what happens to stress when radius is doubled youngs modulus

12 Static Equilibrium and Elasticity

12.three Stress, Strain, and Rubberband Modulus

Learning Objectives

Past the end of this section, you will exist able to:

  • Explicate the concepts of stress and strain in describing elastic deformations of materials
  • Describe the types of elastic deformation of objects and materials

A model of a rigid body is an idealized instance of an object that does not deform nether the actions of external forces. It is very useful when analyzing mechanical systems—and many physical objects are indeed rigid to a nifty extent. The extent to which an object tin be perceived as rigid depends on the physical properties of the material from which it is fabricated. For instance, a ping-pong brawl made of plastic is brittle, and a lawn tennis brawl fabricated of rubber is elastic when acted upon past squashing forces. Still, nether other circumstances, both a ping-pong ball and a tennis brawl may bounciness well as rigid bodies. Similarly, someone who designs prosthetic limbs may be able to approximate the mechanics of human limbs past modeling them every bit rigid bodies; however, the actual combination of bones and tissues is an elastic medium.

For the balance of this affiliate, we move from consideration of forces that touch on the motion of an object to those that touch on an object'southward shape. A change in shape due to the awarding of a force is known every bit a deformation . Fifty-fifty very small forces are known to cause some deformation. Deformation is experienced by objects or concrete media under the action of external forces—for instance, this may be squashing, squeezing, ripping, twisting, shearing, or pulling the objects apart. In the language of physics, two terms describe the forces on objects undergoing deformation: stress and strain.

Stress is a quantity that describes the magnitude of forces that crusade deformation. Stress is generally defined as force per unit area. When forces pull on an object and crusade its elongation, like the stretching of an elastic band, we call such stress a tensile stress. When forces cause a pinch of an object, nosotros call it a compressive stress. When an object is beingness squeezed from all sides, similar a submarine in the depths of an ocean, we call this kind of stress a bulk stress (or book stress). In other situations, the acting forces may be neither tensile nor compressive, and still produce a noticeable deformation. For example, suppose you lot concord a book tightly between the palms of your hands, then with one hand you press-and-pull on the forepart cover away from you, while with the other manus you lot press-and-pull on the back comprehend toward you. In such a case, when deforming forces act tangentially to the object'south surface, nosotros call them 'shear' forces and the stress they cause is chosen shear stress.

The SI unit of stress is the pascal (Pa). When one newton of force presses on a unit surface area of i meter squared, the resulting stress is one pascal:

\[\text{one pascal}=1.0\,\text{Pa}=\frac{1.0\,\text{N}}{1.0\,{\text{m}}^{2}}.\]

In the British system of units, the unit of stress is 'psi,' which stands for 'pound per foursquare inch'

\[({\text{lb/in}}^{2}).\]

Another unit that is often used for bulk stress is the atm (atmosphere). Conversion factors are

\[\begin{array}{c}1\,\text{psi}=6895\,\text{Pa}\enspace\text{and}\enspace1\,\text{Pa}=1.450\,×\,{10}^{-4}\text{psi}\\ 1\,\text{atm}=1.013\,×\,{10}^{5}\text{Pa}=14.7\,\text{psi.}\end{array}\]

An object or medium nether stress becomes deformed. The quantity that describes this deformation is called strain. Strain is given every bit a partial change in either length (under tensile stress) or book (under bulk stress) or geometry (nether shear stress). Therefore, strain is a dimensionless number. Strain under a tensile stress is called tensile strain, strain nether bulk stress is called bulk strain (or book strain), and that caused by shear stress is called shear strain.

The greater the stress, the greater the strain; still, the relation betwixt strain and stress does not demand to be linear. Just when stress is sufficiently low is the deformation it causes in direct proportion to the stress value. The proportionality constant in this relation is called the elastic modulus. In the linear limit of low stress values, the general relation between stress and strain is

\[\text{stress}=\text{(elastic modulus)}\,×\,\text{strain.}\]

As nosotros can meet from dimensional analysis of this relation, the elastic modulus has the same physical unit as stress because strain is dimensionless.

