Module 11 Elasticity

2005 Pearson Education South Asia Pte Ltd

MECHANICS

FE1001 Physics I NTU - College of Engineering

1. Units, Physical Quantities and Vectors

2. Motion Along A Straight Line3. Motion in 2 or 3 Dimensions4. Newton’s Law of Motion

5. Applying Newton’s Laws6. Work and Kinetic Energy7. Potential Energy and Energy Conservation


MECHANICS

FE1001 Physics I NTU - College of Engineering

8. Momentum, Impulse, and Collisions

9. Rotation of Rigid Bodies10. Dynamics of Rotational

Motion11. Equilibrium and Elasticity12. Gravitation13. Periodic Motion14. Fluid Mechanics

11. Equilibrium and Elasticity


Chapter Objectives

• Introduce the concepts of stress, strain and elastic modulus.

• Define Hooke’s Law that help us to predict what deformations will occur when forces are applied to a rigid body.



Chapter Outline

1. Conditions for Equilibrium2. Center of Gravity 3. Solving Rigid-Body Equilibrium Problems4. Stress, Strain, and Elastic Moduli5. Elasticity and Plasticity



11.1 Conditions for Equilibrium

• The equivalent statement for an extended body is that the center of mass of the body has zero acceleration if the vector sum of all external forces acting on the body is zero.

• This is called the first condition for equilibrium.• In vector and component forms,

0F =∑ur

0 0 0 (11.1)x y zF F F= = =∑ ∑ ∑where the sum includes external forces only




• Second condition for an extended body to be equilibrium is that the body must have no tendency to rotate.

• A rigid body is not rotating about a certain point has zero angular momentum about that point,

• Therefore when due to all the external forces acting on the body is zero.

• This is the second condition of equilibrium. 0L =ur

τ∑r

0 (11.2)τ =∑r




• The sum of the torques due to all external forces acting on the body, with respect to any specified point, must be zero.

• When a rigid body is at rest, a body is said to be in static equilibrium.



11.2 Center of Gravity

• When the entire force of gravity (weight) is concentrated at a point, it is called the center of gravity.

• For a collection of particles with mass and coordinates , the coordinates of the center of mass are given by

1 2, ,...m m( ) ( )1 1 1 2 2 2, , , , , ,...x y z x y z

, and cm cm cmx y z

1 21 2

1 2

... (11.4)

...

iii

cmi

i

m rm r m r

rm m m

+ += =+ +

∑∑

rr r

r




• In the figure, the gravitational torque about any point can be found by assuming that all the weight of the body acts at its center of gravity (cg).

• The cg is identical to the center of mass (cm) if is the same at all points on the body.

gur




• The torque vector of the weight with respect to is

iτr

iwur

i ii i ir w r m gτ = × = ×r r ur r ur

• The total torque due to the gravitational force on all particles is

1 21 2 ...i

r m g r m gτ τ= = × + × +∑r r r ur r ur

( )1 21 2 ...m r m r g= + + ×r r ur

iii

m r g⎛ ⎞

= ×⎜ ⎟⎜ ⎟⎝ ⎠∑

r ur




• When we multiply and divide by the total mass of the body, we get

1 21 2

1 2

......

iii

ii

m rm r m r

M g M gm m m

τ + += × = ×+ +

∑∑

rr r

r ur ur

• Thus (11.5)cm cmr M g r wτ = × = ×

r r ur r ur

• If has the same value at all points on a body, its center of gravity is identical to its center of mass.

gur




• When a body acted on by a gravity is supported at a single point, the center of gravity is always at or directly above or below the point of suspension.

• If it is elsewhere, the weight would have a torque with respect to the point of suspension, and the body could not be in rotational equilibrium.

Finding and Using the Center of Gravity




• Figure (a) shows the center of gravity is within the area bounded by the supports, and the car is in equilibrium.

• Figure (b) and (c) shows the car and the truck and it will tip over because their center of gravity lie outside the area of support.





• The lower the center of gravity and the larger the area of support, the more difficult it is t overturn a body.




