fundamentals of machine elements sm ch 02

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Page 2-1

Chapter 2 • Load, Stress and Strain

2.1 The stepped shaft A-B-C shown in sketch a is loaded with the forces P1 and/or P2. Note

that P1 gives a tensile stress σ in B-C and σ /4 in A-B and that P2 gives a bending stress σ

at B and 1.5σ at A. What is the critical sectiona) If only P1 is applied?

b) If only P2 is applied?

c) If both P1 and P2 are applied?

Notes: This solution will ignore advanced concepts such as stress concentrations that would exist

at the fillet, since this material is covered later in the text. Similarly, the location in the crosssection where the maximum stress occurs could be identified with the information in Chapter 4,

but will be left for the student.

Solution:

If only P1 is applied, the critical section is the length B-C , since the tensile stress is highest in thissection.

If only P2 is applied, the largest stress occurs at A.

If both P1 and P2 are applied, the maximum stress in section A-B occurs at point A and is

σA=σ /4+1.5σ=1.75σThe maximum stress in section B-C occurs at point B and is

σB=σ+σ=2σTherefore, the critical section is B-C , in particular point B.

2.2 The stepped shaft in Problem 2.1 (sketch a) has loads P1 and P2. Find the load

classification if P1’s variation is sinuous and P2 is the load from a weight

a) If only P1 is applied.b) If only P2 is applied.

c) If both P1 and P2 are applied.

Notes: This solution uses the terminology on pages 30-32.

Solution:

If only P1 is applied, then the load is cyclic with respect to time (see page 30), and is a

normal load (see page 31).

If only P2 is applied, the load is sustained with respect to time, and is both a shear loadand a bending load. However, for long columns, bending is usually more important than shear, soit is reasonable to classify P2 as a bending load.

If both P1 and P2 are applied, then the load is cyclic with respect to time (although the

mean stress is not zero), and the loading is combined (see page 32).

2.3 The sign convention for bending moments can be derived from the differential equation

for the elastic line:





Page 2-3

P y = 0 = R y − gma ; R y = gma∑

2.5 Sketch b shows the forces acting on a rectangle. Is the rectangle in equilibrium?

Notes: To satisfy equilibrium, Equations (2.3) must be satisfied.

Solution:

Taking the sum of forces in the x-direction,

ΣF x=30N-20N-10N=0

Therefore, equilibrium in the x-direction is satisfied. Taking equilibrium in the y-direction gives

ΣF y=-18N+20N-20N+18N=0

Therefore, equilibrium in the y-direction is satisfied. Taking moments about point A,

Σ M A=(20N)(7cm)+(10N)(5cm)-18N(7cm)=64Ncm

Therefore, moment equilibrium is not satisfied, and the rectangle is not in equilibrium.

2.6 Sketch c shows the forces acting on a triangle. Is the triangle in equilibrium?

Notes: To satisfy equilibrium, Equations (2.3) must be satisfied.

Solution:


ΣF x=65.37N-40N(cos10°)-30N(cos30°)=0.00


ΣF y=39.76N-17.81N-40N(sin10°)-30N(sin30°)=0.00

Therefore, equilibrium in the y-direction is satisfied. Taking moments about point A,

Σ M A=(39.76N)(2.5cm)-40N(cos10°)(5cm)+30N(sin30°)(6.5cm)=0.00Ncm

Therefore, the triangle is in equilibrium.



Page 2-4

2.7 Sketch d shows a cube with side lengths a and eight forces acting at the corners. Is the

cube in equilibrium?

Notes: For three dimensional equilibrium, Equations (2.1) and (2.2) must be satisfied.

Solution:


ΣF x=P+P-P-P=0


ΣF y=P+P-P-P=0

Therefore, equilibrium in the y-direction is satisfied. There are no forces in the z-direction, so

equilibrium is automatically satisfied. Taking moments about the z-axis,

Σ M oz=Pa+Pa-Pa-Pa=0

Therefore, equilibrium about the z-axis is satisfied. Taking moments about the x-axis,

Σ M oz=Pa-Pa=0Therefore, equilibrium about the x-axis is satisfied. Taking moments about the y-axis,

Σ M oy=Pa-Pa=0

Therefore, the cube is in equilibrium.

2.8 A downhill skier stands on a slope with a 5° inclination. The coefficient of friction

between the skis and the snow is 0.10 when the ski is stationary and 0.07 when the ski

starts to slide. Is the skier in static equilibrium when he is stationary and when he slides

down the slope?

Notes: For equilibrium, the force Equations in (2.3) needs to be satisfied.



Page 2-5

Solution:

A free-body diagram of the skier is shown tothe right. The skier’s weight is mag, and the

slope has a normal force as a reaction and a

friction force which develops tangent to the

slope. Summing forces in the direction normal

to the slope yields:ΣF = N -magcos5°; N =magcos5°

Taking forces tangent to the slope,

ΣF =magsin5°-µ N =magsin5°-µ magcos5°; µ=tan5°=0.08749

If the skier is stationary, the coefficient of friction is 0.10, and there is enough friction to keep the

skier stationary. If the skier is moving, then the coefficient of friction is 0.07, so the friction force

developed is not enough to maintain equilibrium. The skier will accelerate.

2.9 Given the components shown in sketches e and f, draw the free-body diagram of eachcomponent and calculate the forces.

Notes: For equilibrium, the force Equations in (2.3) needs to be satisfied. Assume there is nofriction in sketch (e). In sketch f, moment equilibrium shows that the only possible force in each

member is an axial force.

Solution:

The free body diagram for sketch e is shown to the left. Flat surfaces give normal reaction forcesas shown. From Equation (2.3), and summing forces in the y-direction,

ΣF y=0=-W + N 1cos30°+ N 2cos60°; N 1=W /cos30°- N 2(cos60°/cos30°)

Summing forces in the x-direction,

ΣF x=0= N 1sin30°- N 2sin60°=(W /cos30°- N 2(cos60°/cos30°))sin30°- N 2sin60°; N 2=0.5W

Substituting into the x-direction equilibrium equation gives N 1=0.866W .

The free body diagram for sketch f is shown to the right. For equilibrium in the x-direction,

ΣF x=0=P-P2sin30°; P2=2P=2.4kN

Taking equilibrium in the y-direciton.

ΣF y=-P1+P2cos30°=-P1+2Pcos30°; P1=1.732P=2.08kN



Page 2-6

2.10 A 5-m-long bar is loaded as shown in sketch g. The bar cross section is constant along its

length. Draw the shear and moment diagrams and locate the critical section.

Notes: In such problems, one must first obtain the reactions before generating the shear and

moment diagrams. Also, it is useful to consider moment equilibrium first. In this problem, the

shear and moment diagrams are obtained through the method of sections.

Solution:

The reaction forces have been added to the figure in red. Summing moments about the origin

gives

Σ M o=0=10kN(2m)- Ay(3m)+3.2kN(5m); Ay=12kN

Summing forces in the y-direction yieldsΣF y=0=-Oy+10kN- Ay+3.2kN; Oy=10kN-12kN+3.2kN=1.2kN

The method of sections described on page 39 is used to obtain the shear and moment diagrams.

This proceeds as follows:

a) 0≤ x<2m: ΣF y=0=-Oy+V ; V =Oy=1.2kN

Σ M O=0= M -V ( x); M =Vx=(1.2kN) x

Note: at x=2m, M =2.4kNm

b) 2m≤ x<3m: ΣF y=0=-Oy+10kN+V ;V =Oy-10kN=-8.8kN

Σ M O=0= M -V ( x)-10kN(2m); M =20kNm-(8.8kN) x

c)3m≤ x: ΣF y=0=-Oy+10kN-12kN+V ;V =3.2kN

Σ M O=0= M -V ( x)-10kN(2m)+12kN(3m);

M =-16kNm+3.2 x

The shear and moment diagrams are plotted as follows:

From these shear and moment diagrams, it can be seen that the critical section is at point A.

2.11 A beam is supported by a wire in one end and by a pin in the other end. The beam is

loaded by a force parallel with the wire. What is the direction of the reaction force in the

pin?

Notes: This merely requires an understanding of the equations of equilibrium. One of the

fundamental notions is that “you can’t push on a rope”, so that the wire can only support an axial,

tensile load.



Page 2-7

Solution:Force equilibrium gives that the force in the pin has to be parallel with the wire since the wire can

only transmit an axial force. Therefore, the force in the pin has to be in the same direction as in

the wire.

2.12 Sketch h shows a simple bridge. The midpoint of the bridge is held up by a wire leading

to the top hinge of the two equal beams. When a truck runs over the bridge, the force in

the wire is measured to be 500,000 N. Determine the forces in the beams and the

horizontal force component in the middle of the bridge.

Notes: This problem can be easily solved by drawing a free body diagram of one of the beams

and then applying the equations of equilibrium.

Solution:

A free body diagram of the left beam is shown to the right. It is assumed that each beam takesone-half the load, as shown. From moment equilibrium, about the pin, it can be seen that

Σ M =0=(250,000N)(lcos45°)-PH(lsin45°); PH=250,000N

Therefore, the load carried by the beam is

Pbeam = 250kN ( )2 + 250kN ( )

2 = 353.6kN

Taking force equilibrium in the horizontal direction gives

ΣF h=0= Rx-PH; Rx=PH=250kN

This is the horizontal force component in the middle of the bridge.

2.13 Sketch i shows a 0.05-m-diameter steel shaft supported by self-aligning bearings (which

can provide radial but not bending loads on the shaft). A gear causes each force to be

applied as shown. All length dimensions are in meters. The shaft can be considered

weightless. Determine the forces at A and B and the maximum bending stress. Draw

shear and moment diagrams.



