skiladæmi 10Due: 11:59pm on Wednesday, November 11, 2015

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Alternative Exercise 11.125

A bar with cross­sectional area A is subjected to equal and opposite tensile forces at its ends. Consider a plane throughthe bar making an angle with a plane at right angles to the bar (the figure ).

Part A

What is the tensile (normal) stress at this plane in terms of , , and ?

ANSWER:

Correct

Part B

What is the shear (tangential) stress at the plane in terms of , , and ?

ANSWER:

Correct

Part C

For what value of is the tensile stress a maximum?

ANSWER:

F θ

F A θ

(θ)FA

cos2

F A θ

cos(θ)sin(θ)FA

θ

= 0 θ ∘

Correct

Part D

For what value of is the shear stress a maximum?

ANSWER:

Correct

Alternative Exercise 11.116

A moonshiner produces pure ethanol (ethyl alcohol) late at night and stores it in a stainless steel tank in the form of acylinder 0.310 in diameter with a tight­fitting piston at the top. The total volume of the tank is 260 . In an attempt tosqueeze a little more into the tank, the moonshiner piles lead breaks of total mass 1430 on top of the piston.

Part A

What additional volume of ethanol can the moonshiner squeeze into the tank? (Assume that the wall of the tank isperfectly rigid.)

ANSWER:

Correct

± A Wire under Stress

A steel wire of length 2.01 with circular cross section must stretch no more than 0.200 when a tensile (stretching)force of 380 is applied to each end of the wire.

Part A

What minimum diameter is required for the wire?

Express your answer in millimeters. Take Young's modulus for steel to be = 2.00×1011 .

Hint 1. How to approach the problem

Recall that Young's modulus is defined as

.

Compute the strain in terms of quantities given in the problem introduction, and write the stress in terms of givenquantities and the unknown wire diameter. Use these along with the given value of Young's modulus for steel tosolve for the diameter of the wire.

Hint 2. Calculate the tensile strain

θ

= 45 θ ∘

m Lkg

= 5.31×10−2 V L

m cmN

dmin

Y Pa

Y = tensile stresstensile strain

Calculate the tensile strain on the wire.

Hint 1. Definition of tensile strain

Tensile strain is defined as the ratio of the elongation to the original length of a material that isunder stress:

.

ANSWER:

Correct

Hint 3. Definition of tensile stress

Tensile stress is defined as the force perpendicular to the surface of a material divided by the cross­sectionalarea of the surface:

.

Hint 4. Relation between the area and the diameter

The relation between the area and the diameter of a circle is

.

ANSWER:

Correct

Note that you were asked for the minimum diameter. Where does this figure?

The extension is directly proportional to the stress, i.e., the force per unit area. One way to decrease the stress isto increase the surface area over which the stretching force is applied. So any diameter (and so area) greater thanthe one you calculated would serve to keep the extension within the tolerance specified (i.e., the maximumallowable extension).

± Young's Modulus

Learning Goal:

To understand the meaning of Young's modulus, to perform some real­life calculations related to stretching steel, a commonconstruction material, and to introduce the concept of breaking stress.

Hooke's law states that for springs and other "elastic" objects,

where is the magnitude of the stretching force, is the corresponding elongation of the spring from equilibrium, and is a constant that depends on the geometry and the material of the spring. If the deformations are small enough, mostmaterials, in fact, behave like springs: Their deformation is directly proportional to the external force. Therefore, it may beuseful to operate with an expression that is similar to Hooke's law but describes the properties of various materials, as

ΔL L0

tensile strain = ΔLL0

tensile strain = 9.95×10−4

tensile stress = F⊥

A

A d

A = π d2

4

= 1.56 dmin mm

F = kΔxF Δx k

L

opposed to objects such as springs. Such an expression does exist. Consider, for instance, a bar of initial length andcross­sectional area stressed by a force of magnitude . As a result, the bar stretches by .

Let us define two new terms:

Tensile stress is the ratio of the stretching force to thecross­sectional area:

.Tensile strain is the ratio of the elongation of the rod to theinitial length of the bar:

.It turns out that the ratio of the tensile stress to the tensile strainis a constant as long as the tensile stress is not too large. Thatconstant, which is an inherent property of a material, is calledYoung's modulus and is given by

Part A

What is the SI unit of Young's modulus?

Hint 1. Look at the dimensions

If you look at the dimensions of Young's modulus, you will see that they are equivalent to the dimension ofpressure. Use the SI unit of pressure.

