailer 19.613 kN

2.

The correction factor, from equation (3-15), is

1

1 Mtractor 3 Mtrailer

0.857

3.

Eliminating E from equations (3.9a) and (3.10) and solving for the horizontal force on the ball Fi yields 1 2 Fi i = Mtrailer vi 2 2 1

Fi

Mtrailer vi i

2

Fi 55.1 kN


3-7-1

PROBLEM 3-7Statement: The piston of an internal-combustion engine is connected to its connecting rod with a "wrist pin." Find the force on the wrist pin if the 0.5-kg piston has an acceleration of 2 500 g. Mass of piston Acceleration of piston Mpiston 0.5 kg a piston 2500 g

Given:

Assumptions: The force on the wrist pin due to the weight of the piston is very small compared with the acceleration force. Solution: 1. See Mathcad file P0307. a piston 2.452 10 4

The acceleration in m/s is

m sec2

2.

Using Newton's Second Law expressed in equation (3.1a), the force on the wrist pin is Fwristpin Mpiston a piston Fwristpin 12.258 kN


3-8-1


_____

A cam-follower system similar to that shown in Figure 3-15 has a mass m = 1 kg, a spring constant k = 1000 N/m, and a damping coefficient d = 19.4 N-s/m. Find the undamped and damped natural frequencies of this system. cps := 2 rad sec Mass1 1

Units: Given:

M := 1 kg,

Spring constant1

k := 1000 N m

Damping coefficient d := 19.4 N s m Solution: 1.


Calculate the undamped natural frequency using equation 3.4. k M rad sec

n :=2.

n = 31.6

n = 5.03 cps

Calculate the undamped natural frequency using equation 3.7.2

d :=

k M

d 2 M

d = 30.1

rad sec

d = 4.79 cps


3-9-1

PROBLEM 3-9Statement: A ViseGrip plier-wrench is drawn to scale in Figure P3-3. Scale the drawing for dimensions. Find the forces acting on each pin and member of the assembly for an assumed clamping force of P = 4000 N in the position shown. What force F is required to keep it in the clamped position shown? Clamping force Dimensions P 4.00 kN a 50.0 mm b 55.0 mm c 39.5 mm d 22.0 mm e 28.0 mm f 26.9 mm g 2.8 mm h 21.2 mm

Given:

21.0 deg 129.2 deg

Assumptions: Links 3 and 4 are in a toggle position, i.e., the pin that joins links 3 and 4 is in line with the pins that join 1 with 4 and 2 with 3. Solution: 1. See Figure 3-9 and Mathcad file P0309.

The FBDs of the assembly and each individual link are shown in Figure 3-9. The dimensions, as scaled from Figure P3-3 in the text, are given above and are shown on the link FBDs.F 4 P 1

3 F55.0 = b 50.0 = a 22.0 = d

2 P

F14

F

39.5 = c

129.2

1

4 F34 P

F41

F21

28.0 = e

P

2.8 = g

F43 3 F23 F

F12 F32 2

21.2 = h

26.9 = f

FIGURE 3-9Free Body Diagrams for Problem 3-9

2.

Looking first at Part 3, we see that it is a three-force body. Therefore, the lines of action of the three forces must intersect at a point. But, since Parts 3 and 4 are in a toggle position, F43 and F23 are colinear, which means that their x- and y-components must be equal and opposite, leading to the conclussion that F = 0.


3-9-2

3.

Now, looking at Part 1, we see that (for F = 0) it is also a three-force body, as is Part 2. In fact, the forces on Part 1 are identical to those on Part 2. Solving for the unknown reactions on Parts 1 and 2, Fx: Fy: F41 cos( 180 deg ) F21 cos( 180 deg) = 0 F41 sin( 180 deg ) F21 sin( 180 deg) P = 0 (a) (b)

Solving equation (a) for F21 F21 = F41 cos( 180 deg ) cos( 180 deg) (c)

Substituting equation (c) into (b) F41 sin( 180 deg ) Solving equation (d) for F41 F41 sin( 180 deg ) P cos( 180 deg ) cos( 180 deg) sin( 180 deg) F41 cos( 180 deg ) cos( 180 deg) sin( 180 deg) P = 0 (d)