Page 2-8


moment diagrams. Also, it is useful to consider moment equilibrium first. In this problem, theshear and moment diagrams are obtained through direct integration, as suggested by Equations

(2.4) and (2.5). The maximum stress is given by σ= Mc / I where I is the moment of inertia, given

in Table 4.1 on page 148.

Solution:Taking moment equilibrium about point A yields:

Σ M A=0=(1kN)(0.2m)- By(0.5m)-(0.5kN)(0.64m); By=-0.24kN=-240N

Vertical force equilibrium gives Ay:

ΣF y=0= Ay-1kN+ By+0.5kN; Ay=1kN- By-0.5kN=0.74kN=740N

The shear and moment diagrams are constructed directly from the applied load, using Equations

(2.4) and (2.5). Since the loads are simple, this approach allows simple integration. The diagrams

are as follows:

As can be seen, the maximum moment is at x=0.2 and has the value of

M max=(740N)(0.2m)=148Nm. This is the area under the shear force curve, indicated in red.

The moment of inertia of a round cross section is (see Table 4.1 on page 148, or the inside back

cover):

I =πd

4

64=

π 0.05m( )4

64= 3.068× 10

−7 m4

The maximum bending stress in the shaft is:

σ = Mc

I = 148 Nm( ) 0.025m( )

3.068× 10−7

m4 = 12.06 MPa

2.14 Sketch j shows a simply supported bar loaded with a force P at a position one-third of the

length from one of the support points. Determine the shear force and bending moment in

the bar. Also, draw the shear and moment diagrams.



shear and moment diagrams are obtained through direct integration, as suggested by Equations

(2.4) and (2.5).



Page 2-9

Solution:

Moment equilibrium about point A yields:

Σ M A=0=P(l /3)- By(l); By=P /3

Vertical force equilibrium gives:

ΣF y=0= Ay-P+ By; Ay=P- By=P-P /3=2P /3

The shear and moment diagrams are obtained through direct integration and are shown below.

2.15 Sketch k shows a simply supported bar loaded by two equally large forces P at a distance

l /4 from its ends. Determine the shear force and bending moment in the bar, and find the

critical section with the largest bending moment. Also, draw the shear and moment

diagrams.



shear and moment diagrams are obtained through direct integration, as suggested by Equations(2.4) and (2.5).

Solution:

Moment equilibrium about point A yields:

Σ M A=0=P(l /4)+P(3l /4)- By(l); By=P

Vertical force equilibrium gives

ΣF y=0= Ay-P-P+ By; Ay=P


The largest bending moment occurs between the two applied forces and has the magnitude of

M =Pl /4.

2.16 The bar shown in sketch l is loaded by the force P. Determine the shear force and

bending moment in the bar. Also, draw the shear and moment diagrams.



Page 2-10

Notes: In such problems, one must first obtain the reactions before generating the shear andmoment diagrams. Also, it is useful to consider moment equilibrium first. In this problem, the


(2.4) and (2.5).

Solution:Moment equilibrium about point A yields:

Σ M A=0=Pl+ Byl; By=-P


ΣF y=0=-P+ Ay+ By; Ay=2P


2.17 Sketch m shows a simply supported bar with a constant load per unit length w0 imposed

over its entire length. Determine the shear force and bending moment as functions of x.Draw a graph of these functions. Also, find the critical section with the largest bending

moment.




(2.4) and (2.5).

Solution:

For the purposes of equilibrium, the distributed load can be replaced by a point load of magnitude

w0l at the midpoint of the span. Therefore, moment equilibrium about point A yields:

M A = 0 = w0l l

2

− B yl∑ ; B y =

w0l

2Force equilibrium in the vertical direction gives:

F y = 0 = A y − w0l∑ + B y ; A y = w0 l

2The shear and moment diagrams are obtained through direct integration and are shown below.



Page 2-11

It can be seen that the maximum moment occurs at mid-span. The magnitude of this force is the

area under either triangle on the shear curve, as suggested by Equation (2.5). This gives

M max =1

2

wol

2

l

2

=

w 0l2

8

2.18 Sketch n shows a simply supported beam loaded with a ramp function over its entire

length, the largest value being 2P / l. Determine the shear force and the bending moment

and the critical section with the largest bending moment. Also, draw the shear and

moment diagrams.

Notes: One must first obtain the reactions before generating the shear and moment diagrams.

Also, it is useful to consider moment equilibrium first. In this problem, the shear and moment

diagrams are obtained through the method of sections.

Solution:The loading is described by

q x( ) = −2P

l

x

l

= −

2Px

l2

In analyzing equilibrium, the loading is replaced by a force of magnitude P at a location of

x=2l /3. Therefore, moment equilibrium about point A gives:

M A = 0 = P2 l

3

− B y l( ); B y =

2P

3∑


F y = 0 = A y − P + B y ; A y = P − B y = P

3∑

Taking a section at a location x in the span,

F y = 0 = A y −1

2

2Px

l2

x( )∑ −V ; V =

P

3−

Px2

l2

M A = 0 =1

2

2Px

l2

x( )

2

3 x

∑ + Vx − M ;

M =2Px

3

3l2+

P

3−

Px2

l2

x =

Px

3−

Px3

3l2

The shear and moment diagrams are shown below:



Page 2-12

The maximum moment occurs when V =0, which occurs at:

0 = P

3−

Px2

l2 ; x =

l

3Substituting this into the moment equation gives

M max = P

3

l

3

−

P

3l2

l

3

3

= Pl

3 3−

Pl

9 3=

2Pl

9 3

2.19 Draw the shear and moment diagrams, and determine the critical section for the loading

conditions shown in sketch o. Neglect the weight of the bar. Let w0=100kN/m, P=5kN,

M =2x104N-m, and l=300mm.


Also, it is useful to consider moment equilibrium first. In this problem, the shear and moment

diagrams are obtained through singularity functions. Table 2.2 on page 43 is useful for writingthe load distribution in terms of singularity functions.

Solution:

For equilibrium reasons, the distributed loads can be replaced by point loads as shown below:

Taking moment equilibrium about point A gives:

M A = 0 = 5kN ( )0.200m

3

+10kN 0.15m( )+ 5kN 0.233 m( ) − B y 0.2m( )+ 5kN 0.3m( ) + 20kNm∑

or By=22.6kN. Taking force equilibrium in the vertical direction gives:

F y = 0 = A y − 5kN −10kN − 5kN + B y − 5kN ; A y = −2.4kN ∑Therefore, the load distribution can be written in terms of singularity functions as (see Table 2.2

on page 43):

−q x( ) = 2.4kN x−1 − 1kN / m x

1 − x − 0.1 m1[ ] + 22.6 kN x − 0.2 m

−1

+1kN / m x − 0.2m 1 − x − 0.3 m 1[ ]− 5kN x −0.3m −1 + 20kNm x − 0.3m −2

Therefore, the shear distribution is given by Equation (2.4) as

V x( )= 2.4kN x0 −

1kN / m

2 x

2 − x − 0.1m2 − x − 0.2m

2 + x − 0.3m2[ ]

+22.6kN x − 0.2m0 − 5kN x − 0.3m

0 + 20kNm x − 0.3m−1

And the moment distribution is given by Equation (2.5) as



Page 2-13

M x( ) = 2.4 kN x1 −

1kN / m

6 x

3 − x − 0.1 m3 − x − 0.2 m

3 + x − 0.3m3[ ]

+22.6kN x − 0.2 m1 − 5kN x − 0.3m

1 + 20kNm x − 0.3m0

The shear and moment diagrams are sketched below:

2.20 Redo problem 2.19 while considering the weight of the bar per unit length to be

w0=5N/m.

Notes: This problem is useful after Problem 2.19 has been completed, since the statics then

suggests the answer - that the beam weight is negligible.

Solution:

Taking moment equilibrium about point A gives:

M A = 0 = 5kN ( )0.200m

3

− 5 N / m( ) 0.3m( ) 0.15m( )∑+10kN 0.15m( ) + 5kN 0.233m( ) − B y 0.2m( ) + 5kN 0.3m( ) + 20kNm

or By=22.601kN. Taking force equilibrium in the vertical direction gives:

F y = 0 = A y − 5kN −10kN − 5kN + B y − 5kN ; A y = −2.4004kN ∑These reactions are so close to the reactions in Problem 2.19 that the difference in the shear andmoment diagrams is negligible. The derivation is identical to Problem 2.19, but the load function

has two terms associated with the weight.

2.21 Find the expressions for the shear force, bending

moment, and axial load P for the curved member shown

in sketch p. Neglect the weight of the bar. Let P=10N

and r =1m.

Notes: The reactions must be determined before the problem canbe solved. The method of sections can be used to obtain a

solution for this problem.

Solution:

Taking moments about the base of the curved member gives:

Σ M =0= M 0-(Pcos30°)r -(Psin30°)r ; M 0=1.366Pr =13.7Nm

Taking vertical force equilibrium:

ΣF y=0= Ry-Psin30°; Ry=Psin30°=5N

Taking horizontal force equilibrium,

ΣF x=0= Rx+Pcos30°; Rx=-8.66N

Taking a section of the member at an angle θ gives the free body diagram shown below.

Equilibrium in the x-direction gives:



Page 2-14

ΣF x=0= Rx+V cosθ-PsinθEquilibrium in the y-direction gives:

ΣF y=0= Ry-V sinθ-PcosθThese two equations yield

P= Rxsinθ+ Rycosθ=(-8.66N)sinθ+(5N)cosθ

V =- Rxcosθ+ Rysinθ=(8.66N)cosθ+(5N)sinθMoment equilibrium about the curved member end gives:

Σ M =0= M 0+ Rx(r sinθ)- Ry(r -r cosθ)+ M ; M =- M 0-rRxsinθ+rRy(1-cosθ)

M =-13.7Nm-(1m)(-8.66N)sinθ+(1m)(5N)(1-cosθ)=-13.7Nm+8.66Nmsinθ+5Nm(1-cosθ)

2.22 Sketch q shows a sinusoidal distributed force applied to a beam. Determine the shear

force and bending moment for each section of the beam.