We tin can too see from (Figure) that when an object is characterized by a large value of elastic modulus, the effect of stress is small. On the other hand, a minor rubberband modulus means that stress produces large strain and noticeable deformation. For example, a stress on a safe band produces larger strain (deformation) than the same stress on a steel band of the aforementioned dimensions because the rubberband modulus for rubber is two orders of magnitude smaller than the elastic modulus for steel.

The elastic modulus for tensile stress is called Young'southward modulus; that for the bulk stress is called the bulk modulus; and that for shear stress is called the shear modulus. Note that the relation between stress and strain is an observed relation, measured in the laboratory. Elastic moduli for various materials are measured under various physical weather, such every bit varying temperature, and collected in engineering data tables for reference ((Figure)). These tables are valuable references for manufacture and for anyone involved in engineering or construction. In the next department, we talk over strain-stress relations beyond the linear limit represented by (Figure), in the full range of stress values upward to a fracture point. In the remainder of this section, we study the linear limit expressed by (Figure).

Approximate Rubberband Moduli for Selected Materials
Fabric Immature's modulus

\[×\,{10}^{10}\text{Pa}\]

Bulk modulus

\[×\,{10}^{10}\text{Pa}\]

Shear modulus

\[×\,{10}^{10}\text{Pa}\]

Aluminum 7.0 7.5 ii.5
Bone (tension) 1.6 0.8 8.0
Bone (compression) 0.9
Brass ix.0 6.0 3.5
Brick 1.5
Physical two.0
Copper 11.0 14.0 four.4
Crown glass 6.0 five.0 2.5
Granite iv.5 iv.v ii.0
Hair (human being) 1.0
Hardwood 1.five i.0
Iron 21.0 16.0 7.7
Pb 1.6 4.1 0.6
Marble half dozen.0 7.0 2.0
Nickel 21.0 17.0 7.8
Polystyrene 3.0
Silk vi.0
Spider thread 3.0
Steel 20.0 16.0 7.5
Acetone 0.07
Ethanol 0.09
Glycerin 0.45
Mercury ii.5
Water 0.22

Tensile or Compressive Stress, Strain, and Young'southward Modulus

Tension or compression occurs when ii antiparallel forces of equal magnitude human activity on an object along simply one of its dimensions, in such a way that the object does not move. One mode to envision such a state of affairs is illustrated in (Effigy). A rod segment is either stretched or squeezed by a pair of forces acting forth its length and perpendicular to its cross-section. The internet result of such forces is that the rod changes its length from the original length

\[{L}_{0}\]

that it had before the forces appeared, to a new length 50 that it has nether the activity of the forces. This modify in length

\[\text{Δ}L=L-{L}_{0}\]

may be either elongation (when Fifty is larger than the original length

\[{L}_{0})\]

or contraction (when Fifty is smaller than the original length

\[{L}_{0}).\]

Tensile stress and strain occur when the forces are stretching an object, causing its elongation, and the length change

\[\text{Δ}L\]

is positive. Compressive stress and strain occur when the forces are contracting an object, causing its shortening, and the length change

\[\text{Δ}L\]

is negative.

In either of these situations, we define stress as the ratio of the deforming force

\[{F}_{\perp }\]

to the cantankerous-sectional area A of the object existence deformed. The symbol

\[{F}_{\perp }\]

that we reserve for the deforming forcefulness means that this strength acts perpendicularly to the cross-section of the object. Forces that deed parallel to the cross-section practice not modify the length of an object. The definition of the tensile stress is

\[\text{tensile stress}=\frac{{F}_{\perp }}{A}.\]

Tensile strain is the measure of the deformation of an object under tensile stress and is defined as the fractional modify of the object's length when the object experiences tensile stress

\[\text{tensile strain}=\frac{\text{Δ}L}{{L}_{0}}.\]

Compressive stress and strain are defined by the same formulas, (Figure) and (Figure), respectively. The only difference from the tensile situation is that for compressive stress and strain, we take absolute values of the right-manus sides in (Figure) and (Figure).