F21

F41 cos( 180 deg ) cos( 180 deg)

F41 5.1 kN F21 7.5 kN

Checking moment balance on Part 1, F41 sin( ) c F21 sin( 90 deg) d P a 0 kN m The result is, within the accuracy of the scaled dimensions, zero as it must be. 4. The x and y components of the pin forces on Part 1 are F41x F41 cos( 180 deg ) F41y F41 sin( 180 deg ) F21x F21 cos( 180 deg) F21y F21 sin( 180 deg) 5. The forces on the pins at the ends of Part 4 are F14 F41 F34 F14 6. The forces on the pins at the ends of Part 3 are F43 F34 F23 F43 F43 5.1 kN F23 5.1 kN F14 5.1 kN F34 5.1 kN F41x 4.749 kN F41y 1.823 kN F21x 4.749 kN F21y 5.823 kN


3-9-3

7.

The forces on the pins at the ends of Part 2 are F12 F21 F32 F23 Checking moment equilibrium on Part 2, F12 ( e cos( 90 deg) g sin( 90 deg) ) 0 kN m F32 ( h cos( ) f sin( ) ) which is zero, as it must be. F12 7.5 kN F32 5.1 kN


3-10-1

PROBLEM 3-10Statement: An overhung diving board is shown in Figure P3-4a. Find the reaction forces and construct the shear and moment diagrams for this board with a 100 kg person standing at the free end. Determine the maximum shear force, maximum moment and their locations. Beam length Distance to support Mass at free end L 2000 mm a 700 mm M 100 kg2000 = L R1 P

Given:

Assumptions: The weight of the beam is negligible compared to the applied load and so can be ignored.700 = a

R2

Solution:

See Figure 3-10 and Mathcad file P0310. FIGURE 3-10AFree Body Diagram for Problem 3-10

1. From inspection of Figure 3-10, write the load function equation q(x) = -R1-1 + R2-1 - P-1 2. Integrate this equation from - to x to obtain shear, V(x) V(x) = -R10 + R20 - P0 3. Integrate this equation from - to x to obtain moment, M(x) M(x) = -R11 + R21 - P1 4. Determine the magnitude of the force, P P M g P 980.7 N

5. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L, where both are zero. At x = L+, V = M = 0 R1 P La a V = R1 R2 P = 0 R1 1821 N R2 2802 N M = R1 L R2 ( L a ) = 0

R2 P R1 6. Define the range for x

x 0 in 0.005 L L

7. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z, and a value of one when it is greater than or equal to z. S ( x z) if ( x z 1 0 ) 8. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. V ( x) R1 S ( x 0 in) R2 S ( x a ) P S ( x L) M ( x) R1 S ( x 0 in) x R2 S ( x a ) ( x a ) P S ( x L) ( x L)

MACHINE DESIGN - An Integrated Approach, 4th Ed.9. Plot the shear and moment diagrams. Shear Diagram1000

3-10-2

0 V ( x) N 1000

2000

0

500

1000 x mm

1500

2000

Moment Diagram

0

375 M ( x) Nm

750

1125

1500

0

500

1000 x mm

1500

2000

FIGURE 3-10BShear and Moment Diagrams for Problem 3-10

10. The maximum value of the shear force ocuurs throughout the distance from x = 0 to x = a and is R1 1821 N 11. Find the maximum value of the bending moment by determining the value of x where the shear is zero. Inspection of the shear diagram shows that this occurs at x = a. Mmax M ( a ) Mmax 1275 N m


3-11-1

PROBLEM 3-11Statement: Determine the impact force and dynamic deflection that will result when the 100-kg person in Problem 3-10 jumps up 250 mm and lands back on the board. Assume that the board weighs 29 kg and deflects 131 mm statically when the person stands on it. Find the reaction forces and construct the shear and moment diagrams for this dynamic loading. Determine the maximum shear force, maximum moment and their locations along the length of the board. Beam length Distance to support Mass of person Mass of board Static deflection Height of jump L 2000 mm a 700 mm mpers 100 kg mboard 29 kgR1 2000 = L Fi

Given:

st 131 mmh 250 mm700 = a

R2

Assumptions: Equation (3.15) can be used to approximate a mass correction factor. Solution: See Figure 3-11 and Mathcad file P0311.