Example 11.1 Walking the plank

A uniform wooden plank of length L = 6.0m and mass M = 90kg rests on top of two sawhorses separated by D = 1.5m, located equal distances from the center of the plank (Fig 11.4). Your cousin Throckmorton tries to stand on the right-hand end of the plank. If the plank is to remain at rest, how massive can Throckmorton be?



Example 11.1 (SOLN)

IdentifyIf the system of plank and Throckmorton is just in balance, the center of gravity of his system will be directly over the right-hand sawhorse (just barely within the area bounded by the 2 supports). The target variable is Throcky’s mass.



Example 11.1 (SOLN)

Set upWe take the origin at C, the geometrical center and center of gravity of the uniform plank, and take the positive x-axis to point horizontally to the right. Then the x-coordinates of the centers of gravity of the plank (mass M) and Throcky (unknown mass) are and respectively. We will use Eqs(11.3) to locate the center of gravity of system of plank and Throcky.

0px =/ 2 3.0Tx l m= =



Example 11.1 (SOLN)

ExecuteFrom the first of Equs(11.3),

( ) ( )0 / 2

2cgM m L m L

xM m M m

+= =

+ +

Setting this equal to D/2, the x-coordinate of the right hand sawhorse, we have

2 2m L D

M m=

+

( )mL M m D= +

( ) 1.590 30

6.0 1.5D m

m M kg kgL D m m

= = =− −



Example 11.1 (SOLN)

EvaluateTo check our result, let’s repeat the calculation with a different choice of origin. Now we take the origin to be at S, the position of the right hand sawhorse, so that . The centers of gravity of the plank and Throcky are now at and respectively, so

0cgx =

/ 2px D=− ( ) ( )/ 2 / 2Tx L D= −

( ) ( ) ( )[ ]/ 2 / 2 / 20cg

M D m L Dx

M m

− + −= =

+

( ) ( )/ 2

30/ 2 / 2

MD Dm M kg

L D L D= = =

− −



Example 11.1 (SOLN)

EvaluateThe mass doesn’t depend on our arbitrary choice of origin.

A 60 kg child could only stand halfway between the right hand sawhorse and the end of the plank. Can you see why?



11.3 Solving Rigid-Body Equilibrium Problems

• There are 2 key principles of rigid-body equilibrium.

1. The vector sum of forces on the body must be zero.

2. The sum of torques about any point must be zero.0 and 0x yF F= =∑ ∑

0 (11.6)z

τ =∑




IDENTIFY• The 1st and 2nd condition for equilibrium are useful

whenever there is a rigid body that is not rotating and accelerating in space.

Problem-Solving Strategy (Equilibrium of a Rigid Body)

SET UP• Draw a sketch of the physical situation, including

dimension, and select the body in equilibrium to be analyzed.




SET UP• Draw a free body diagram showing the forces acting on the selected body and no

others• Do not include forces exerted by this body on other bodies.• Choose a coordinate system.• In choosing a point to compute torques, note that is a force has a line of action that

goes through a particular point, the torque of the force with respect to that point is zero.





EXECUTE• Write equations expressing the equilibrium conditions.• Remember that are

always separate equations; never add x- and y-components in a single equation.


0, 0, and 0x y zF F τ= = =∑ ∑ ∑

EVALUATE• Check your answer whether it is making any

physical sense.



Example 11.2 Weight distribution for a car

An auto magazine reports that a certain sports car has 53% of its weight on the front wheels and 47% on its rear wheels, with a 5.46m wheelbase. This means that the total normal force on the front wheels is 0.53w and that on the rear wheels is 0.47w, where w is the total weight. The wheelbase is the distance between front and rear axles. How far in front of the rear axle is the car’s center of gravity?



Example 11.2 Weight distribution for a car



Example 11.2 (SOLN)

IdentifyWe can use the two conditions for equilibrium since the car is assumed to be at rest. The conditions also apply when the car is driving in a straight line at constant speed, since the net force and net torque on the car are also zero in that situation. The target variable is the coordinate of the car’s center of gravity.