Notes: The reactions must be determined before the problem can be solved. This solution uses

singularity functions to obtain the shear and bending moment diagrams.

Solution:The sinusoidal distribution is symmetric, so we know that the statically equivalent load can be

placed at the center of the beam, and has a magnitude of

P = w0 sinπ xl

dx

0

l∫ = −

lw0

πcos

π xl

x=0

x= l

=2 lw0

πTaking moment equilibrium about point A,

M A = 0 = P

l

2

− B y l( ); B y =

P

2 =

lw0

π∑Vertical force equilibrium gives

F y = A y − P + B y ; A y = P − B y =2lw0

π−∑ lw0

π=

lw0

πThe load function is given by

−q x( )= − lw0

π x

−1 + w0 sinπ xl

−

lw0

π x − l

−1

Equation (2.4) gives

V x( )= − q x( )dx∫ = −lw0

π x

0 − lw0

πcos

π xl

+

lw0

π−

lw0

π x − l

0


M x( ) = V x( )dx∫ = − lw0

π x 1 − l2 w0

π2 sin π xl

+ lw0 x

π− lw0

π x − l 1

2.23 Find the length c that gives the smallest maximum bending moment for the load

distribution shown in sketch r.



Page 2-15

Notes: The reactions must be determined before the problem can be solved. This solution usesmethod of sections to obtain the shear and bending moment diagrams. Note that large negative

bending stresses are still objectionable. For small values of c, note that the moment in the center

part will be large; for large values of c, the moment towards the edges will be larger. Therefore,

the solution requires that the absolute value of the largest negative moment equal the largest

positive moment.

Solution:

Because of symmetry, both R1 and R2 will have the value of w0l /4. Note that the shear and

bending moment diagrams will be symmetric. Therefore, using the method of sections between c

and l /2, and considering only moment equilibrium,

M x = 0 = w0 x

2

2 l

x

3

−

w0l

4 x − c( ) + M ; M =

w0l

4 x − c( ) −∑

w0 x3

6 lAt x=c, this gives a value of

M x=c = w0l

4c − c( ) −

w0c3

6l= −

w0c3

6lwhile at x=l /2,

M x= l / 2 = w0 l

4

l

2− c

−

w0l

2

3

6l=

w0l2

8−

w0lc

4−

w0l2

48=

5w0 l2

48−

w0lc

4

Using the requirement that M x=l/2=- M x=c gives:

w0c3

6l=

5w0 l2

48−

w0lc

4;

c3

6l=

5l2

48−

lc

4This is solved numerically as c=0.380l.

2.24 Draw the shear and moment diagrams for the load distribution shown in sketch s and the

load intensity of q( x)=< x>n when n=2 and 3. Also, calculate the reaction forces.

Notes: The statics is difficult in this problem, since the load distribution requires one to obtain the

centroid through integration. Once the reactions are found, the shear and moment diagrams can be

found through the use of singularity functions.



Page 2-16

Solution:

To obtain the forces, the load distribution is replaced by a concentrated force. The magnitude of the force is given by:

P = kxn

dx0

1

∫ = k

n + 1 x

n+1

0

l

= kl

n+1

n + 1

The position of the concentrated force for this statically equivalent system is given by:

x =

xkxndx

0

l

∫

kxndx

0

l

∫ =

kxn+1

dx

0

l

∫

kxndx

0

l

∫ =

k

n + 2ln+2

k

n + 1ln+1

= n + 1

n + 2l

Moment equilibrium about point 1 yields R2 as follows:

M = 0∑ = R2l − Px; R2 = Px

l=

1

l

kln+1

n +1

n + 1

n + 2l

=

kln+1

n + 2

Taking vertical force equilibrium,

F y = 0 = R1 + R2 − P; R1 = P − R2 = kl

n+1

n +1 − kl

n+1

n + 2 = kl

n+1

n +1( ) n + 2( )∑Therefore, the load distribution function for n=2 is given by

q x( ) = − R1 x−1 + k x

2 − k x − l2 − 2kl x − l

1 − kl2

x − l0 − R2 x − l

−1

Using Equation (2.4), the shear in terms of singularity functions is

−V x( ) = − R1 x0 +

k

3 x

3 − x − l3( ) − kl x − 1

2 − kl2

x − l1 − R2 x − l

0

Equation (2.5) gives the moment distribution in terms of singularity functions as

− M x( ) = − R1 x1 +

k

12 x

4 − x − l4( ) −

kl

3 x −1

3 − kl 2

2 x − l

2 − R2 x − l1

These are sketched below.

2.25 Draw the shear and moment diagrams and give the reaction forces for the load

distribution shown in sketch t.



Page 2-17


Also, it is useful to consider moment equilibrium first. In this problem, the shear and momentdiagrams are obtained through direct integration, as suggested by Equations (2.4) and (2.5).

Solution:

Due to symmetry, the reaction forces are equal, and have the value Ay= By=w0a. The shear and

moment diagrams are obtained through direct integration and are as follows:

2.26 The loading acting on a beam consists of a distributed load of 2500 N/m over the entire

beam, and three concentrated loads, each of 5000N, located at l /4, l /2, and 3l /4. The beam

length is 8m. Choose the type of beam support and the position. Also, draw the shear and

moment diagrams.

Notes: There is much freedom in this problem to choose the support. This solution uses a beam

which is simply supported at its ends, so that the problem is illustrated as follows.

Solution:From symmetry, the reactions are simply calculated as Ay= By=17.5kN. Therefore, the shear and

moment diagrams, obtained through direct integration, are as follows:

2.27 The maximum load for a bridge crossing a river occurs when a fully loaded truck passes

over. The maximum load is 550,000N distributed over 22m (the farthest distance betweenthe wheels of a vehicle). The weight of the 100m long bridge is 210 tons. Choose the

position of the bridge pillars and the type of support.



Page 2-18

Solution:This is an open-ended problem. Since the number of pillars is not restricted, any number can be

chosen, but due to accuracy in the building process it should be assumed that each pillar carries

its own load independent of the other pillars. Thus the bridge works as if it is hinged at each

pillar. Choosing the distance between pillars to be 20 or 25m with the support type hinged at one

end of the bridge and simply supported, e.g., on rollers, on the other pillars to accommodate thethermal expansion due to temperature variation.

2.28 The simply supported bar shown in sketch u has P1=5kN, P2=8kN, w0=4kN/m, and

l=12m. Use singularity functions to determine the shear force and bending moments as

functions of x. Also, draw your results.


Also, it is useful to consider moment equilibrium first. Table 2.2 on page 43 helps in writing the

load in terms of singularity functions.

Solution:

Summing moments about A gives:

M A = 0 = P1l

4

−

w0l

2

l

2

+ P2

3l

4

−

w0l

8

3 l

4+

l

12

+ B y l( )∑

B y = − 5kN 4

+ 4kN / m( ) 12m( )4

− 3 8kN ( )4

+ 5 4kN / m( ) 12m( )48

= 9.75kN

Vertical force equilibrium gives

F y∑ = 0 = A y + P1 − w0l

2+ P2 −

w0l

8+ B y = 0 ; A y = 7.25kN

Using the information in Table 2.2 on page 43, the load using singularity functions is:

−q x( )= A y x−1 + P1 x −

l

4

−1+ P2 x −

3l

4

−1+ B y x − l

−1 − w0 x − l

4

0

+ 4w0

l x −

3l

4

1

− 4w0

l x − l

1

Using Equation (2.4), the shear in terms of singularity functions is

V x( )= −q x( )dx∫

= A y x0

+ P1 x − l

4

0

+ P2 x −3l

4

0

+ B y x − l0

− w0 x − l

4

1

+2w0

l x −3l

4

2

−2w0

l x − l2

Equation (2.5) gives the moment distribution in terms of singularity functions as

M x( ) = V x( )dx∫

= A y x1 + P1 x −

l

4

1

+ P2 x −3 l

4

1

+ B y x − l1 −

w0

2 x −

l

4

2

+2w0

3l x −

3l

4

3

−2w0

3l x − l

3

These are sketched below.



Page 2-19

2.29 Use singularity functions for the force system shown in sketch v to determine the load

intensity, the shear force, and the bending moment. From a force analysis determine thereaction forces R1 and R2. Also, draw the shear and moment diagrams.


Also, it is useful to consider moment equilibrium first. Table 2.2 on page 43 helps in writing theload in terms of singularity functions.

Solution:

Taking moment equilibrium about point O,

M O∑ = 0 = 40 lb( ) 4in( ) − 30lb( ) 8in( )− R2 14in( )+ 60lb( ) 18in( ); R2 = 71.43lb

Taking force equilibrium in the vertical direction,

ΣF y=0= R1-40lb+30lb+ R2-60lb; R1=-1.43lb

Using the information in Table 2.2 on page 43, the load distribution can be written in terms of

singularity functions as:

−q x( )= −1.43lb x−1 − 40lb x − 4in

−1 + 30lb x − 8in−1 + 71.43lb x −14in

−1 − 60lb x −18in−1

Using Equation (2.4), the shear distribution in terms of singularity functions is

V x( )= −1.43lb x0 − 40lb x − 4in

0 + 30lb x −8in0 + 71.43lb x −14in

0 − 60lb x −18in0

Using Equation (2.5), the moment distribution in terms of singularity functions is:

M x( ) = −1.43lb x1 − 40lb x − 4in

1 + 30lb x − 8in1 + 71.43lb x −14in

1 − 60lb x −18in1

The shear and moment diagrams are sketched below.