FIGURE 3-11AFree Body Diagram for Problem 3-11

1. The person is the moving object and the board is the struck object. The mass ratio to be used in equation (3.15) for the correction factor is massratio mpers mboard massratio 3.448

2. From equation (3.15), the correction factor is

1

1 1 3 massratio

0.912

3. The weight of the moving mass is

Wpers mpers g

Wpers 0.981 kN

4. The dynamic force is found by solving equation (3.14) for Fi. Fi Wpers 1

1

2 h

st Fi Wpers 3.12

Fi 3.056 kN

From this we see that the dynamic force ratio is

5. From inspection of Figure 3-11, write the load function equation q(x) = -R1-1 + R2-1 - Fi-1 6. Integrate this equation from - to x to obtain shear, V(x) V(x) = -R10 + R20 - Fi0 7. Integrate this equation from - to x to obtain moment, M(x) M(x) = -R11 + R21 - Fi1


3-11-2

8. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L, where both are zero. At x = L+, V = M = 0 R1 Fi L a a V = R1 R2 Fi = 0 R1 5676 N R2 8733 N M = R1 L R2 ( L a ) = 0

R2 Fi R1 9. Define the range for x

x 0 in 0.005 L L

10. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z, and a value of one when it is greater than or equal to z. S ( x z) if ( x z 1 0 ) 11. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. V ( x) R1 S ( x 0 in) R2 S ( x a ) Fi S ( x L) M ( x) R1 S ( x 0 in) x R2 S ( x a ) ( x a ) Fi S ( x L) ( x L) 12. Plot the shear and moment diagrams. Shear Diagram4 2 0 2 4 6

Moment Diagram0

1 M ( x)

V ( x) kN

2 kN m 3

0

0.5

1 x m

1.5

2

4

0

0.5

1 x m

1.5

2


13. The maximum value of the shear force ocuurs throughout the distance from x = 0 to x = a and is R1 5676 N 14. Find the maximum value of the bending moment by determining the value of x where the shear is zero. Inspection of the shear diagram shows that this occurs at x = a. Mmax M ( a ) Mmax 3973 N m


3-12-1

PROBLEM 3-12Statement: An overhung diving board is shown in Figure P3-4b. Find the reaction forces and construct the shear and moment diagrams for this board with a 100 kg person standing at the free end. Determine the maximum shear force, maximum moment and their locations. Beam length Mass at free end L 1300 mm M 100 kg2000 1300 = L P

Given:

Assumptions: 1. The weight of the beam is negligible compared to the applied load and so can be ignored. Solution: See Figure 3-12 and Mathcad file P0312.

M1 700

R1

1. From inspection of Figure 3-12, write the load function equation q(x) = -M1-2 + R1-1 - P-1 2. Integrate this equation from - to x to obtain shear, V(x) V(x) = -M1-1 + R10 - P0 3. Integrate this equation from - to x to obtain moment, M(x) M(x) = -M10 + R11 - P1 4. Determine the magnitude of the force, P P M g


P 980.7 N

5. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L, where both are zero. At x = L+, V = M = 0 R1 P M1 R1 L 6. Define the range for x V = R1 P = 0 R1 981 N M1 1275 m N M = M1 R1 L = 0

x 0 in 0.005 L L

7. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z, and a value of one when it is greater than or equal to z. S ( x z) if ( x z 1 0 ) 8. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. V ( x) R1 S ( x 0 mm) P S ( x L) M ( x) M1 S ( x 0 mm) R1 S ( x 0 mm) ( x 0 mm) P S ( x L) ( x L) 9. Plot the shear and moment diagrams.


