uwo mechanical components design for mechatronic systems (mse 3380) sample problems - solutions v3

MSE 3380 UWO Sample Problems (v3)

Fernando Freitas Alves [email protected] Jan 21, 2015 1/25

1. [Stress analysis] A sign of dimensions 2.0 m 1.2 m is supported by a hollow circular

pole having outer diameter 220 mm and inner diameter 180 mm as shown in the left and

centre figures. The sign is offset 0.5 m from the centerline of the pole and its lower edge

is 6.0 m above the ground.

(a) (b) (c)

A wind pressure against the sign produces a resultant force, , that acts at the midpoint of

the sign (right figure) and is equal to the pressure, , times the area, sign, over which it

acts:

= sign = (2.0 kPa)(2.0 m 1.2 m) = 4.8 kN .

The line of action of is at height = 6.6 m above the ground and at distance = 1.5 m

from the centerline of the pole. Neglecting stress concentrations and the weight of all

components, determine:

(a) the moment of inertia, , polar moment of inertia, , cross-sectional of the pole, pole,

and the relevant first moment of area for a hollow-semicircle given by,

=2

3(23 1

3).

(b) the equivalent force-couple system (, , , ,, ) at the centroid of the section

of interest.

(c) the normal and shear stress components acting on the element at .

(d) the normal and shear stress components acting on the element at .

(e) the principal stresses, maximum shear stress and corresponding normal stress at

(max, min, max) geometrically using Mohrs circle.

(f) the principal stresses, maximum shear stress and corresponding normal stress at

(max, min, max) using any method. Sketch two stress elements (at ) that illustrate

the principal and maximum shear stress states (explicitly identify their orientations).

Bonus: (+1 point) Determine if the pole will yield at using the maximum-distortion-

energy (von Mises) criterion given = 60 MPa for the steel used.



SOLUTION

(a) Let the pole symbol be and the sign symbol be . Thus, we have the second moment

of area (or moment of inertia) with respect to the -axis to their centroids

= 2d

= 2dd 2

2

2

2

= d 2

2

2d 2

2

=

3

12 , where = 2.0 m and = 1.2 m

= 0.288 m4

= 2dext

ext

2dintint

= ( sin )2(dd)ext

0

2

0

( sin )2(dd)int

0

2

0

= sin2 d2

0

3dext

int

=

4(ext4 int

4 ) ,

where ext =220

2 mm = 0.110 m and ext =

180

2 mm = 0.090 m

63.5 106 m4 = 0.635 dm4

and with respect to the -axis to their centroids

= 2d

= 2dd 2

2

2

2

= 2d 2

2

d 2

2

=312

, where = 2.0 m and = 1.2 m

= 0.8 m4



= 2dext

ext

2dintint

= ( cos )2(dd)ext

0

2

0

( cos )2(dd)int

0

2

0

= cos2 d2

0

3dext

int

=

4(ext4 int

4 ) ,

where ext =220

2 mm = 0.110 m and ext =

180

2 mm = 0.090 m

63.5 106 m4 = 0.635 dm4 =

Their polar second moment of area (or polar moment of inertia) are

= 2d

= (2 + 2)d

= 2d

+ 2d

= +

= (0.8 + 0.288) m4

= 1.088 m4

= 2dext

ext

2dintint

= 2(dd)ext

0

2

0

2(dd)int

0

2

0

= d2

0

3dext

int

=

2(ext4 int

4 )

= 2 = 2

2 (63.5 106 m4)

= 127 106 m4 = 1.27 dm4



The cross-sectional area of the pole is

pole = = ext int

= (ext2 int

2 ) ,

where ext =220

2 mm = 0.110 m and ext =

180

2 mm = 0.090 m

12.6 103 m2 = 1.26 dm2

The relevant first moment of area for a hollow-semicircle is given by

= ext()dext

int()dint

,

where () is the thickness in the -axis

= ext

ext

int

int ,

where is the top (or bottom) portion of the members

cross-sectional area, defined from the section where ()

is measured, and is the distance to the centroid of ,

measured from the neutral axis

= [ ( sin )(dd)

ext0

0

ext ] ext

[ ( sin )(dd)

int0

0

int ] int

= sin d

0

2dext

int

=2

3(ext3 int

3 ) ,

where ext =220

2 mm = 0.110 m and ext =

180

2 mm = 0.090 m

401 106 m3 = 401 cm3

int

ext

ext int

Centroid of int

Centroid of ext

int()

ext()