Page 2-20

2.30 Use singularity functions for the force system shown in sketch w to determine the load

intensity, the shear force, and the bending moment. Draw the shear and moment

diagrams. Also, from a force analysis, determine the reaction forces R1 and R2.

Notes: This problem is very similar to problem 12.29. One must first obtain the reactions before

generating the shear and moment diagrams. Also, it is useful to consider moment equilibrium

first. Table 2.2 on page 43 helps in writing the load in terms of singularity functions.

Solution:

Taking moment equilibrium about point O,Σ M O=0=(35lb)(3in)+(40lb)(7in)+(60lb)(8in)- R2(18in); R2=48.06lb

Taking force equilibrium in the vertical direction,

ΣF y=0= R1-35lb-40lb-60lb+R2; R1=86.94lb



−q x( )= 86.94lb x−1 − 35lb x − 3in

−1 − 40lb x − 7in−1 − 60lb x − 8in

−1 + 48.06lb x −18in−1


V x( )= 86.94lb x0 − 35lb x − 3in

0 − 40lb x − 7in0 − 60lb x − 8in

0 + 48.06lb x −18in0


M x( ) = 86.94lb x1 − 35lb x − 3in

1 − 40lb x − 7in1 − 60lb x − 8in

1 + 48.06lb x −18in1


2.31 Draw a free-body diagram of the forces acting on the simply supported bar shown in

sketch x, with w0=6kN/m and l=12m. Use singularity functions to draw the shear force

and bending moment diagrams.



Page 2-21

Notes: One must first obtain the reactions before generating the shear and moment diagrams. Thestatics is greatly simplified by the symmetry of the problem. Table 2.2 on page 43 helps in

writing the load in terms of singularity functions.

Solution:

Because of symmetry,

A y = B y =1

2

2w0l

3

=

w0l

3=

6kN / m( ) 12m( )

3= 24kN



−q x( )= 24kN x −1 − 3w0l

x 1+ 3w0l

x − l3

1

+ 3w0l

x − 2 l3

1

+ 24kN x − l −1 − 3w0l

x − l 1


V x( )= 24kN x 0 −

3w0

2 l x

2 + 3w0

2 l x −

l

3

2

+ 3w0

2 l x −

2 l

3

2

+ 24kN x − l 0 −

3w0

2 l x − l

2


M x( ) = 24kN x1 −

w0

2l x

3 + w0

2l x −

l

3

3

+ w0

2l x −

2l

3

3

+ 24kN x − l1 −

w0

2l x − l

3

= 24kN x1 −

1kN

4m2 x

3 +1kN

4m2 x −

l

3

3

+1kN

4m2 x −

2 l

3

3

+ 24kN x − l1−

1kN

4m2 x − l

3


2.32 Sketch y shows a simply supported bar with w0=5kN/m and l=12m. Draw a free-body

diagram of the forces acting along the bar as well as the coordinates used. Use singularity

functions to determine the shear force and the bending moment.



Page 2-22

Notes: One must first obtain the reactions before generating the shear and moment diagrams. The

statics is greatly simplified by the symmetry in the problem. Table 2.2 on page 43 helps in writing

the load in terms of singularity functions.

Solution:

Because of symmetry,

A y = B y =1

2

w0l

2

=

w 0l

4=

5kN / m( ) 12m( )

4= 15kN



−q x( )= 15kN x −1 − 5 kN m

x 0 + 5

kN 6m

2 x 1 − 5

kN 3m

2 x − 6m 1 + 5

kN 6m

2 x −12m 1 + 5kN / m x − 12 0 + 15 x −12 −1


V x( )= 15kN x 0 − 5

kN

m x 1 +

5kN

12m2

x 2 −

5kN

6m2

x − 6m 2 +

5kN

12m2

x −12m 2 + 5kN / m x −12 1 + 15 x −12 0


M x( ) = 15kN x1 −

5kN

2m x

2 +5kN

36m2 x

3 −5kN

18m2 x − 6m

3 +5kN

36m2 x −12m

3 +5kN

2m x −12

2 +15 x −121


2.33 Redo the case study 2.1 problem if the 400-N force is evenly distributed along the width

of the roller and a unit step is used to represent the loading. The width of the roller is 30 mm. Do

both (a) and (b) portions of the case study while considering a unit step representation.

Notes: The approach is very similar to that in the case study, except that the singularity functionsare slightly different to reflect periodic distributed loads instead of periodic point loads.

Solution:

The new load distribution function is



Page 2-23

−q x( )= −800 N x −1 + 400 N

0.03m x − 0.060m 0 −

400 N

0.03m x − 0.090m 0 +

400 N

0.03m x − 0.13m 0 −

400 N

0.03m x − 0.16m 0

+ 400 N

0.03m x − 0.20m 0 −

400 N

0.03m x − 0.23m 0 +

400 N

0.03m x − 0.27m 0 −

400 N

0.03m x − 0.30m 0 − 800 N x − 0.36m −1


V x( )= −800 N x0 + x − 0.36m

0

[ ]+

400 N

0.03m x − 0.060m

1 − x − 0.090m1 + x − 0.13m

1

[− x − 0.16m

1 + x − 0.20m1 − x − 0.23m

1 + x − 0.27m1 − x − 0.30m

1]Using Equation (2.5), the moment distribution in terms of singularity functions is:

M x( ) = −800 N x1 + x − 0.36m

1[ ]+400 N

0.06m x − 0.060m

2 − x − 0.090m2 + x − 0.13m

2[

− x − 0.16m2 + x − 0.20m

2 − x − 0.23m2 + x − 0.27m

2 − x − 0.30m2]


The maximum moment can be evaluated at x=0.18m as

M 0.18m( ) = −800 N 0.18m1[ ]+

400 N

0.06m0.18m − 0.060m

2 − 0.18m − 0.090m2[

+ 0.18m − 0.13m2 − 0.18m − 0.16m

2] = −88 Nm

The general expressions are similar to those in the case study, and are:

−q x( )= − nP

2 x

−1 + x − 2a − n −1( )b−1[ ]+

P

c x − a − b i −1( )−

c

2

0

− x − a − b i−1( ) + c

2

0

i=1

n

∑

V x( )= − nP

2 x

0 + x − 2a − n −1( )b0[ ]+

P

c x − a − b i −1( ) −

c

2

1

− x − a − b i−1( ) + c

2

1

i=1

n

∑

M x( ) = − nP

2 x

1 + x − 2a − n −1( )b1[ ]+

P

2c x − a − b i − 1( ) −

c

2

2

− x − a− b i−1( ) + c

2

2

i=1

n

∑

where n is the number of rollers, a is the distance from bearing to first roller, b is the center

distance between rollers and c is the roller width.

2.34 A simply supported beam has a parabolic load distribution beginning at x=0. Use

singularity functions to draw the shear and moment diagrams.

Notes: The maximum value is not specified, so it is given the value of qmax. Equilibrium is slightly

more complex in this case, since the geometry has no simple centroid expression.



Page 2-24

Solution:

Moment equilibrium about point A gives

M A = 0 =∑ qmax x2

l2 x( )dx − P2l ;P2 =0

l∫

1

l

qmax x2

l2 x( )dx = qmaxl

40

l∫

Force equlibrium allows determination of P1 as:

F A = 0 =∑ P1 + P2 − qmax x

2

l2 dx; P1 = −qmaxl

4 + qmaxl3 = qmaxl

120l∫

Therefore, the load function is

−q x( )= qmaxl

12 x

−1 − x

2

l2 + l

4 x − l

−1


V x( )= qmaxl

12 x

0 − x

3

3l2 +

l

4 x − l

0


M x( ) = qmax

l

12 x

1 − x

4

12 l2+

l

4 x − l

1


2.35 Draw a free-body diagram of the forces acting on the simply suported beam shown in

sketch x and used in Problem 2.31. The central section (l /3< x<2l /3) has the load acting

upward (in the negative y direction) instead of as shown. Use singularity functions to

determine the shear force and bending moments and draw the diagram.

Notes: This is straightforward if Problem 2.31 has already been completed. One must first obtain

the reactions before generating the shear and moment diagrams. The statics is greatly simplified

by the symmetry of the problem. Table 2.2 on page 43 helps in writing the load in terms of

singularity functions.

Solution:



Page 2-25

One can see that the areas where the load acts downwards and upwards are equal, so there is no

net load applied. Because of symmetry, Ra= Rb=0. Using the information in Table 2.2 on page 43,the load distribution can be written in terms of singularity functions as:

−q x( )=3w0

l x

1 −3w0

l x −

l

3

1

− 2w0 x − l

3

0

+ 2w0 x −2l

3

0

−3w0

l x −

2 l

3

1

+3w0

l x − l

1


V x( )=3w0

2l x

2 − x − l

3

2

− x −2l

3

2

+ x − l2

− 2w0 x − l

3

1

− x −2l

3

1


M x( ) = w0

2l x

3 − x − l

3

3

− x −2l

3

3

+ x − l3

− w0 x − l

3

2

− x −2l

3

2


2.36 An extra concentrated force with an intensity of 60kN is applied downward at the center

of the simply supported bar shown in sketch y. Draw a free-body diagram of the forces

acting on the bar. Assume l=12m and w0=5kN/m. Use singularity functions to determinethe shear force and bending moments and draw the diagrams.

Notes: This problem is straightforward if Problem 2.32 has been completed. One must first obtainthe reactions before generating the shear and moment diagrams. The statics is greatly simplified

by the symmetry in the problem. Table 2.2 on page 43 helps in writing the load in terms of singularity functions.

Solution:Because of symmetry,

A y = B y =1

2

w0l

2

+

1

2P = 45kN



Page 2-26


singularity functions. Here we decide to ignore the terms acting at the right end of the beam, sincethese do not affect the shear and moment diagrams.