ANSWER:

Correct

Part B

Consider a metal bar of initial length and cross­sectional area . The Young's modulus of the material of the bar is . Find the "spring constant" of such a bar for low values of tensile strain.

Express your answer in terms of , , and .

Hint 1. Use the definition of Young's modulus

Consider the equation defining . Then isolate and compare the result with Hooke's law: .

ANSWER:

Correct

Part C

LA F ΔL

stress = FA

strain = ΔLL

Y = .F/A

ΔL/L

Pa

L AY k

Y L A

Y F F = kΔx

= k Y AL

L k

Ten identical steel wires have equal lengths and equal "spring constants" . The wires are connected end to end, sothat the resultant wire has length . What is the "spring constant" of the resulting wire?

Hint 1. The spring constant

Use the expression for the spring constant determined in Part B. From the expression derived in the Part B, youcan determine what happens to the spring constant when the length of the spring increases.

ANSWER:

Correct

Part D

Ten identical steel wires have equal lengths and equal "spring constants" . The wires are slightly twisted together,so that the resultant wire has length and its cross­sectional area is ten times that of the individual wire. What is the"spring constant" of the resulting wire?

Hint 1. The spring constant

Use the expression for the spring constant determined in Part B. From the expression derived in Part B, you candetermine what happens to the spring constant when the area of the spring increases.

ANSWER:

Correct

Part E

Ten identical steel wires have equal lengths and equal "spring constants" . The Young's modulus of each wire is .The wires are connected end to end, so that the resultant wire has length . What is the Young's modulus of theresulting wire?

ANSWER:

L k10L

0.1k

k

10k

100k

L kL

0.1k

k

10k

100k

L k Y10L

0.1Y

Y

10Y

100Y

Correct

Part F

Ten identical steel wires have equal lengths and equal "spring constants" . The Young's modulus of each wire is .The wires are slightly twisted together, so that the resultant wire has length and is ten times as thick as theindividual wire. What is the Young's modulus of the resulting wire?

ANSWER:

Correct

By rearranging the wires, we create a new object with new mechanical properties. However, Young's modulusdepends on the material, which remains unchanged. To change the Young's modulus, one would have to changethe properties of the material itself, for instance by heating or cooling it.

Part G

Consider a steel guitar string of initial length meter and cross­sectional area square millimeters.The Young's modulus of the steel is pascals. How far ( ) would such a string stretch under atension of 1500 newtons?

Use two significant figures in your answer. Express your answer in millimeters.

ANSWER:

Correct

Steel is a very strong material. For these numeric values, you may assume that Hooke's law holds. However, forgreater values of tensile strain, the material no longer behaves elastically. If the strain and stress are largeenough, the material deteriorates. The final part of this problem illustrates this point and gives you a sense of the"stretching limit" of steel.

Part H

Although human beings have been able to fly hundreds of thousands of miles into outer space, getting inside the earthhas proven much more difficult. The deepest mines ever drilled are only about 10 miles deep. To illustrate thedifficulties associated with such drilling, consider the following: The density of steel is about 7900 kilograms per cubicmeter, and its breaking stress, defined as the maximum stress the material can bear without deteriorating, is about

pascals. What is the maximum length of a steel cable that can be lowered into a mine? Assume that themagnitude of the acceleration due to gravity remains constant at 9.8 meters per second per second.

Use two significant figures in your answer, expressed in kilometers.

Hint 1. Why does the cable break?

The cable breaks because of the stress exerted on it by its own weight. At the moment that the breaking stressis reached, the stress at the top of the cable reaches its maximum, and the material begins to deteriorate.

L k YL

0.1Y

Y

10Y

100Y

L = 1.00 A = 0.500Y = 2.0 × 1011 ΔL

= 15 ΔL mm

2.0 × 109

Introduce an arbitrary cross­sectional area of the cable (which will cancel out of the final answer). The mass ofthe cable below the top point can be found as the product of its volume and its density. Use this to find the forceat the top that will lead to the breaking stress.

Hint 2. Find the stress in the cable

Assume that the cable has cross­sectional area and length . The density is . The maximum stress in thecable is at the very top, where it has to support its own weight. What is this maximum stress?

Express your answer in terms of , , and , the magnitude of the acceleration due to gravity. Recall thatthe stress is the force per unit area, so the area will not appear in your expression.