Example 11.2 (SOLN)

Set upFigure 11.6 shows a free-body diagram for the car, along with the x- and t-axes and our convention that counterclockwise torques are positive. We have drawn the weight w as acting at the center of gravity, and the distance we want is . It is the lever arm of the weight with respect to the rear axle R, so it is reasonable to take torques with respect to R. Note that the torque due to the weight is negative because it tends to cause a clockwise rotation about R.

cgL



Example 11.2 (SOLN)

ExecuteYou can see from Fig. 11.6b that the first condition for equilibrium is satisfied: because there aren’t any x-components of force and because . The force equation doesn’t involve the target variable , so we must solve for it using the torque equation for point R:

0xF =∑0yF =∑

( )0.47 0.53 0w w w+ + − =cgL

( ) ( )0.47 0 0.53 2.46 0R cgw wL mτ = − + =∑1.30cgL m=



Example 11.2 A heroic act

Sir Lancelot is trying to rescue the Lady Elayne from Castle Von Doom by climbing a uniform ladder that is 5.0m long and weighs 180N. Lancelot, who weighs 800N, stops a third of the way up the ladder. The bottom of the ladder rests on a horizontal stone ledge and leans across the moat in equilibrium against a vertical wall that is frictionless because of a thick layer of moss. The ladder makes an angle of 53.1 with the horizontal, conveniently forming a 3-4-5 right triangle. A) Find the normal and friction forces on the ladder at its base. B) Find the minimum coefficient of static friction needed to prevent slipping at the base. C) Find the magnitude and direction of the contact force on the ladder at the base.

°



Example 11.3 A heroic act



Example 11.3 (SOLN)

IdentifyThe system of ladder and Lancelot is stationary, so we can use the 2 conditions for equilibrium. These conditions by themselves are enough to solve part (a). In part (b), we also need the relationship given in Section 5.3 between the static friction force, the coefficient of static friction, and the normal force. The contact force asked for in part (c) is the vector sum of the normal force and the direction force acting at the base of the ladder, which w e find in part (a)



Example 11.3 (SOLN)

Set upFigure 11.7b shows the free-body diagram for the system of the ladder and Lancelot. We choose the x- and y-directions as shown and take counterclockwise torques to be positive. The ladder is described as ‘uniform,’ so its center of gravity is at its geometrical center. Lancelot’s weight acts at a point on the ladder one-third of the way from the base toward the wall.



Example 11.3 (SOLN)

Set upThe frictionless wall exerts only a normal force at the top of the ladder. The forces at the base are the upward normal force and the static friction force , which must point to the right to prevent slipping; the magnitudes and are the target variables in part (a). From Eq.(5.6), these magnitudes are related by , where is the coefficient of static friction (the target variable is part (b))

1n

2n sf

2n sf2s sf nμ≤

sμ



Example 11.3 (SOLN)

Executea) From Eq. (11.6), the first condition for equilibrium gives

( )1 0x sF f n= + − =∑( ) ( )2 800 180 0yF n N N= + − + − =∑

These are 2 equations for the three unknown _ and _ and _. The first equation tells us that the 2 horizontal forces must be equal and opposite, and the 2nd equation gives

2 980n N=



Example 11.3 (SOLN)

ExecuteThe ground pushes up with a force of 980N to balance the total (downward) weight (800N + 180N).

We don’t yet have enough equations, but now we can use the 2nd condition for equilibrium. We can take torques about any point we choose. The smart choice is the point that will give us the fewest terms and fewest unknowns in the torques equation. With this in mind, we choose point B, at the base of the ladder. The two forces _ and _ have no torque about that point. From Fig 11.7b we see that the lever arm for the ladder’s weight is 1.5m, the lever arm for Lancelot’s weight is 1.0m, and the lever arm for _ is 4.0m. The torque equation for point B is



Example 11.3 (SOLN)

Execute

Solving for , we get = 268N. We now substitute this back into the equation to get