−q x( )= 45kN x−1 − 5

kN

m x

0 +5kN

6m2 x1 − 60kN x − 6m

−1 −5kN

3m2 x − 6m1


V x( )= 45kN x0 − 5

kN

m x

1 + 5kN

12m2 x

2 − 60kN x − 6m0 − 5kN

6m2 x − 6m

2


M x( ) = 45kN x1 −

5kN

2m x

2 +5kN

36m2 x

3 − 60kN x − 6m1−

5kN

18m2 x − 6m

3


2.37 Use singularity functions to determine the shear and bending moments for the loading

situation shown in sketch z.

Notes: One must first obtain the reactions before generating the shear and moment diagrams.Table 2.2 on page 43 helps in writing the load in terms of singularity functions.

Solution:

Moment equilibrium about point A gives:

M A = 0 = 2Pl + w0l

2

l

3

+ 2w0 l( ) 2 l( ) +∑ w0 l

2

11l

3

+ P 3l( ) − B y 4l( ); B y =

5P

4+

3w0l

2Force equilibrium gives:

F y = 0 = A y + B y − 3P − 3w0l ; A y = 3P + 3w0l∑ − B y =3w0 l

2

+7P

4Therefore, the load distribution is given by

−q x( ) = − A y x−1 + w0 x

0 − w 0

l x 1 + 2 P x − l

−1 +3w0

l x − l

1

−4w0

l x − 2 l

1 + P x − 3l−1 +

3w0

l x − 3l

1

−q x( ) = 3w0l

2+ 7P

4

x

−1 + w0

ll x

0 − x 1 + 3 x − l

1 − 4 x − 2l 1 +3 x − 3l

1[ ]+ P 2 x − l−1 + x − 3l

−1[ ]



Page 2-27

Note that we have ignored the terms which become active at x=4l, since these do not affect the

diagram or calculations on the beam. The shear distribution is obtained from Equation (2.4) as:

V x( )= 3w0 l

2+ 7P

4

x

0 + w0

2l2l x

1 − x 2 + 3 x − l

2 − 4 x − 2l 2 + 3 x −3l

2[ ]+ P 2 x − l 0 + x − 3l

0[ ]Using Equation (2.5), the moment distribution in terms of singularity functions is:

M x( ) =

3w0l

2 +

7P

4

x

1

+

w0

6l3

l x

2

− x

3

+3

x − l

3

−4

x −2

l

3

+3

x −3

l

3

[ ]+ P 2

x − l

1

+ x −3

l

1

[ ]

2.38 Obtain the shear force and moment expressions using singularity functions for a pinned

beam at both ends with loading conditions described in sketch aa. Draw a free-body

diagram of the forces acting on the bar. Draw the shear force and moment along the

length of the bar and give tabular results. Assume w0=4kN/m, P=3kN, and l=12m.

Notes: One must first obtain the reactions before generating the shear and moment diagrams.Table 2.2 on page 43 helps in writing the load in terms of singularity functions.

Solution:

Moment equilibrium about point A gives:

M A = 0 = P l

3−

w0l

6

2l

9

∑ + B yl − P

2l

3

+

w0l

6

7l

9

; B y = −

5l

9

w0

6

+

P

3= −3.444kN

Force equilibrium gives:

F y = 0∑ = A y + B y + P −P + w0 l

6 − w0l

6 ; A y = − B y = 3.444kN

Therefore, the load distribution is given by

−q x( )= A y x−1 −

3w0

l x

1 − x − l

3

1

+ x − 2l

3

1

+ w0 x −

l

3

0

+ x − 2l

3

0

+ P x −

l

3

−1− x −

2l

3

−1

Note that we have ignored the terms which become active at x=l, since these do not affect the

diagram or calculations on the beam. The shear distribution is obtained from Equation (2.4) as:

V x( )= A y x 0 −

3w0

2l x

2 − x − l

3

2

+ x − 2l

3

2

+ w0 x − l

3

1

+ x − 2l

3

1

+ P x − l

3

0

− x − 2l

3

0


M x( ) = A y x 1 −

w 0

2l

x 3 − x −

l

3

3

+ x − 2l

3

3

+ w0

2

x − l

3

2

+ x − 2l

3

2

+ P x − l

3

1

− x − 2l

3

1

The shear and moment diagrams are sketched below:



Page 2-28

2.39 A steel bar is loaded by a tensile force P=25kN. The cross section of the bar is circular

with a radius of 7mm. What is the normal stress in the bar?

Notes: Equation (2.7) is needed to solve this problem.

Solution:The cross sectional area of the bar is:

A = πr 2 = π 0.007m( )

2 = 1.539×10−4

m2


σ = P

A=

25kN

1.539 ×10−4

m2 = 162 MPa

2.40 A stainless steel bar of square cross section has a tensile force of P=15kN acting on it.

Calculate how large the side l of the cross-sectional area must be to give a tensile stress in

the bar of 120MPa.


Solution:

The cross sectional area of the bar is l2. Therefore, from Equation (2.7),

σ = P

A; A = l

2 = P

σ; l =

P

σ=

15kN

120 MPa= 0.0112m = 11.2 mm

2.41 What is the maximum length lmax that a copper wire can have if its weight should not give

a higher stress than 75 MPa when it is hanging vertically? The density of copper is 8900

kg/m3, and the density of air is so small relative to that of copper that it may be neglected.

The acceleration of gravity is 9.81 m/s2.


Solution:

If the wire has a cross sectional area of A, then its volume is V = Almax. The weight of this wire is

P=ρgAlmax

From Equation (2.7),



Page 2-29

σ = P

A=

ρgAlmax

A= ρglmax ; lmax =

σρg

=75 MPa

8900kg / m3( ) 9.81m / s

2( )= 859m

2.42 A machine weighing 5 tons will be lifted by a steel rod with an ultimate tensile strength

of 860 MPa. A safety factor of 5 is to be used. Determine the diameter needed for thesteel rod.

Notes: Equations (1.1) and (2.7) are needed to solve this problem. Note that a ton is 1000 kg

when the problem is stated in metric units.

Solution:

The design stress is determined from Equation (1.1) as:

ns =σall

σd

; σd =σall

ns

=860 MPa

5=172 MPa

The applied load is P=(5000kg)(9.81m/s2)=49.05kN. Therefore, the required diameter iscalculated from Equation (2.7) as:

σ = P

A=

P

πd 2

4

; d =4P

πσ=

4 49.05kN ( )

π 172 MPa( )= 0.0191m =19.1mm

2.43 The largest measured ocean depth (near the Philippines in the Pacific) is 11 km. What

ultimate strength is required for a steel wire to reach the bottom without being torn apart

by its own weight? The density of steel is 7860kg/m3 and that of water is 1000kg/m

3.

Notes: The approach is very similar to problem 2.41, but now the buoyancy force is subtracted

from the load.

Solution:

If the wire has a cross sectional area of A, then its volume is V = Almax. The load on the top of the

wire is the weight of the wire minus the buoyancy force which tends to support the wire.

Therefore,

P=(ρsteel-ρwater)gAlmax


σ = P

A=

ρsteel − ρwater ( )gAlmax

A= ρsteel − ρwater ( )glmax

= 7860kg / m3 −1000kg / m

3( ) 9.81m / s2( ) 11000m( ) = 740 MPa

Reviewing the properties of steel on the inside front cover, there is no steel that can reach to thisdepth without failing at the surface.

2.44 In Problem 2.43 a steel wire is used to measure the ocean depth. Find a way to do the

measurements without risking breaking the wire due to tensile stresses being too high.



Page 2-30

Notes: It is very helpful to complete Problem 2.43 before attempting this problem. This is an

open-ended problem, with many possible solutions. Students should be encouraged to developtheir own creative solutions to this problem.

Solution:

The obvious solution is to make the wire of a polymer or other material with exactly the same

density as water and with the same compressibility as water. Each part of the wire is then buoyantand the tensile stress is zero along the whole wire. Another way is to use a steel wire but attach

floats at regular intervals, this producing buoyancy of the wire parts.

2.45 A string on a guitar is made of nylon and has a cross-sectional diameter of 0.6mm. It is

tightened to a force P=15N. What is the stress in the string?


Solution:The cross sectional area of the string is:

A =πd

2

4=

π 0.0006m( )2

4= 2.827×10

−7m

2


σ = P

A=

15 N

2.827 ×10−7

m2 = 53.1 MPa

2.46 Determine the normal and shear stresses in sections A and B of sketch bb. The cross-

sectional area of the rod is 0.025m2. Ignore bending and torsional effects.

Notes: To solve this problem, one must be capable of solving for the reactions at A, be familiar

with the method of sections, and be able to use Equation (2.7).

Solution:

From equilibrium, the reactions at the wall include the forces Rx=0, Ry=10kNcos30°=8.66kN, and Rz=10kNsin10°=5kN. Taking a section at A, and applying Equation (2.7),

σ = − P y

A= −

8.66kN

0.025m2 = −346.4kPa

The maximum shear stress is discussed in Chapter 5, but we can estimate the average shear stressas

τave = P z

A=

−5.0kN

0.025m2 = −200kPa



Page 2-31

The negative signs are consistent with the sign convention shown on page 33. At location B, there

is no normal force, so σ=0. Since we are told to ignore torsional effects, we can estimate the

average shear stress as:

τave = P

A=

10kN

0.025m2 = 400kPa

2.47 Determine the normal and shear stresses due to axial and shear forces at sections A and B

in sketch cc. The cross sectional area of the rod is 0.025m2.

Notes: To solve this problem, one must be capable of solving for the reactions at A, be familiar

with the method of sections, and be able to use Equation (2.7). Also, use θ=30°. Finally, calculate

the average shear stress, not the maximum for a round cross section as is discussed in Chapter 4.