3-12-2

Shear Diagram

1000 800 600 400 200 0

V ( x) N

0

0.5

1 x m

1.5

2

Moment Diagram

0 300 600 900 1200 1500

M ( x) Nm

0

0.5

1 x m

1.5

2


10. The maximum value of the shear force ocuurs throughout the distance from x = 0 to x = L and is R1 981 N 11. Find the maximum value of the bending moment by determining the value of x where the shear is zero. Inspection of the shear diagram shows that this occurs at x = 0. Mmax M ( 0 mm) Mmax 1275 N m


3-13-1

PROBLEM 3-13Statement: Determine the impact force and dynamic deflection that will result when a 100-kg person jumps up 25 cm and lands back on the board. Assume the board weighs 19 kg and deflects 8.5 cm statically when the person stands on it. Find the reaction forces and construct the shear and moment diagrams for this dynamic loading. Determine the maximum shear force, maximum moment, and their locations along the length of the board. Total board length Supported length Mass of board Static board deflection Mass of person Height of jump b 2000 mm a 700 mm mboard 19 kg2000 1300 = L Fi

Given:

stat 85 mmmperson 100 kg h 250 mmM1 700 R1

Assumptions: 1. The board can be modelled as a cantilever beam with maximum shear and moment at the edge of the support. Solution: 1. See Figure 3-13 and Mathcad file P0313.


The person impacts the board upon landing. Thus, the board is the struck object and the person is the striking object. To determine the force exerted by the person we will first need to know the impact correction factor from equation (3.15).

1

1 mboard 3 mperson

0.94

(1)

2.

We can now use equation (3.14) to determine the impact force, Fi, Fi mperson g 1

1

2 h

stat

Fi 3.487 kN

(2)

3.

Write an equation for the load function in terms of equations 3.17 and integrate the resulting function twice using equations 3.18 to obtain the shear and moment functions. Note use of the unit doublet function to represent the moment at the wall. For the beam in Figure 3-13, q(x) = -M1-2 + R1-1 - Fi-1 V(x) = -M1-1 + R10 - Fi0 + C1 M(x) = -M10 + R11 - Fi1 + C1x+ C2 The reaction moment M1 at the wall is in the z direction and the forces R1 and Fi are in the y direction in equation (4). All moments in equation (5) are in the z direction. (3) (4) (5)

4. 5.

Because the reactions have been included in the loading function, the shear and moment diagrams both close to zero at each end of the beam, making C1 = C2 = 0. The reaction force R1 and the reaction moment M1 can be calculated from equations (4) and (5) respectively

by substituting the boundary conditions x = l+, V = 0, M = 0. Note that we can substitute l for l+ since their difference is vanishingly small.

MACHINE DESIGN - An Integrated Approach, 4th Ed.Unsupported beam length l b a l 1300 mm

3-13-2

V(l) = -M1-1 + R10 - Fi0 = 0 V = R1 Fi = 0 R1 Fi M(l) = -M10 + R11 - Fi1 = 0 M = M1 R1 l Fi ( l l) = 0 M1 R1 l M1 4533 N m (7) R1 3.487 kN (6)

6. To generate the shear and moment functions over the length of the beam, equations (4) and (5) must be evaluated for a range of values of x from 0 to l, after substituting the above values of C1, C2, R1, and M1 in them. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than the dummy variable z, and a value of one when it is greater than or equal to z. It will have the same effect as the singularity function. Range of x Unit step function x 0 in 0.005 l l S ( x z) if ( x z 1 0 )

Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. V ( x) R1 S ( x 0 in) ( x 0 ) Fi S ( x l) ( x l)0 0 0 1 1

(8)

M ( x) M1 S ( x 0 in) ( x 0 ) R1 S ( x 0 in) ( x 0 ) Fi S ( x l ) ( x l) Plot the shear and moment diagrams (see below). Shear Diagram0 1 2

Moment Diagram

3 V ( x) kN 2 M ( x)

kN m 3 4 5

1

0

0

0.5 x m

1

0

0.5 x m

1

1.5

7. The graphs show that the shear force and the moment are both largest at x = 0. The function values of these points can be calculated from equations (4) and (5) respectively by substituting x = 0 and evaluating the singularity functions: Vmax = V(0) = R10 - Fi0 = R1 (9)