( ) ( )( ) ( )( )1 4.0 180 1.5 800 1.0B n m N m N mτ = − −∑( ) ( )2 0 0 0sn f+ + =

1n 1n

0xF =∑

268sf N=



Example 11.3 (SOLN)

Execute

b) The static friction force cannot exceed , so the minimum coefficient of static friction to prevent slipping is

sf 2snμ

( )min2

2680.27

980s

sf Nn N

μ = = =

c) The components of the contact force at the base are the static friction force , and the normal force , so

BFur

sf 2n

$ ( ) ( )$2 268 980B sF f i n j N i N j= + = +ur $ $



Example 11.3 (SOLN)

Execute

The magnitude and direction of (Fig.11.7c) are then

( ) ( )2 2268 980 1020BF N N N= + =

BFur

980arctan 75

268NN

θ = = °



Example 11.3 (SOLN)

Evaluate

As Fig 11.7c shows, the contact force is not directed along the length of the ladder. You may be surprised by this, but there’s really no good reason why the two directions should be the same. Can you show that if were directed along the ladder, there would be a net counterclockwise torque with respect to the top of the ladder, and equilibrium would be impossible?

BFur

BFur



Example 11.3 (SOLN)

Evaluate

Here are a few final comments. First, as Lancelot climbs higher on the ladder, the lever arm and torque of his weight about B increase; this increases the values of , and . At the top, his lever arm would be nearly 3m, giving minimum coefficient of static friction of nearly 7.0. The values of would not be this large for present day aluminum ladder on a wood floor, which is why such ladders are usually equipped with non-slip rubber pads. Lancelot’s medieval ladder is unlikely to have this feature, and the ladder is likely to slip as he climbs.

1n

sf ( )minsμ

sμ



Example 11.3 (SOLN)

Evaluate

Second, a larger ladder angle would decrease the lever arms with respect to B of the weights of the ladder and Lancelot and increase the lever arm of , all of which would decrease the required friction force. The R.D Werner Ladder Co. recommends that its ladders be used at an angle of 75 . Why not 90 ?

1n

° °



Example 11.3 (SOLN)

Evaluate

Finally, if we had assumed friction on the wall as well as on the floor, the problem would be impossible to solve by using the equilibrium conditions alone. Such a problem is said to be statically indeterminate. The difficulty is that it’s no longer adequate to treat the body as being perfectly rigid. Another simple example of such a problem is a four-legged table; there is no way to use the equilibrium conditions along to find the force on each separate leg. In a sense, four legs are one too many; three, properly located, are sufficient for stability.



Example 11.4 Another rescue attempt

After Lancelot falls into the moat, Sir Gawain tosses a grappling hook through an open upstairs window. In Fig. 11.8a he is resting partway up the rope. Gawain weighs 700N, his body makes an angle of 60 with the wall, and his center of gravity is 0.85m from his feet. The rope force acts 1.30m from his feet, and the rope makes an angle of 20 with the vertical. Find the tension in the rope and forces exerted on his feet by the moss-free region of the wall.

°

°



Example 11.4 Another rescue attempt



Example 11.4 (SOLN)

Identify and Set up

We use the condition for equilibrium as in the previous eg. Figure 11.8b shows the free-body diagram for Gawain. The forces on his feet include a normal component n and a frictional component f. The rope tension is T. The angles needed to find the torques are shown.



Example 11.4 (SOLN)

Execute

Because two unknown forces act at point B as Gawain’s feet, we take torques about B, obtaining a toque equation in which T is the only unknown. The easiest way to find the torque of each force is to use , where is the angle between the position vector (From B to a point of application of force) and the force vector . Taking counterclockwise torques as positive, we obtain the torque equation

sinrFτ φ=φ r

r

Fur

( ) ( )1.30 sin140B m Tτ = °∑( )( )( ) ( ) ( )0.85 700 sin 60 0 0 0m N n f− ° + + =



Example 11.4 (SOLN)

Execute

We solve this equation for T; the result is

617T N=

Note the rope tension is less than Gawain’s weight.