Solution:

Taking a section at AA, there is an axial force and a shear force. While one can work in the x and

y directions, the answer is more easily obtained by applying equilibrium to the coordinate systemwhich is defined by the section. The free body diagram of the forces (moments are ignored forthis problem) at section AA are:

Therefore, taking force equilibrium in the direction normal to the surface:

ΣF =0=Pn-Psinθ; Pn=Psinθ=(10kN)(sin30°)=5kN

Therefore the normal stress is:

σ = Pn

A= 5kN

0.025m2 = 200kPa

Taking force equilibrium in the direction of the surface yields:

ΣF =0=Ps-Pcosθ; Ps=Pcosθ=(10kN)(cos30°)=8.66kN

Therefore the average shear stress is

τ = Ps

A=

8.66kN

0.025m2 = 346.4kPa



Page 2-32

At section BB, force equilibrium gives Pn=0 and Ps=10kN. Therefore the normal stress is σ=0

and the shear stress is

τ = Ps

A=

10kN

0.025 m2 = 400kPa

2.48 For the stress element given in Figure 2.13 determine how large the shear stress will

become if all three normal stresses are doubled.

Notes: This problem is deceptively simple.

Solution:

The normal stresses and shear stresses are independent of each other; if the normal stresses are

doubled, there is no effect on the shear stresses, so they remain the same.

2.49 The normal stresses in the x, y, and z directions are all equal to 100MPa, and the shear

stresses are all zero. Find the stresses acting on a plane having the normal in the direction of thespace diagonal

1

3,

1

3,

1

3

.

Notes: This problem uses Equations (B.4) through (B.9).

Solution:Refer to the directions in the new plane as x’, etc. These stresses are given by Equations (B.4)

through (B.9) in Appendix B. As long as the directions are orthogonal,

σ ′ x =1

3σ x + σ y + σ z( )+ 0 + 0 + 0 = 100 MPa

σ ′ y =1

3σ x + σ y + σ z( )+ 0 + 0 + 0 = 100 MPa

σ ′ z =1

3σ x + σ y + σ z( )+ 0 + 0 + 0 = 100 MPa

The shear stresses are:

τ ′ z , ′ x =σ3

cos ′ x , z( )+ cos ′ x , x( )+ cos ′ x , y( )[ ]+ 0 + 0 + 0

If x’ is perpendicular to y’ and z’, then this gives τz’,x’=0. Similarly, τx’,y’=τy’,z’=0.

2.50 If the stresses are such that σx=-σy and σz=τxy=τxz=τyz=0, find the shear stress acting on

the plane diagonal between the x and y axes and parallel to the z direction.

Notes: Equation (B.8) solves this problem.

Solution:

In this new orientation, note that



Page 2-33

cos ′ x , x( ) =1

2, c o s′ x , y( ) =

1

2, c o s ′ x , z( ) = 0

cos ′ y , x( ) = −1

2, c o s ′ y , y( ) =

1

2, c o s ′ y , z( ) = 0

cos ′ z , x( ) = 0 , c o s ′ z , y( ) = 0, cos ′ z , z( ) =1

Using Equation (B.8) on page 906 yields the following, keeping only non-zero terms:

τ ′ x ′ y =σ x

2−

1

2

+

σ y

2

1

2

+ 0 + 0+ 0 + 0 = −

σ x

2+

σ y

2

Since σx=-σy, one can write that τx’y’=-σx=σy.

2.51 A stressed element with shear stress τyx acting in the x-direction on the surface has the

normal in the y direction. Determine the shear stress acting in the y direction on the

surface having a normal in the x direction.

Notes: This merely requires an understanding of equilibrium.

Solution:

Refer to Figure 2.14 (b) to visualize the stresses. Applying moment equilibrium about the z-axes

yields:

M = 0 =∑ τ xydA1dx

22( )− τ yxdA2

dy

22( )

Since dA1=dydz and dA2=dxdz,τ xydxdydz − τ yxdxdydz = 0 ; τ xy = τ yx

2.52 Sketch dd shows a distributed load on a

semi-infinite plane. The stress in polar

coordinates based on the plane stressassumption is

σr = −2w0 cosθ

πr σθ=τrθ=τθr=0

Determine the expressions σx, σy, and τxy

in terms of r and θ.

Notes: Since x, y and r ,θ are orthogonal coordinates, Equations (2.12) and (2.14) can be used to

solve the problem.

Solution:

The derivation on pages 51 and 52 applies equally well if the desired stress in Equation (2.12) is

σx in terms of σr, σθ, and τrθ. Therefore,

σ x = 2τr θ sinθ cosθ + σr cos2 θ +σθ sin 2θ = 0 −2w0 cosθ

πr

cos2θ + 0 = −

2w0 cos3θ

πr For σy, use θ’=90°-θ and apply Equation (2.12):



Page 2-34

σ y = 2τr θ sin ′θ cos ′θ + σ r cos2 ′θ + σθ sin

2 ′θ

= 2τr θ sinθ cosθ + σr sin2 ′θ +σ θ cos

2 ′θ = 0 − 2w0 cosθ

πr

sin

2 θ + 0 = −2w0 cosθ sin

2θπr

τxy is given by Equation (2.14) as

τ xy = τr θ cos2θ −σ r − σ

θ2 s i n 2θ = − σr

sin2θ2 = −σr sinθ cosθ

2.53 Sketch ee shows loading of an extremely thin and infinitely long plane. Determine the

angle θ needed so that the stress element will have no shear stress. For plate thickness t w,

modulus of elasticity E and Poisson’s ratio ν, find the reduction in thickness of the plate.

Notes: Equation (2.14) gives the required shear stress equation. The thickness reduction is given

by Hooke’s Law, the proper form of which is given in Table B.2 on page 920.

Solution:

The applied stress state is σx, σy, τxy=0. Equation (2.14) gives the shear stress at an orientation φas:

τφ = τ xy cos2φ −σ x −σ y

2sin2 φ = 0 ; s i n 2φ = 0; φ = 0

From Table B.2 on page 920, Hooke’s Law for a biaxial stress state gives:

ε z =σ z

E −

ν E

σ x +σ y( ) = − ν E

σ x +σ y( )From Equation (2.33),

ε z = − ν E

σ x + σ y( ) =∆ t w

t w; ∆t w = −

νt w

E σ x + σ y( )

2.54 The stress state in a machine element is given by

S =

167 40 −30

40 25 12

−30 12 −25

where all the values are in megapascals. Determine the stress tensor if the coordinates are

rotated such that xà y= x’; yà- x= y’; and zà z= z’.

Notes: Equation (B.4) through (B.9) in Appendix B are needed for this problem.

Solution:

Note for this problem, cos( x’, x)=0, cos( x’, y)=1, cos( x’, z)=0, cos( y’, x)=-1, cos( y’, y)=0,

cos( y’, z)=0, cos( z’, x)=0, cos( z’, y)=0, cos( z’, z)=1. From Equation (B.5) on page 905,



Page 2-35

σx’=(σx)0+σy(1)+σz(0)+2τxy(0)(1)+2τyz(1)(0)+2τzx(0)(0)=σy

Therefore, σx’=25MPa. Similarly, σy’=σx=167MPa, σz’=σz=-25MPa, τx’y’=-τxy=-40MPa, τy’z’=-

τzx=30MPa, τz’x’=τyz=12MPa. Therefore, the stress tensor in the rotated coordinate system is:

S =

25 −40 12

−40 167 30

12 30 −25

2.55 A stress tensor is given by

S =

178 −83 0

−83 12 0

0 0 0

where all values are in megapascals. The stress state is applied to a machine element

made of AISI 1020 steel. Calculate the principal normal stresses and the principal shear

stresses. Alsom find the stress tensor if the coordinate system is rotated such that

xà z= x’; yà y= y’; and zà- x= z’.

Notes: Equations (2.16) and (2.18) are used in this problem in addition to (B.4) through (B.9) as

in Problem 12.54.

Solution:

Note that τxz=τyz=σz=0, so this is a plane stress field. Therefore, Equation (2.16) gives

σ1,σ2 =σ x +σ y

2± τ xy

2 +σ x − σ y( )

2

4= 178 MPa +12 MPa

2± −83 MPa( )

2 + 178 MPa −12 MPa( )2

4

or σ1=212.4MPa and σ2=-22.38MPa. From Equation (2.18), the principal shear stress is:

τ1 ,τ2 = ± τ xy

2

+

σ x −σ y( )2

4 = ± −83 MPa( )

2

+

178 MPa −12 MPa( )2

4or τ1=117.4MPa and τ2=-117.4MPa. Note for this case cos( x’, x)=0, cos( x’, y)=0, cos( x’, z)=1,

cos( y’, x)=0, cos( y’, y)=1, cos( y’, z)=0, cos( z’, x)=-1, cos( z’, y)=0, cos( z’, z)=0. Therefore, from

Equations (B.4) to (B.9), σx’=σz=0, σy’=σy=12MPa, σz’=σz=178MPa, τx’y’=τyz=0, τy’z’=-

τxy=83MPa, τz’x’=-τzx=0. Therefore the stress tensor is

S =

0 0 0

0 12 83

0 83 178

2.56 A thin, square steel plate has normals to the sides in the x and y directions. A tensile

stress σ acts in the x direction, and a compressive stress -σ acts in the y direction.

Determine the normal and shear stresses on the diagonal of the square.

Notes: Equations (2.20) and (2.21) give the stresses at the desired angle.

Solution:

Note that there are no shear stresses on the sides of the square, so the applied normal stresses are

the principal stresses, and also this is a plane stress case. Therefore, σ1=σ and σ2=-σ. Also, the



Page 2-36

diagonals of the square occur at angles φ=45° and 135° from the reference coordinate system.