ANSWER:

ANSWER:

CorrectThis is only about 16 miles, and we have assumed that no extra load is attached. By the way, this length is smallenough to justify the assumption of virtually constant acceleration due to gravity. When making such assumptions,one should always check their validity after obtaining a result.

Problem 11.76

Two identical, uniform beams weighing 260 each are connected at one end by a frictionless hinge. A light horizontalcrossbar attached at the midpoints of the beams maintains an angle of 53.0 between the beams. The beams aresuspended from the ceiling by vertical wires such that they form a " ", as shown in the figure .

Part A

What force does the crossbar exert on each beam?

ANSWER:

A L ρ

ρ L g

maximum stress = ρLg

26 km

N∘

V

= 130 F N

Correct

Part B

Is the crossbar under tension or compression?

ANSWER:

Correct

Part C

What is the magnitude of the force that the hinge at point A exerts on each beam?

ANSWER:

Correct

Part D

What is the direction of the force that the hinge at point A exerts on the right­hand beam?

ANSWER:

Correct

Part E

What is the direction of the force that the hinge at point A exerts on the left­hand beam?

ANSWER:

Correct

Understanding Bernoulli's Equation

Bernoulli's equation is a simple relation that can give useful insight into the balance among fluid pressure, flow speed, andelevation. It applies exclusively to ideal fluids with steady flow, that is, fluids with a constant density and no internal friction

tension

compression

= 130 F N

= 180 with the direction to the right ϕ ∘

= 0 with the direction to the right ϕ ∘

forces, whose flow patterns do not change with time. Despite its limitations, however, Bernoulli's equation is an essentialtool in understanding the behavior of fluids in many practical applications, from plumbing systems to the flight of airplanes.

For a fluid element of density that flows along a streamline, Bernoulli's equation states that

,where is the pressure, is the flow speed, is the height, is the acceleration due to gravity, and subscripts 1 and 2refer to any two points along the streamline. The physical interpretation of Bernoulli's equation becomes clearer if werearrange the terms of the equation as follows:

.

The term on the left­hand side represents the total work done on a unit volume of fluid by the pressure forces ofthe surrounding fluid to move that volume of fluid from point 1 to point 2. The two terms on the right­hand side represent,respectively, the change in potential energy, , and the change in kinetic energy, , of the unit

volume during its flow from point 1 to point 2. In other words, Bernoulli's equation states that the work done on a unit volumeof fluid by the surrounding fluid is equal to the sum of the change in potential and kinetic energy per unit volume that occursduring the flow. This is nothing more than the statement of conservation of mechanical energy for an ideal fluid flowing alonga streamline.

Part A

Consider the portion of a flow tube shown in the figure. Point1 and point 2 are at the same height. An ideal fluid enters theflow tube at point 1 and moves steadily toward point 2. If thecross section of the flow tube at point 1 is greater than that atpoint 2, what can you say about the pressure at point 2?

Hint 1. How to approach the problem

Apply Bernoulli's equation to point 1 and to point 2. Since the points are both at the same height, their elevationscancel out in the equation and you are left with a relation between pressure and flow speeds. Even though theproblem does not give direct information on the flow speed along the flow tube, it does tell you that the crosssection of the flow tube decreases as the fluid flows toward point 2. Apply the continuity equation to points 1 and2 and determine whether the flow speed at point 2 is greater than or smaller than the flow speed at point 1. Withthat information and Bernoulli's equation, you will be able to determine the pressure at point 2 with respect to thepressure at point 1.

Hint 2. Apply Bernoulli's equation

Apply Bernoulli's equation to point 1 and to point 2 to complete the expression below. Here and are thepressure and flow speed, respectively, and subscripts 1 and 2 refer to point 1 and point 2. Also, use forelevation with the appropriate subscript, and use for the density of the fluid.

Express your answer in terms of some or all of the variables , , , , , , and .

Hint 1. Flow along a horizontal streamline

Along a horizontal streamline, the change in potential energy of the flowing fluid is zero. In other words,when applying Bernoulli's equation to any two points of the streamline, and they cancel out.

ρ

+ρg + ρ = +ρg + ρp1 h112

v21 p2 h2

12

v22

p v h g

− = ρg( − )+ ρ( − )p1 p2 h2 h112

v22 v2

1

−p1 p2

ρg( − )h2 h1 ρ( − )12 v2

2 v21

p vh

ρ

p1 v1 h1 p2 v2 h2 ρ

=h1 h2

ANSWER:

Hint 3. Determine with respect to

By applying the continuity equation, determine which of the following is true.