To find the components of force n and f at Gawain’s feet, we use the first equilibrium condition. From Fig 11.8b, is negative and is positive, so and . Then

xT yT sin 20xT T=− °cos20yT T= °

( )sin 20 0xF n T= + − ° =∑( )cos20 700 0yF f T N= + ° + − =∑



Example 11.4 (SOLN)

Execute

When we substitute the value of T into these equations, we get

211 and 120n N f N= =EvaluateAs a check, note the sum of vertical friction force and the vertical component of tension equals Gawain’s 700N weight.( )cos20 580T N° =( )120f N=



Example 11.5 Equilibrium and pumping iron

Figure 11.9a shows a human arm lifting a dumbbell, and Fig 11.9b is a free body diagram for the forearm. The forearm is in equilibrium under the action of the weight w of the dumbbell, the tension T in the tendon connected to the biceps muscle, and the force E exerted on the forearm by the upper arm at the elbow joint. For clarity the point A where the tendon is attached is drawn farther away from the elbow than its actual position. The weight w and the angle are given; we want to fine the tendon tension and the two components of force at the elbow (three unknown scalar quantities in all). We neglect the weight of the forearm itself.

θ



Example 11.5 Equilibrium and pumping iron



Example 11.5 (SOLN)

Identify and Set Up

The system is at rest, so once again we use the conditions for equilibrium. As shown in Fig 11.9b, we represent the tendon force in terms of its components and , using the given angle and the unknown magnitude T:

xT

yT θ

cosxT T θ=sinyT T θ=



We also represent the force at the elbow in terms of its components and . We’ll guess that the directions of these components are as shown in Fig 11.9b; there’s no need to agonizes over this guess, since the results for and will tell us the actual directions. Our target variable and of the force at the elbow.

Example 11.5 (SOLN)

xE

xE

yE

Identify and Set Up

xE

yE

yE



The simplest way to find the tension T is to take torques about the elbow joint. The resulting torque equation does not contain , or because the lines of action of all these forces pass through this point. The torque equation is then simply

Example 11.5 (SOLN)

Execute

xE yE xT

0E yLw DTτ = − =∑From this we find

and siny

Lw LwT T

D D θ= =



To find and , we use the first conditions for equilibrium, and

Example 11.5 (SOLN)

Execute

xE yE0xF =∑ 0yF =∑( ) 0x x xF T E= + − =∑

cos cos cotsinx xLw Lw

E T TD D

θ θ θθ

= = = =

Lw D LwD h h

= =



The negative sign shows that our guess for the direction of , shown in Fig 11.9b was wrong; it is actually vertically downward.

Example 11.5 (SOLN)

Execute

( ) 0y y yF T E w= + + − =∑( )

yL D wLw

E wD D

−= − = −

yE



We can check our results by find and in a different way that uses two more torques equations. We take torques about the tendon attach point, A:

Example 11.5 (SOLN)

Evaluate

xE yE

( ) ( )0 and A y y

L D wL D w DE E

Dτ

−= − + = = −∑

Finally, we take torques about point B in the figure:

0 and B x xLw

Lw hE Eh

τ = − = =∑



We chose points A and B because the tendon tension T has zero torques about either of these points. Notice how much we have simplified these calculations by choosing the point of calculating torques so as to eliminate one or more of the unknown quantities.In our alternative determination of and , we didn’t explicitly use the first condition for equilibrium. As a check compute and to verify that they really are zero. Consistency checks are always a good idea!

Example 11.5 (SOLN)

Evaluate

xE yE

xF∑ yF∑



As a specific example, suppose w = 200N , D = 0.05m; L = 0.30m and .Then from , we find

Example 11.5 (SOLN)

Evaluate

80θ = ° tan /h Dθ =

( )( )tan 0.500 5.67 0.28h D m mθ= = =

From the previous general results we find

( )( )( )( )0.30 200

1220sin 0.050 0.98

m NLwT N

D mθ= = =

( ) ( )( )0.30 0.050 200

0.050yL D w m m N

ED m

− −= − = −

1000N=−



Example 11.5 (SOLN)

Evaluate

( )( )0.30 200210

0.28xm NLw

E Nh m

= = =

The magnitude of the force at the elbow is

2 2 1020x yE E E N= + =

In view of the magnitudes of our results, neglecting the weight of the forearm itself, which may be 20N or so, will cause only relatively small errors in our results.