σ45° =σ1 +σ2

2+

σ1 − σ2

2cos2φ =

σ −σ2

+σ − −σ( )

2cos90° =0

σ135° =σ1 + σ2

2

+σ1 −σ2

2

cos2φ =σ − σ

2

+σ − −σ( )

2

cos270 °= 0


τ45° = −σ1 − σ 2

2sin2φ = −

σ − −σ( )

2sin90° = − σ

τ135° = −σ1 − σ2

2s i n 2φ = −

σ − −σ( )

2s in270° =σ

2.57 A thin, rectangular brass plate has normals to the sides in the x and y directions. A tensile

stress σ acts on the four sides. Determine the principal normal and shear stresses.

Notes: The principal normal stresses are given by (2.16) and the principal shear stresses are given

by Equation (2.18).

Solution:

From Equation (2.16), the principal normal stresses are:

σ1,σ2 =σ x +σ y

2± τ xy

2 +σ x − σ y( )

2

4=

σ + σ2

± 0 + 0 = σ

From Equation (2.18), the principal shear stresses are:

τ1 ,τ2 = ± τ xy2 +

σ x −σ y( )2

4= 0

2.58 Given the thin, rectangular brass plate in Problem 2.57, but with the stress in y-direction

being σy=-σ instead of +σ, determine the principal normal and shear stresses and their

directions.

Notes: The solution approach is the same as Problem 2.57.

Solution:

From Equation (2.16), the principal normal stresses are:

σ1,σ2 =σ x +σ y

2± τ xy

2 +σ x − σ y( )

2

4=

σ − σ2

± 0 +σ +σ

2

2

= ±σ

From Equation (2.18), the principal shear stresses are:

τ1 ,τ2 = ± τ xy2 +

σ x −σ y( )2

4= ± 0+

σ − −σ( )( )2

4= ±σ

2.59 For the following stress states, draw the appropriate Mohr’s circle, determine the

principal stresses and their directions, and show the stress elements:



Page 2-37

a) σx=30, σy=-20, and τxy=10

b) σx=30, σy=-30, and τxy=10

c) σx=50, σy=-50, and τxy=0

d) σx=σy=-10

All stresses are in megapascals.

Notes: The Mohr’s Circle approach described on pages 55-56 is used to solve this problem.

Solution:

a) For this stress state, the center of the circle is placed at (5,0) per Equation (2.22). Using a point

on the circle of (σy,τxy)=(-20,10), the circle is drawn as follows:

The radius of the circle is given by Equation (2.23) as

r =30 − −20( )

2

2

+ 10( )2 = 26.93

Therefore the principal stresses are 5±26.93, so σ1=31.93 and σ2=-21.93. The angle of the stress

element is given by Equation (2.15) as

t a n 2φσ =2τ xy

σ x

− σ y

=20

50= 0.4; φσ = 10.9°

b) For this stress state, the center of the circle is placed at (0,0) per Equation (2.22). Using a point



r =σ x −σ y

2

2

+ τ xy2 =

30 − −30( )

2

2

+ 10( )2 = 31.62





Page 2-38

t a n 2φσ =2τ xy

σ x − σ y

=20

60= 0.333; φσ = 9.22°

c) For this stress state, the center of the circle is placed at (0,0) per Equation (2.22). Using a point



r = σ x −σ y

2

2+ τ xy

2 = 50 − −50( )2

2+ 0( )

2 = 50

Therefore the principal stresses are 0±50, so σ1=50 and σ2=-50. The angle of the stress element is

given by Equation (2.15) as

t a n 2φσ =2τ xy

σ x − σ y

=0

100= 0; φσ = 0°

d) For this stress state, the center of the circle is placed at (-10,0) per Equation (2.22). Using a

point on the circle of (σy,τxy)=(-10,0), the circle is drawn as follows:


r =σ x −σ y

2

2

+ τ xy2 =

−10 − −10( )

2

2

+ 0( )2 = 0

Mohr’s circle for this case is a point, the element is under hydrostatic pressure. The angle to theprincipal stresses is undefined, since all orientations give the same stress state. The principal

stresses are -10±0, so σ1=-10 and σ2=-10.

2.60 Repeat Problem 2.59 for

a) σx=55, σy=-15, and τxy=40

b) σx=0, σy=30, and τxy=20



Page 2-39

c) σx=-20, σy=40, and τxy=-40

d) σx=30, σy=0, and τxy=-20

All stresses are in megapascals.

Notes: The Mohr’s Circle approach described on pages 55-56 is used to solve this problem.

Solution:a) For this stress state, the center of the circle is placed at (20,0) per Equation (2.22). Using a

point on the circle of (σy,τxy)=(-15,40), the circle is drawn as follows:


r =σ x −σ y

2

2

+ τ xy2 =

55 − −15( )

2

2

+ 40( )2 = 53.15



t a n 2φσ =2τ xy

σ x − σ y

=80

70= 1.1428; φσ = 24.4°

b) For this stress state, the center of the circle is placed at (15,0) per Equation (2.22). Using apoint on the circle of (σy,τxy)=(30,20), the circle is drawn as follows:


r =σ x −σ y

2

2

+ τ xy2 =

0 − 30( )

2

2

+ 20( )2 = 25

Therefore the principal stresses are 15±25, so σ1=40 and σ2=-10. The angle of the stress element

is given by Equation (2.15) as



Page 2-40

t a n 2φσ =2τ xy

σ x − σ y

=40

−30= −1.33333; φσ = −26.56°

c) For this stress state, the center of the circle is placed at (10,0) per Equation (2.22). Using a

point on the circle of (σy,τxy)=(40,-40), the circle is drawn as follows:


r = σ x −σ y

2

2+ τ xy

2 = −20 − 40

2

2+ −40( )

2 = 50



t a n 2φσ =2τ xy

σ x − σ y

=−80

−60= 1.3333; φσ = 26.56°

d) For this stress state, the center of the circle is placed at (15,0) per Equation (2.22). Using a

point on the circle of (σy,τxy)=(0,-20), the circle is drawn as follows:


r =σ x −σ y

2

2

+ τ xy2 =

30 − 0

2

2

+ −20( )2 = 25



t a n 2φσ =2τ xy

σ x − σ y

=−40

−30= 1.3333; φσ = 26.56°



Page 2-41

2.61 Given the state of stresses shown in the two parts of sketch ff determine the principal

stresses and their directions by using a Mohr’s circle and the stress equations. Show the

stress elements. All stresses in sketch ff are in megapascals.

Notes: Equations (2.13) and (2.14) are needed to solve this problem.

Solution:a) Note that σ45°=28MPa and τ45°=-20MPa. From Equation (2.13),

σ45° = 28 MPa =σ x + σ y

2+

σ x −σ y

2cos2φ +τ xy sin2φ =

σ x +σ y

2+ 30 MPa;σ x + σ y = −4 MPa


τ45° = −20 MPa = τ xy c o s 9 0° −σ x −σ y

2sin90° = −

σ x − σ y

2; σ x − σ y = 40 MPa

These two equations yield σx=18MPa and σy=-22MPa. The center of the Mohr’s circle is placed,

per Equation (2.22), at ((σx+σy)/2,0)=(-2MPa,0). Using a point on the circle of (-22MPa, 30MPa),

the circle is drawn as follows:

The radius of the circle is given by Equation (2.23) as:

r =σ x −σ y

2

2

+ τ xy2 =18 MPa + 22 MPa

2

2

+ 30 MPa( )2

= 36.0 MPa

Therefore the principal stresses are σ1=-2MPa+36MPa=34MPa and σ2=-2MPa-36MPa=-38MPa.

b) For this case, the known values are σφ=40MPa, σy=60MPa, τφ=20MPa and φ=-30°. Therefore,

Equations (2.13) and (2.14) give the following two simultaneous algebraic equations:

σ−30° = 40 MPa =σ x + σ y

2+

σ x − σ y

2cos2 φ + τ xy s i n 2φ

=σ x +60 MPa

2+

σ x − 60 MPa

2cos −60°( ) + τ xy sin −60°( ); 0.75σ x − 0.866τ xy = 25 MPa

and

τ−30° = 20 MPa = τ xy c o s 2φ −σ x − σ y

2

sin2 φ = τ xy cos −60°( ) −σ x − 60 MPa

2

sin −60°( );

0.5τ xy + 0.433σ x = 46.0 MPa

These equations are solved as σx=69.8MPa and τxy=31.57MPa. The center of the Mohr’s circle is

placed, per Equation (2.22), at ((σx+σy)/2,0)=(64.9MPa,0). Using a point on the circle of

(-60MPa, 31.57MPa), the circle is drawn as follows:



Page 2-42

From the circle, the radius is given by:

r = 64.9 MPa − 60 MPa( )2 + 31.57 MPa( )

2 = 31.95 MPa

Therefore the principal stresses are σ1=64.9MPa+31.95MPa=96.8MPa and σ2=64.9MPa-

31.95MPa=33MPa.

2.62 Given the normal and shear stresses σx=12 ksi, σy=6 ksi and τxy=-4ksi, draw the Mohr’scircle diagram and the principal normal and shear stresses on the x- y axis. Determine thetriaxial stresses and give the corresponding Mohr’s circle diagram. Also, determine the

octahedral stresses.

Notes: Mohr’s circle is obtained in the same fashion as in Problems 2.59 and 2.60. The triaxial

stresss are ordered according to σ1≥σ2≥σ3, and it is known that this is a plane stress state so one of

these triaxial normal stresses is zero. The three dimensional Mohr’s circle diagram is then easily

drawn. The octahedral stresses are obtained from Equations (2.26) and (2.27).