Hint 1. The continuity equation

The continuity equation expresses conservation of mass for incompressible fluids flowing in a tube. Itsays that the amount of fluid flowing through a cross section of the tube in a time interval must be the same for all cross sections, or

.

Therefore, the flow speed must increase when the cross section of the flow tube decreases, and viceversa.

ANSWER:

ANSWER:

Correct

Thus, by combining the continuity equation and Bernoulli's equation, one can characterize the flow of an idealfluid.When the cross section of the flow tube decreases, the flow speed increases, and therefore the pressuredecreases. In other words, if , then and .

Part B

As you found out in the previous part, Bernoulli's equation tells us that a fluid element that flows through a flow tubewith decreasing cross section moves toward a region of lower pressure. Physically, the pressure drop experienced bythe fluid element between points 1 and 2 acts on the fluid element as a net force that causes the fluid to __________.

Hint 1. Effects from conservation of mass

Recall that, if the cross section of the flow tube varies, the flow speed must change to conserve mass.This means that there is a nonzero net force acting on the fluid that causes the fluid to increase or decreasespeed depending on whether the fluid is flowing through a portion of the tube with a smaller or larger crosssection.

= + ρp112

v21 +p2

ρv22

2

v2 v1

ΔV A Δt

= =ΔVΔt

A1v1 A2v2

>v2 v1

=v2 v1

<v2 v1

The pressure at point 2 is lower than the pressure at point 1.

equal to the pressure at point 1.

higher than the pressure at point 1.

<A2 A1 >v2 v1 <p2 p1

A v

ANSWER:

Correct

Part C

Now assume that point 2 is at height with respect to point 1, as shown in the figure. The ends of the flow tube havethe same areas as the ends of the horizontal flow tube shownin Part A. Since the cross section of the flow tube isdecreasing, Bernoulli's equation tells us that a fluid elementflowing toward point 2 from point 1 moves toward a region oflower pressure. In this case, what is the pressure dropexperienced by the fluid element?

Hint 1. How to approach the problem

Apply Bernoulli's equation to point 1 and to point 2, as you did in Part A. Note that this time you must take intoaccount the difference in elevation between points 1 and 2. Do you need to add this additional term to the otherterm representing the pressure drop between the two ends of the flow tube or do you subtract it?

ANSWER:

Correct

Part D

From a physical point of view, how do you explain the fact that the pressure drop at the ends of the elevated flow tubefrom Part C is larger than the pressure drop occurring in the similar but purely horizontal flow from Part A?

Hint 1. Physical meaning of the pressure drop in a tube

As explained in the introduction, the difference in pressure between the ends of a flow tube representsthe total work done on a unit volume of fluid by the pressure forces of the surrounding fluid to move that volumeof fluid from one end to the other end of the flow tube.

decrease in speed

increase in speed

remain in equilibrium

h

The pressure drop is smaller than the pressure drop occurring in a purely horizontal flow.

equal to the pressure drop occurring in a purely horizontal flow.

larger than the pressure drop occurring in a purely horizontal flow.

−p1 p2

ANSWER:

CorrectIn the case of purely horizontal flow, the difference in pressure between the two ends of the flow tube had tobalance only the increase in kinetic energy resulting from the acceleration of the fluid. In an elevated flow tube, thedifference in pressure must also balance the increase in potential energy of the fluid; therefore a higher pressure isneeded for the flow to occur.

Water Flowing from a Tank

Water flows steadily from an open tank as shown in the figure.The elevation of point 1 is 10.0 meters, and the elevation of points2 and 3 is 2.00 meters. The cross­sectional area at point 2 is0.0480 square meters; at point 3, where the water is discharged, itis 0.0160 square meters. The cross­sectional area of the tank isvery large compared with the cross­sectional area of the pipe.

Part A

Assuming that Bernoulli's equation applies, compute the discharge rate .

Express your answer in cubic meters per second.

Hint 1. How to approach the problem

The discharge rate is the rate at which a given volume of water flows across the exit of the pipe per unit time. Itis also defined as volume flow rate, and it depends on both the cross­sectional area of the pipe at the exit andthe fluid speed at that point.

Hint 2. The volume flow rate

Consider a steadily moving incompressible fluid, and let denote the cross­sectional area of a flow tube. Thevolume of fluid flowing across the cross section of area at speed during a small interval of time isgiven by . Therefore, the rate at which fluid volumes cross a portion of the flow tube is

.