MACHINE DESIGN - An Integrated Approach, 4th Ed.Vmax R1 Vmax 3.49 kN

3-13-3

M.max = M(0) = -M10 + R11 - Fi1 = -M1 Mmax M1 Mmax 4533 N m

(10)


3-14-1

PROBLEM 3-14Statement: Figure P3-5 shows a child's toy called a pogo stick. The child stands on the pads, applying half her weight on each side. She jumps off the ground, holding the pads up against her feet, and bounces along with the spring cushioning the impact and storing energy to help each rebound. Find the natural frequency of the system, the static deflection of the spring with the child standing still, and the dynamic force and deflection when the child lands after jumping 2 in off the ground.2

Units: Given:

blob :=

lbf sec in

Child's weight Spring constant Pogo stick weight Height of drop

Wc := 60 lbf k := 100 lbf in Wp := 5 lbf h := 2 in1

Assumptions: 1. An approximate energy method will be acceptable. 2. The correction factor for energy dissipation will be applied. Solution: See Figure 3-14 and Mathcad file P0314.

1. Find the natural frequency of the (child/spring) system. Mass of child (striker) m := Wc g Wp g k m m = 0.155 blob

Fi /2

Fi /2

Mass of stick (struck)

mb :=

mb = 0.013 blob

Natural frequency

:=

= 25.367

rad sec

f :=

2

Pf = 4.037 Hz FIGURE 3-14Free Body Diagram for Problem 3-14

2. The static deflection of the spring with the child standing still is Static deflection of spring

st :=

Wc k

st = 0.6 in

3. Determine the mass ratio correction factor from equation (3.15): Correction factor

:=1+

1 mb 3 m

= 0.973

4. Using equation (3.14), determine the dynamic force.

Fi := Wc 1 +

1+

2 h

st

Fi = 224 lbf


3-15-1

PROBLEM 3-15Statement: A pen plotter imparts a constant acceleration of 2.5 m/sec2 to the pen assembly, which travels in a straight line across the paper. The moving pen assembly weighs 0.5 kg. The plotter weighs 5 kg. What coefficient of friction is needed between the plotter feet and the table top on which it sits to prevent the plotter from moving when the pen accelerates? Acceleration of pen ass'y a 2.5 m sec Mass of pen ass'y mpen 0.5 kg Mass of plotter Solution: 1. See Mathcad file P0315. mplot 5 kg2

Given:

The force imparted to the pen assembly by the internal drive mechanism must be reacted at the table top by the plotter feet. The horizontal force at the feet will be equal to the force on the pen assembly and must be less than or equal to the maximum friction force, which is the product of the coefficient of friction and the normal force, which is the weight of the plotter. Horizontal driving force on pen ass'y Weight of plotter Minimum coefficient of friction Fpen mpen a Wplot mplot g Fpen 1.25 N Wplot 49.033 N

Fpen Wplot

0.025


3-16-1

PROBLEM 3-16Statement: A track to guide bowling balls is designed with two round rods as shown in Figure P3-6. The rods are not parallel to one another but have a small angle between them. The balls roll on the rods until they fall between them and drop onto another track. The angle between the rods is varied to cause the ball to drop at different locations. Each rod's unsupported length is 30 in and the angle between them is 3.2 deg. The balls are 4.5 in dia and weigh 2.5 lb. The center distance between the 1-in-dia rods is 4.2 in at the narrow end. Find the distance from the narrow end at which the ball drops through and determine the worst-case shear and moment maximum for the rods as the ball rolls a distance from the narrow end that is 98% of the distance to drop. Assume that the rods are simply supported at each end and have zero deflection under the applied loading. (Note that assuming zero deflection is unrealistic. This assumption will be relaxed in the next chapter after deflection has been discussed.) Unsupported rod length Half-angle between rods Bowling ball diameter L 30 in 1.6 deg D 4.5 in Bowling ball weight Rod diameter Half width of rod gap W 2.5 lbf d 1.0 in c 2.1 in

Given:

Solution: 1.


Calculate the distance bet

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