Solution:

The center of the Mohr’s circle is placed, per Equation (2.22), at ((σx+σy)/2,0)=(9ksi,0). Using a

point on the circle of (σy,τxy)=(6ksi,-4ksi), the circle is drawn in the left sketch as follows:

The radius of the circle is calculated from the circle as:

r = 9ksi − 6ksi( )2 + 4ksi( )

2 = 5ksi

Therefore the principal stresses are σ1=9ksi+5ksi=14ksi and σ2=9ksi-5ksi=4ksi. The third

principal stress is zero since this is a plane stress circumstance. The three dimensional Mohr’s

circle can be drawn using these stresses as shown in green. Note that the radii of the circles are

easily calculated to give the principal shear stresses (see Equation (2.25)) as τ1/2=5ksi, τ2/3=2ksi

and τ1/3=7ksi. Therefore, from Equation (2.26), the octahedral normal stress is



Page 2-43

σoct =σ 1 + σ 2 + σ3

3=

14ksi + 4ksi + 0

3= 6ksi

The octahedral shear stress is given by Equation (2.27) as

τoct =2

3τ1 / 2

2 + τ2 /32 + τ1 / 3

2[ ]1 / 2

=2

35ksi( )

2 + 2ksi( )2 + 7ksi( )

2 = 5.888ksi

2.63 Given the normal and shear stresses σx=16 ksi, σy=9 ksi and τxy=5ksi, draw the Mohr’s

circle diagram and the principal normal and shear stresses on the x- y axis. Determine the

triaxial stresses and give the corresponding Mohr’s circle diagram. Also, determine the



stress are ordered according to σ1≥σ2≥σ3, and it is known that this is a plane stress state so one of



Solution:

The center of the Mohr’s circle is placed, per Equation (2.22), at ((σx+σy)/2,0)=(12.5ksi,0). Using

a point on the circle of (σy,τxy)=(9ksi,5ksi), the circle is drawn in the left sketch as follows:


r = 12.5ksi − 9ksi( )2 + 5ksi( )

2 = 6.10ksi

Therefore the principal stresses are σ1=σc+r =12.5ksi+6.1ksi=18.6ksi and σ2=σc-r =12.5ksi-

6.1ksi=6.4ksi. The third principal stress is zero since this is a plane stress circumstance. The three

dimensional Mohr’s circle can be drawn using these stresses as shown in green. Note that the

radii of the circles are easily calculated to give the principal shear stresses (see Equation (2.25))

as τ1/2=6.1ksi, τ2/3=3.2ksi and τ1/3=9.3ksi. Therefore, from Equation (2.26), the octahedral normal

stress is

σoct =σ 1 + σ 2 + σ3

3=

18.6ksi + 6.4ksi + 0

3= 8.33ksi


τoct =2

3τ1 / 2

2 + τ2 /32 + τ1 / 3

2[ ]1 / 2

=2

36.1ksi( )

2 + 3.2ksi( )2 + 9.3ksi( )

2 = 7.71ksi

2.64 Given the normal and shear stresses σx=10 ksi, σy=24 ksi and τxy=-6ksi, draw the Mohr’s

circle diagram and the principal normal and shear stresses on the x- y axis. Determine the



Page 2-44

triaxial stresses and give the corresponding Mohr’s circle diagram. Also, determine the






Solution:

The center of the Mohr’s circle is placed, per Equation (2.22), at ((σx+σy)/2,0)=(17ksi,0). Using a

point on the circle of (σy,τxy)=(24ksi,-6ksi), the circle is drawn in the left sketch as follows:


r = 24ksi− 17ksi( )2 + 6ksi( )

2 = 9.22ksi

Therefore the principal stresses are σ1=σc+r =17ksi+9.22ksi=26.22ksi and σ2=σc-r =17ksi-

9.22ksi=7.78ksi. The third principal stress is zero since this is a plane stress circumstance. The

three dimensional Mohr’s circle can be drawn using these stresses as shown in green. Note that

the radii of the circles are easily calculated to give the principal shear stresses (see Equation

(2.25)) as τ1/2=9.22ksi, τ2/3=3.89ksi and τ1/3=13.11ksi. Therefore, from Equation (2.26), the

octahedral normal stress is

σoct =σ 1 + σ 2 + σ3

3=

26.22ksi + 7.78ksi + 0

3= 11.3ksi


τoct =2

3τ1 / 2

2 + τ2 /32 + τ1 / 3

2[ ]1 / 2

=2

39.22ksi( )

2 + 3.89ksi( )2 + 13.11ksi( )

2 = 11.0ksi

2.65 In a three-dimensional stress field used to describe the stresses below the surface of a ball

in a ball bearing, the principal stresses are given as σ1=300MPa, σ2=200MPa and σ3=-

2800MPa. Calculate and draw the Mohr’s circle diagram.

Notes: Since the principal stresses are given, the circle can be drawn directly.

Solution:Note from Figure 2.19 that the principle stresses are the locations where the individual circles

intersect with the σ-axis. Therefore, the circle can be drawn as follows:



Page 2-45

2.66 Given the normal and shear stresses σx=-10ksi, σy=15ksi, and τxy=5ksi, determine or draw

the following:

a) Two-dimensional Mohr’s circle diagram

b) Normal principal stresses on x-y axisc) Shear principal stress on x-y axis

d) Three-dimensional Mohr’s circle diagram and corresponding principal normaland shear stresses.

e) Octahedral normal and shear stresses.





Solution:

The center of the Mohr’s circle is placed, per Equation (2.22), at ((σx+σy)/2,0)=(2.5ksi,0). Using a

point on the circle of (σy,τxy)=(15ksi,5ksi), the circle is drawn in the left sketch as follows:


r = 15ksi − 2.5ksi( )2 + 5ksi( )

2 = 13.5ksi

Therefore the principal stresses are σ1=σc+r =2.5ksi+13.5ksi=16ksi and σ3=σc-r =2.5ksi-13.5ksi=

-11ksi. The second principal stress is zero since this is a plane stress circumstance, and we’ve

numbered the principal stresses according to σ1≥σ2≥σ3. The three dimensional Mohr’s circle can

be drawn using these stresses as shown below:



Page 2-46

Note that the radii of the circles are easily calculated to give the principal shear stresses (see

Equation (2.25)) as τ1/2=8ksi, τ2/3=5.5ksi and τ1/3=13.5ksi. Therefore, from Equation (2.26), the

octahedral normal stress is

σoct =σ 1 + σ 2 + σ3

3=

16ksi + 0 − 11ksi

3= 1.67ksi


τoct =

2

3 τ1 / 22

+ τ2 /32

+ τ1 / 32

[ ]1 / 2

=

2

3 8ksi( )2

+ 5.5ksi( )2

+ 13.5ksi( )2

= 11.1ksi

2.67 A thin plate is loaded with stresses σx=σ, σy=-σ, σz=0 (perpendicular to the plate),

τxy=τyx=σ /21/2 and τxz=τzx=τyz=τzy=0. Calculate the octahedral normal and shear stresses.

Notes: Equations (2.26) and (2.28) are evaluated to solve the problem.

Solution:

Equation (2.26) gives the octahedral normal stress as:

σoct =

σ x + σ y + σ z

3 =

σ − σ + 0

3 =0

The octahedral shear stress is given by Equation (2.28) as:

τoct =1

3σ x − σ y( )

2+ σ y − σ z( )

2+ σ z − σ x( )

2 + 6 τ xy2 + τ yz

2 + τ xz2( )

=1

3σ + σ( )

2 + −σ( )2 + −σ( )

2 + 6σ 2

2+ 0 + 0

=

1

39σ 2 = σ

2.68 The strain tensor in a machine element is

T =

0.0012 −0.0001 0.0007

−0.0001 0.0003 0.0002

0.0007 0.002 −0.0008

Find the strain in the x, y, and z directions, in the direction of space diagonal

1

3,

1

3,

1

3

, and in the direction εx, εy, εz.

Notes: This problem requires evaluation of direction cosines from Equations (B.4) through (B.9),and evaluation of Equation (B.10) to get the desired strains.



Solution:

Note that the normal strains are εx=0.0012, εy=0.0003, and εz=-0.0008. From Equations (B.4) to

(B.9), cos( z’, x)=0.8147, cos( z’, y)=0.2037, and cos( z’, z)=-0.5431. Therefore, Equation (B.10)

gives εz’=0.0002245. The strains in direction1

3,

1

3,

1

3

is given by

cos( z”, x)=cos( y”, x)=cos( x”, x)=0.577. Equation (B.10) then gives εz”=0.0005.

2.69 The strain tensor is

T =

0.0023 0.0006 0

0.0006 0.0005 0

0 0 0

Calculate the maximum shear strain and the principal strains.

Notes: This is a plane strain situation, so Equations (2.38) and (2.39) are needed to solve this

problem.

Solution:Note that since εz=γ zy=γ zx=0, Equation (2.39) gives:

γ max = ±21

2γ xy

2

+ε x − ε y

2

2

= ±2 0.0006( )2 +

0.0023 − 0.0005

2

2

= ±0.00216

From Equation (2.38), the principal strains are:

ε1 ,ε2 =ε x + ε y

2±

1

2γ xy

2

+ε x − ε y

2

2

=ε x + ε y

2±

γ max

2=

0.0023 + 0.0005

2±

0.00216

2

or ε1=0.00248 and ε2=0.00032.

2.70 For the stresses in a stress element it is possible to find coordinate directions where the

shear stresses are zero and the normal stresses are parallel with the new coordinate

directions. Is it possible to determine the stress state in a machine element just by usingtwo strain gages?

Solution:

It is possible only if the element is in a state of plane stress. Since strain gages are placed on

surfaces, this is possible, unless the machine element is being compressed, such as if it was in apressure vessel.

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fundamentals of machine elements sm ch 02