Hint 3. Find the fluid speed at the end of the pipe

A greater amount of work is needed to balance the

increase in potential energy from the elevation change.

decrease in potential energy from the elevation change.

larger increase in kinetic energy.

larger decrease in kinetic energy.

dVdt

AΔV A v ΔtAvΔt

= AvdV

dt

Assuming that Bernoulli's equation applies, find the speed of the water at point 3. Recall that the area of thetank is very large compared to the cross­sectional area of the pipe, and consequently, the velocity of water at apoint on the surface of the water in the tank may be considered to be zero.

Express your answer in meters per second to three significant figures.

Hint 1. Apply Bernoulli's principle

Let be the atmospheric pressure and the density of water. Consider the entire volume of water as asingle flow tube and apply Bernoulli's principle to point 3 and to point 1. Complete the expression below,where is the fluid speed at point 3.

Express your answer in terms of , , and the free­fall acceleration due to gravity.

Hint 1. Bernoulli's principle

For the steady flow of an incompressible fluid with no internal friction, the pressure and the flowspeed at depth below the surface are linked by an important relationship, known asBernoulli's principle. In particular, at any point at depth along a flow tube, the following relationholds:

where is the density of the fluid and is the accereleration due to gravity.

Since Bernoulli's principle is valid at any point along a flow tube, it takes the form

when applied to two distinct points along a flow tube. The subscripts 1 and 2 refer to such points.

ANSWER:

ANSWER:

ANSWER:

Correct

Part B

What is the gauge pressure at point 2?

Express your answer in pascals.

Hint 1. Definition of gauge pressure

v3

pa ρ

v3

pa ρ g

pv H

H

p+ρgH + ρ = constant,12

v2

ρ g

+ρg + ρ = +ρg + ρp1 H112

v21 p2 H2

12

v22

= +2ρg+ ρpa12

v23 + 10ρgpa

= 12.5 v3 m/s

= 0.200 dVdt

/sm3

Gauge pressure is defined as the excess pressure above atmospheric pressure. Let be the atmosphericpressure and the total pressure of a fluid. Then the gauge pressure is .

Hint 2. How to approach the problem

You can relate the fluid pressure at point 2 with the atmospheric pressure by applying Bernoulli's principle topoint 2 and point 1, or alternatively to point 2 and point 3. To determine the fluid speed at point 2 you can usethe continuity equation, using the fluid speed at the exit of the pipe found in Part A.

Hint 3. Apply Bernoulli's principle

Consider the entire volume of water as a single flow tube. Let and be respectively the pressure and thefluid speed at point 2. Let the atmospheric pressure be and the density of water . Apply Bernoulli's principleto point 1 and point 2 and complete the expression below.

Express your answer in terms of , , and , the free­fall acceleration.

Hint 1. Bernoulli's principle

For the steady flow of an incompressible fluid with no internal friction, the pressure and the flow speed at depth below the surface are linked by an important relationship, known as Bernoulli's principle. Inparticular, at any point at depth along a flow tube, the following relation holds:

where is the density of the fluid and is the accereleration due to gravity.

Since Bernoulli's principle is valid at any point along a flow tube, it takes the form

when applied to two distinct points along a flow tube. The subscripts 1 and 2 refer to such points.

ANSWER:

Hint 4. Find the fluid speed at point 2

Find , the speed of the water at point 2.

Express your answer in meters per second to three significant figures.

Hint 1. The continuity equation

In a steadily moving incompressible fluid, the mass of fluid flowing along a flow tube is constant. Inparticular, consider a flow tube between two stationary cross sections with areas and . Let and be the fluid speeds at these sections, respectively. Then conservation of mass takes the form

,

which is known as the continuity equation.

ANSWER:

Hint 5. Density of WaterRecall that the density of water is

pap p − pa

p2 v2pa ρ

v2 ρ g

p vH

H

p+ρgH + ρ = constant,12 v2

ρ g

+ρg + ρ = +ρg + ρp1 H112 v2

1 p2 H212 v2

2

= −p2 pa −0.5ρ + 8gρv22

v2

A1 A2 v1v2

=A1 v1 A2 v2

= 4.17 v2 m/s

1000 kg/ 3

Recall that the density of water is

ANSWER:

Correct

Score Summary:Your score on this assignment is 101%.You received 7.07 out of a possible total of 7 points.

1000 kg/m3

6.98×104 Pa