11.4 Stress, Strain and Elastic Moduli

• Stress is the strength of the forces causing the deformation, on a “force per unit area” basis.

• Strain is the resulting deformation.• When stress and strain are small enough, the 2 are directly

proportional, we called the proportionality constant as elastic modulus.

(11.7)stress

Elastic Modulusstrain

=

• The proportionality of stress and strain is called Hooke’s law.




• The figure shows an object in tension.• The net force on the object is zero, but the object

deforms.• The tensile stress produces a tensile strain.

Tensile and Compressive Stress and Strain




• We defined the tensile stress at the cross section as the ratio of the force to the cross-sectional area A:


F⊥

(11.8)F

Tensile stressA⊥=

• This is a scalar quantity as is the magnitude of the force.

• SI unit is Pascal (1 Pa).

F⊥

21 1 /Pa N m=




• The tensile strain of the object is equal to the fractional change in length, which is the ratio of the elongation to the original length


lΔ 0l

0

0 0 s (11.9)

l l lTensile train

l l− Δ= =

• The corresponding elastic modulus is called Young’s Modulus, denoted by Y:

0

0

/ (11.10)

/ltensile stress F A F

Ytensile strain l l A l

⊥ ⊥= = =Δ Δ




• Figure shows an object in compression.• The compressive stress and compressive strain are

the defined the same way as tensile stress and strain, except that now denotes the distance that the object contracts.


lΔ




• Table shows the approximate elastic moduli of different materials.





• In many situations, bodies can experience both tensile and compressive stresses at the same time.

• Figure (a) shows a beam supported at both ends and is under both compression and tension.

• Figure (b) shows the cross-sectional shape of an I-beam minimizes both stress and weight.






• In many situations, bodies can experience both tensile and compressive stresses at the same time.



Example 11.6 Tensile stress and strain

A steel rod 2.0m long has a cross-sectional area of 0.30cm2. The rod is now hung by one eng from a support structure and a 550-kg milling machine is hung from the rod’s lower end. Determine the stress, the strain and the elongation of the rod.



We use the definitions of stress, strain and Young’s modulus given by Eqs. (11.8),(11.9), and (11.10), respectively and the value of Y for steel from Table 11.1.

Example 11.6 (SOLN)

Identify and Set Up



We find

Example 11.6 (SOLN)

Execute

( )( )28

5 2

550 9.8 /stress 1.8 10

3.0 10

kg m sFPa

A m⊥

−= = = ×

×

84

100

stress 1.8 10strain 9.0 10

20 10

l Pal Y Pa

−Δ ×= = = = ××

( ) ( )( )40elongation strain 9.0 10 2.0l l m−= Δ = × = ×

0.0018 1.8m mm= =



The small size of this elongation, which results from a load of over half a ton, is a testament to the stiffness of steel.

Example 11.6 (SOLN)

Evaluate




• When the stress is a uniform pressure on all sides, the resulting deformation is a volume change.

• We use the terms bulk stress (or volume stress) and bulk strain (or volume strain) to describe these quantities.

• When an object is immersed in a fluid (liquid or gas) at rest, the fluid exerts a force on any part of the object’s surface, the force is perpendicular to the surface.

Bulk stress and Strain




• The pressure p in the fluid is


(11.11)F

pA⊥=

• SI unit is Pascal (1 Pa).• One atmosphere is the approximate average

pressure of the earth’s atmosphere at sea level:

5 21 atmosphere = 1 atm = 1.013 10 Pa = 14.7 lb/in×




• Pressure is a scalar and not a vector quantity.• The figure shows an object under bulk stress.• Without the stress, the cube has volume ; when

the stress is applied, the cube has a smaller volume V.