Neutral axis



(b) According to the illustrations and knowing that = 1.5 m and = 6.6 m, the

equivalent force-couple system (, , , ,, ) at the point of the section is

(0,, 0,, 0, ) = (0 N, 4.8 kN, 0 N, 7.2 kNm, 0 Nm, 31.68 kNm)

= (0, 4.8, 0, 7.2, 0, 31.68) 103 [S. I. ]

x

y z = 0

=

= 0

=

= 0 =



(c) The state of stress at the point is

where

=ext

, given the flexure (bending) due to

=ext

(4.8 kN) (6.6 m) (0.110 m)

(63.5 106 m4)

55 MPa

= 0 , since no force stress this direction

=ext

, given the torsion due to

=ext

(4.8 kN) (1.5 m) (0.110 m)

(127 106 m4)

6.2 MPa



(d) The state of stress at the point is

where

= = 0 , since no force stress these directions

=ext

+

,

given the torsion and the transverse force due to

where = and = 2 (ext int)

= [ext

+

2(ext int)] ,

where ext =220

2 mm = 0.110 m and ext =

180

2 mm = 0.090 m

(4.8 kN)

[(1.5 m) (0.110 m)

(127 106 m4)+

(401 106 m3)

2 (63.5 106 m4) (0.110 m 0.090 m)]

7.0 MPa



(e) Constructing the Mohrs circle, we got that

max 55.7 MPa

min 0.7 MPa

max 28.2 MPa

27.5 MPa (when = )

(f) Once there is no normal stress in the direction studied in the exercise (d), the maximum

shear stress is the same that is already calculated with null corresponding normal stress

and the principal stresses are those that happen when the orientation of the stress

element is 90 2 = 45 due to the Mohrs circle simplification. Thus, we got

max = || 7.0 MPa (at = 45)

min = || 7.0 MPa (at = 135)

max = || 7.0 MPa

= 0 MPa (when = )

max

45 0

max min



Bonus: According to the Distortion Energy Theory, the distortion energy density in terms

of equivalent (von Misses) stress, VM, is given by

=1 +

3VM2

Knowing that the von Mises stress for a two-dimensional plane stress state can be

defined in terms of principal stresses as

VM = max2 maxmin + min2

and once the stresses at correspond to a situation that only shear stress exists, which

means that

max = min =

we got that the yield stress (critical stress value) is

VM = 2 + () + 2 = 3

3

which means that the material will yield at if the shear stress reaches 1 3 0.577

of the yield stress.

Thus, given = 60 MPa,

7.0 MPa 60 MPa

3 35 MPa

respect the maximum-distortion-energy (von Mises) condition and will not yield at .



2. [Stress-based design] A simply-supported beam is required to support the loads shown,

and the grade of steel to be used has all = 170 MPa and all = 100 MPa. Neglecting the

effect of fillets, stress concentrations, out-of-plane stresses due to the loads and weight of

the bream, determine:

(a) the reaction forces at the supports.

(b) the maximum magnitudes of the shear force and bending moment.

(c) the required section modulus, min, due to flexure.

(d) the most appropriate standard SI Wide-Flange beam section that should be used from

those listed in the table below.

Shape Section modulus

[103 mm3]

W200 19.3 162.0

W150 24 167.0

W150 18 120.0

W150 13.5 91.1

W130 28.1 167.0

W130 23.8 140.0

W100 19.3 89.5

(e) if the selected beam will fail due to the maximum shear in the web, max

(assume uniform shear stress in the web; note that the table of properties of standard W

sections is attached).

(f) if the selected beam will fail due to the maximum flexural stress on the outermost

surface of the flange, .

Bonus: (+2 point) Determine if the selected beam will fail due to the maximum principal

stress at the web-flange junction, max. Assume uniform shear stress in the web.