• The bulk strain is as follow:


0

VBulk (volume) strain = (11.12)

VΔ

• When Hooke’s law is obeyed, an increase in pressure (bulk stress) produces a proportional bulk strain (fractional change in volume).

• The corresponding elastic modulus (ratio of stress to strain) is called the bulk modulus.

0

Bulk stress (11.13)

Bulk strain /p

BV VΔ= =−

Δ




• The reciprocal of the bulk modulus is called compressibility and it is denoted by k.


0

0

/1 1 (11.14)

V V Vk

B p V pΔ= =− =−

Δ Δ

• The units of compressibility are or .1Pa− 1atm−




• Table shows the compressibility of liquids.




Example 11.7 Bulk stress and strain

A hydraulic press contains 0.25m3 (250L) of oil. Find the decrease in volume of oil when it is subjected to a pressure increase . The bulk modulus of the oil is B = 5.0 X 109 Pa and its compressibility is k = 1/B = 20 X 10-6 atm-1.

71.6 10p PaΔ = ×



Example 11.7 (SOLN)

Identify and Set Up

We are given both the bulk modulus and the compressibility, so we can use either Eq.(11.13) or Eq.(11.14) to find the volume change . VΔ

Execute

Solving Eq.(11.13), we find

( )( )2 70

9

0.25 1.6 10

5.0 10

m PaV pV

B Pa

×ΔΔ = − = −

×

4 38.0 10 0.80m L−=− × =−



Example 11.7 (SOLN)

Alternatively, we can use Eq.(11.14). Solving for and using the approximate unit conversions given above, we get

VΔ

( )( )( )6 1 30 20 10 0.25 160V kV p atm m atm− −Δ = − Δ = − ×

4 38.0 10 m− −=− ×



Example 11.7 (SOLN)

Evaluate

We get the same result for with either approach as we should Note that is negative, indicating that the volume decreases when the pressure increases. Note that even though the pressure increase is very large, the fractional change in volume is very small:

VΔVΔ

( ) ( )4 3 30/ 8.0 10 / 0.25 0.0032 or 0.32%V V m m−Δ = − × = − −




• Figure shows an object in shear stress.• Forces are applied tangent to opposite surfaces

of the object.

Shear stress and Strain




• We defined the shear stress as the force acting tangent to the surface, divided by the area A on which its acts:


FShear stress = (11.15)

AP

FP

• We defined shear strain as the ratio of the displacement x to the transverse dimension h,

Shear strain = (11.16)xh




• If the forces are small that Hooke’s law is obeyed, the shear strain is proportional to the shear stress.

• The corresponding elastic modulus is called shear modulus, denoted by S


/Shear stress (11.7)

Shear strain /

F A F hS

x h A x= = =P P



Example 11.8 Shear stress and strain

Suppose the object in is the brass base plate of an outdoor sculpture; it experiences shear forces as a result of an earthquake. The frame is 0.80m square and 0.50cm thick. How large a force must be exerted on each of its edges if the displacement x is 0.16mm?



Example 11.8 (SOLN)

Identify and Set up

We first find the shear strain using Eq.(11.16), then determine the shear stress using Eq.(11.17). We can then solve for the target variable , using Eq.(11.15). The values of all the other quantities are given, including the shear modulus of brass (from table 11.1, S = 3.5 X 1010 Pa). Note that h in Fig 11.16 represents the 0.80m length of each side of the square plate, while the area A is the product of the 0.80-m length and the 0.50-cm thickness.

FP



Example 11.8 (SOLN)

ExecuteThe shear strain is

441.6 10

shear strain 2.0 100.80

x mh m

−−×= = = ×

From Eq(11.17), the shear stress equals the shear strain multiplied by the shear modulus S,

( )stress shear strain S= ×

( )( )4 10 62.0 10 3.5 10 7.0 10Pa Pa−= × × = ×



Example 11.8 (SOLN)

ExecuteFrom Eq.(11.15), the force at each edge is the shear stress multiplied by the area of the edge :

( )shear stressF A= ×P

( )( )( )6 67.0 10 0.80 0.0050 2.8 10Pa m m N= × = ×

EvaluateThe required force is more than three tons! Brass has a large shear modulus, which means that it’s intrinsically difficult to deform. Furthermore, the plate is relatively which (0.50cm), so the area A is relatively large and a large force is needed to provide the necessary stress

FP/F AP



11.5 Elasticity and Plasticity

• Figure shows a typical stress-strain diagram for a ductile metal under tension.




• The straight line from O ends at point a and this point is called proportional limit.

• From a to b, Hooke’s law is not obeyed.• Between O to b, the deformation is reversible, and

the forces are conservative.• The energy put into the material to cause the

deformation is recovered when the stress is removed.

• In region Ob, we say the material shows elastic behavior.




• Point b is called the yield point; the stress at the yield point is called elastic limit.

• Beyond b, the material does not come back to its original length. The material has undergone an irreversible deformation and has acquired what we called a permanent set.

• Fracture happens at point d.• b to d is called plastic flow or plastic deformation, it

is irreversible; when the stress is removed, the material does not return to its original state.




• Ductile material is when there is a large amount of plastic deformation takes place between the elastic limit and the fracture point.

• Brittle material is fracture occurs soon after the elastic limit is passed.

• Figure shows a typical stress-strain curve for vulcanized rubber. The curves are different for increasing and decreasing stress, a phenomenon called elastic hysteresis.

• The stress is not proportional to the strain, but the behavior is elastic when the load is removed, the material returns to its original length.







• The stress required to cause actual fracture of a material is called the breaking stress, the ultimate strength, or (for tensile stress) the tensile strength.

• Table shows the approximate breaking stress of materials.



• For a rigid body to be in equilibrium, two conditions must be satisfied. First, the vector sum of forces must be zero. Second, the sum of torques about any point must be zero. To compare the torque due to a force, find the torque of each force component by using its appropriate lever arm and sign and then add these values.

• The torques due to the weight of a body can be obtained by assuming the entire weight to be concentrated at the center of gravity. If has the same value at all points, the coordinates of the center of gravity are the same as those of the center of mass.

Concept Summary

gur



• Hooke’s law states that in elastic deformations, stress (force per unit area) is proportional to strain (fractional deformation). The proportionality constant is called the elastic modulus.• Tensile stress is tensile force per unit area, . The tensile strain is fractional change in length . The elastic modulus is called Young’s modulus Y. Compressive stress and

strain are defined in the same way as tensile stress and strain.

Concept Summary

/F A⊥0/l lΔ



• Pressure in a fluid is force per unit area. Bulk stress is pressure change, and bulk strain is fractional volume change, . The elastic modulus is called the bulk modulus, B. Compressibility, k is the reciprocal of bulk modulus: k=1/B

• Shear stress is force per unit area, , for a force applied tangent to a surface. Shear strain is the displacement x of one side divided by the transverse dimension h. The elastic modulus is called the shear modulus, S.

Concept Summary

pΔ0/V VΔ

/F AP



• The proportional limit is the maximum stress for which stress and strain are proportional. Beyond the proportional limit, Hooke’s law is not valid. The elastic limit is the stress beyond which irreversible deformation occurs. The breaking stress, or ultimate strength, is the stress at which the material breaks.

Concept Summary



Key Equations

0 0 0 (11.1)x y zF F F= = =∑ ∑ ∑0 about any point (11.2)τ =∑

r

1 21 2

1 2

... (11.4)

...

iii

cmi

i

m rm r m r

rm m m

+ += =+ +

∑∑

rr r

r

(11.7)stress

Elastic Modulusstrain

=



Key Equations

0

0

/ (11.10)

/ltensile stress F A F

Ytensile strain l l A l

⊥ ⊥= = =Δ Δ

(11.11)F

pA⊥=

0

Bulk stress (11.13)

Bulk strain /p

BV VΔ= =−

Δ

/Shear stress (11.7)

Shear strain /

F A F hS

x h A x= = =P P

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Module 11 Elasticity