x

6 kN

12 kN



SOLUTION

(a) Defining the reaction forces at each support as illustrated below in the free body

diagram

and once the beam is static, the sum of every bending moment about any point must be

zero. Thus, about we have

= 0 (6 kN) (2 m) + (12 kN) (4 m) + (6 m) = 0

= 10 kN

Also, this static situation requires that the sum of the forces in all directions must be

zero, we have

{ = 0 = 0

= 0 + (6 kN) + (12 kN) + = 0 = 8 kN

x

6 kN

12 kN



(b) Analysing the shear forces across the beam, we have that

() = {

= 8 kN, 0 m < 2 m

+ (6 kN) = 2 kN, 2 m < 4 m

+ (6 kN) + (12 kN) = 10 kN, 4 m < 6 m

Analysing the bending moments across the beam, we have that

() = {

= 8 [kNm], 0 m < 2 m

+ (6 kN)( 2) = 2 + 12 [kNm], 2 m < 4 m

+ (6 kN)( 2) + (12 kN)( 4) = 10 + 60 [kNm], 4 m 6 m

Thus, the maximum shear force max and the maximum bending moment max are

max = 10 kN and max = 20 kNm



(c) Knowing the maximum bending moment at 4 m, the correspondent maximum shear

stress at that point is given by

max =maxmax

where max is the maximum distance of material of the cross section from the neutral

axis. However, we do not have any information about the cross section of the beam, so

we cannot know . However, we can use the definition of the elastic section modulus

=

max

so that, using the previous equation, we have

max =max

=maxmax

Once the maximum bending moment was calculated in the last exercise and maximum

shear stress allowed was given, i.e. max = all = 170 MPa, the required minimum

elastic sections modulus is

=20 103 Nm

170 106N m2 118 103 mm3

(d) The most appropriate standard SI Wide-Flange beam section that should be used from

the table given is the one that have the least value of elastic section modulus that is

greater that min, which is the shape W150 18 . This choice often guarantees the

selection of the least expansive beam in the table, once their price generally increases

with greater section modulus. The chosen beams dimensions are shown above:

153 mm

7.11 mm

102 mm

5.84 mm

1

2

3



(e) We define the maximum shear stress distribution by

max =max

However, if we assume an uniform shear stress on the beams section, we can define

the maximum shear stress by

max =max

where is the section area of the web, i.e. = 2, so that

max =(10 kN)

(0.00584 m)(0.153 m 2 0.00711 m) 12 MPa

Thus, once

max < all = 100 MPa

the beam chosen will endure the loads without any failing on the web.



(f) Now that we have information about the section, we can calculate the real maximum

shear stress at the outermost surface of the flange by

a =maxmax

where the second moment of area can be checked within a table or calculated as

= 1 + 2 + 3 = (1 + 112) + (2 + 22

2) + (3 + 332)

= {1

12(0.102 m)(0.00711 m)3

1

+ (0.102 m)(0.00711 m) {[0.153 m (0.00711 m) 2 ] (0.153 m) 2 }2}

1 1

+1

12(0.00584 m)(0.153 m 2 0.00711 m)3

2

+ {1

12(0.102 m)(0.00711 m)3

3

+ (0.102 m)(0.00711 m) [(0.00711 m) 2 (0.153 m) 2 ]2}

3 3

9.0 106 m4

which does not match exactly with the value 9.20 106 m4 of the table from

Appendix C of [1]. Using the both values, we have

1 (20 103 Nm) [(0.153 m) 2 ]

(9.0 106 m4)

170 MPa

2 =(20 103 Nm) [(0.153 m) 2 ]

(9.20 106 m4)

166 MPa

Once 2 < 1 all = 170 MPa , the chosen beam will (almost) not fail.

Bonus: At the web-flange junction, we have a different = max = max 0.00711 m

= (0.153 m) 2 (0.00711 m) = 0.06939 m, so that

max1 (20 103 Nm) (0.06939 m)

(7.8 106 m4)

75 MPa

max2 =(20 103 Nm) (0.06939 m)

(9.20 106 m4)

77 MPa

Once max2 < max1 < all = 170 MPa , the chosen beam will not fail.

0

[1] F. P. Beer, E. R. Johnston, Jr., J. T. DeWolf, and D. F. Mazurek, Mechanics of Materials, 7th ed.

New York, NY: McGraw-Hill, 2015.



3. [Stress analysis in shafts] A torque = 100 Nm is applied to the shaft , which is

running at constant speed and contains gear . Gear transmits torque to shaft

through gear , which drives the chain sprocket at , transmitting a force as shown.

Sprocket , gear , and gear have pitch diameters of = 150 mm, = 250 mm, and

= 125 mm, respectively. The contact force between the gears is transmitted through the

pressure angle = 20. Assuming no frictional losses and considering the bearings at

, , , and to be simple supports, locate the point on shaft that contains the

maximum normal and maximum shear stresses. Combine these stresses and determine the

maximum principal normal and shear stresses in the shaft.



SOLUTION

Analysing the first gear, , we have the following free body diagram

Thus, equalling the sum of all the bending moments to zero, we have the following

reaction force

= 0

2+ = 0

=

2

=100 Nm

62.5 103 m

= 1.6 kN

and using the geometry of the force , we can calculate its norm by

=cos

=1.6 kN

cos 20

1.7 kN

and the norm of the normal force

= tan

= (1.6 kN) tan 20

0.58 kN

62.5 mm

1

Gear

= 100 Nm



Using the law of conservation of energy with no frictional losses, the reaction force

calculated in the gear is the same reaction force in the gear with opposite orientation

such that the norms are the same

= 1.7 kN

= = 1.6 kN

= 0.58 kN

and the static give the torque by

= 0

2+ = 0

=

2

= (1.6 kN) (125 103 m)

= 200 Nm

This value could also be calculated by using the following relationship between the

diameter (or radius) of both gears

=

=

= (100 Nm) (250 mm

125 mm)

= 200 Nm

125 mm

2

Gear



Assuming that this torque is transmitted without losses to the sprocket , we have

equal magnitude and opposite orientation

=

such that we have the following diagram

Determining the force , we have

=

=

=200 Nm

0.075 m

= 2. 6 kN

0 15 mm 75 mm 125 mm

= 200 Nm

= 200 Nm



Thus, analysing the beam forces in the plane

we have the reaction forces given by the equilibrium of the bending moments in

= 0

( + ) + ( + + ) = 0

= + ( + )

+ +

(2. 6 kN) (0.075 m) + (0.58 kN) (0.325 m)

(0.450 m)

0.87 kN

and the equilibrium in the -axis

= 0

+ = 0

= +

2. 6 kN + 0.58 kN 0.87 kN

2.4 kN

This gives us the following shear force distribution in the -plane

() = {

2.4 kN, 0 mm < 75 mm

0.28 kN, 75 mm < 325 mm

0.87 kN, 325 mm 450 mm

0.58 kN = 2. 6 kN

= 75 mm = 250 mm = 125 mm



and the following bending moments distribution in the same plane

()

= {

2.4 [kNm], 0 mm < 75 mm

( 0.075) 0.28 + 0.20 [kNm], 75 mm < 325 mm

( 0.075) ( 0.325) 0.87 + 0.39 [kNm], 325 mm 450 mm


max 2.4 kN from to and max 180 Nm at

Analysing the state of stress at in the plane ,

using the maximum bending moment max, we can calculate the maximum principal

normal stress in the shaft

max =max( 2 )

=max( 2 )

4 4=2max3

=2 (180 Nm)

(0.015 m)3

34 MPa

max

max

Shaft



and the maximum shear stress

max = max(max

( 2 )

)

= max(max(2

3 3 )

(4 4 )( 2 )

(4 2 ))

= max(8max 3

33)

8 (2.4 103 N) (0.015 m) + 3 (200 Nm)

3 (0.015 m)3

28 MPa

Otherwise, analysing the beam forces in the plane

we have the reaction forces given by the equilibrium of the bending moments in

= 0

+ ( + ) ( + + ) = 0

= + ( + )

+ +

=(2. 6 kN) (0.075 mm) + (1.6 kN) (0.325 m)

(0.450 m)

= 0.71 kN

and the equilibrium in the -axis

= 0

+ = 0

= +

= 2. 6 kN 1.6 kN + 0.71 kN

= 1. 7 kN

= 1.6 kN = 2. 6 kN

= 75 mm = 250 mm = 125 mm



This gives us the following shear force distribution in the -plane

() = {

= 1. 7 kN, 0 mm < 75 mm

= 0. 8 kN, 75 mm < 325 mm

+ = 0.71 kN, 325 mm 450 mm

and the following bending moments distribution in the same axis

()

= {

= 1. 7 [kNm], 0 mm < 75 mm

( 0.075) = 0. 8 + 0.20 [kNm], 75 mm < 325 mm

( 0.075) + ( 0.325) = 0.71 0.32 [kNm], 325 mm 450 mm


max = 1. 7 kN from to and max = 133. 3 Nm at



Analysing the state of stress at in the plane ,

using the maximum bending moment max, we can calculate the maximum principal

normal stress in the shaft

max =max( 2 )

=max( 2 )

4 4=2max3

=2 (133. 3 Nm)

(0.015 m)3

25 MPa

and the maximum shear stress

max = max(max

( 2 )

)

= max(max(2

3 3 )

(4 4 )( 2 )

(4 2 ))

= max(8max 3

33)

8 (1. 7 103 N) (0.015 m) + 3 (200 Nm)

3 (0.015 m)3

26 MPa

max

max

Shaft



Plotting in a 3D view, we got the following state of stress at

max 34 MPa

max 25 MPa

max 28 MPa

max 26 MPa max 26 MPa

max 28 MPa

1. [Stress analysis] A sign of dimensions 2.0 m 1.2 m is supported by a hollow circular pole having outer diameter 220 mm and inner diameter 180 mm as shown in the left and centre figures. The sign is offset 0.5 m from the centerline of the pole and...(a) the moment of inertia, , polar moment of inertia, , cross-sectional of the pole, ,-pole., and the relevant first moment of area for a hollow-semicircle given by, =,2-3.,,-2-3.,-1-3...(b) the equivalent force-couple system ,,,-.,,-.,,,-.,,-.. at the centroid of the section of interest.(c) the normal and shear stress components acting on the element at .(d) the normal and shear stress components acting on the element at .(e) the principal stresses, maximum shear stress and corresponding normal stress at (,-max., ,-min., ,-max.) geometrically using Mohrs circle.(f) the principal stresses, maximum shear stress and corresponding normal stress at (,-max., ,-min., ,-max.) using any method. Sketch two stress elements (at ) that illustrate the principal and maximum shear stress states (explicitly identi...Bonus: (+1 point) Determine if the pole will yield at using the maximum-distortion-energy (von Mises) criterion given ,-.=60 MPa for the steel used.SOLUTION(a) Let the pole symbol be and the sign symbol be . Thus, we have the second moment of area (or moment of inertia) with respect to the -axis to their centroids(b) According to the illustrations and knowing that =1.5 m and =6.6 m, the equivalent force-couple system (,,-.,,-.,,,-.,,-.) at the point of the section is(c) The state of stress at the point is(d) The state of stress at the point is(e) Constructing the Mohrs circle, we got that(f) Once there is no normal stress in the direction studied in the exercise (d), the maximum shear stress is the same that is already calculated with null corresponding normal stress and the principal stresses are those that happen when the orientatio...Bonus: According to the Distortion Energy Theory, the distortion energy density in terms of equivalent (von Misses) stress, ,-VM., is given by

2. [Stress-based design] A simply-supported beam is required to support the loads shown, and the grade of steel to be used has ,-all.=170 MPa and ,-all.=100 MPa. Neglecting the effect of fillets, stress concentrations, out-of-plane stresses due to...(a) the reaction forces at the supports.(b) the maximum magnitudes of the shear force and bending moment.(c) the required section modulus, ,-min., due to flexure.(d) the most appropriate standard SI Wide-Flange beam section that should be used from those listed in the table below.(e) if the selected beam will fail due to the maximum shear in the web, ,-max. (assume uniform shear stress in the web; note that the table of properties of standard W sections is attached).(f) if the selected beam will fail due to the maximum flexural stress on the outermost surface of the flange, ,-..Bonus: (+2 point) Determine if the selected beam will fail due to the maximum principal stress at the web-flange junction, ,-max.. Assume uniform shear stress in the web.SOLUTION(a) Defining the reaction forces at each support as illustrated below in the free body diagram(b) Analysing the shear forces across the beam, we have that(c) Knowing the maximum bending moment at 4 m, the correspondent maximum shear stress at that point is given by(d) The most appropriate standard SI Wide-Flange beam section that should be used from the table given is the one that have the least value of elastic section modulus that is greater that ,-min., which is the shape ,W15018.. This choice often gu...(e) We define the maximum shear stress distribution by(f) Now that we have information about the section, we can calculate the real maximum shear stress at the outermost surface of the flange byBonus: At the web-flange junction, we have a different =,-max-.=,-max.0.00711 m =,,0.153 m.-2.,0.00711 m.=0.06939 m, so that

3. [Stress analysis in shafts] A torque =100 Nm is applied to the shaft , which is running at constant speed and contains gear . Gear transmits torque to shaft through gear , which drives the chain sprocket at , transmitting ...SOLUTION

uwo mechanical components design for mechatronic systems (mse 3380) sample problems - solutions v3

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