Figure A is a schematic drawing of a cylinder with a length L0 that is under the tensile strain. Two forces at the different sides of cylinder increase its length by Delta L. Figure B is a schematic drawing of a cylinder with a length L0 that is under the compressive strain. Two forces at the different sides of cylinder reduce its length by Delta L.
Figure 12.eighteen When an object is in either tension or compression, the net force on it is aught, but the object deforms by changing its original length

\[{L}_{0}.\]

(a) Tension: The rod is elongated by

\[\text{Δ}L.\]

(b) Compression: The rod is contracted by

\[\text{Δ}L.\]

In both cases, the deforming force acts along the length of the rod and perpendicular to its cross-section. In the linear range of low stress, the cantankerous-sectional area of the rod does non change.

Immature's modulus Y is the elastic modulus when deformation is caused by either tensile or compressive stress, and is defined by (Figure). Dividing this equation by tensile strain, nosotros obtain the expression for Young'south modulus:

\[Y=\frac{\text{tensile stress}}{\text{tensile strain}}=\frac{{F}_{\perp }\text{/}\,A}{\text{Δ}L\,\text{/}\,{L}_{0}}=\frac{{F}_{\perp }}{A}\,\frac{{L}_{0}}{\text{Δ}L}.\]

Example

Compressive Stress in a Colonnade

A sculpture weighing 10,000 N rests on a horizontal surface at the superlative of a half-dozen.0-m-alpine vertical pillar (Figure). The pillar's cross-exclusive area is

\[0{\text{.20 m}}^{2}\]

and it is made of granite with a mass density of

\[{2700\,\text{kg/m}}^{3}.\]

Detect the compressive stress at the cantankerous-section located iii.0 g below the top of the colonnade and the value of the compressive strain of the top 3.0-m segment of the pillar.

Picture shows a photograph of Nelson's Column in Trafalgar Square.
Figure 12.xix Nelson's Cavalcade in Trafalgar Square, London, England. (credit: modification of work by Cristian Bortes)

Strategy

Start we observe the weight of the 3.0-m-long peak department of the pillar. The normal force that acts on the cross-department located 3.0 chiliad down from the tiptop is the sum of the pillar's weight and the sculpture's weight. Once nosotros have the normal force, we utilize (Effigy) to find the stress. To find the compressive strain, we find the value of Young'south modulus for granite in (Figure) and invert (Figure).

Solution

The volume of the pillar segment with height

\[h=3.0\,\text{m}\]

and cross-sectional area

\[A=0.20\,{\text{m}}^{2}\]

is

\[V=Ah=(0.20\,{\text{m}}^{2})(3.0\,\text{m})=0.60\,{\text{m}}^{3}.\]

With the density of granite

\[\rho =2.7\,×\,{10}^{3}\,{\text{kg/m}}^{3},\]

the mass of the pillar segment is

\[m=\rho V=(2.7\,×\,{10}^{3}\,{\text{kg/m}}^{3})(0.60\,{\text{m}}^{3})=1.60\,×\,{10}^{3}\text{kg}.\]

The weight of the pillar segment is

\[{w}_{p}=mg=(1.60\,×\,{10}^{3}\text{kg})(9.80\,{\text{m/s}}^{2})=1.568\,×\,{10}^{4}\text{N.}\]

The weight of the sculpture is

\[{w}_{s}=1.0\,×\,{10}^{4}\text{N},\]

and then the normal force on the cantankerous-exclusive surface located 3.0 m below the sculpture is

\[{F}_{\perp }={w}_{p}+{w}_{s}=(1.568+1.0)\,×\,{10}^{4}\text{N}=2.568\,×\,{10}^{4}\text{N.}\]

Therefore, the stress is

\[\text{stress}=\frac{{F}_{\perp }}{A}=\frac{2.568\,×\,{10}^{4}\text{N}}{0.20\,{\text{m}}^{2}}=1.284\,×\,{10}^{5}\text{Pa}=\text{128.4 kPa.}\]

Young's modulus for granite is

\[Y=4.5\,×\,{10}^{10}\text{Pa}=4.5\,×\,{10}^{7}\text{kPa}.\]

Therefore, the compressive strain at this position is

\[\text{strain}=\frac{\text{stress}}{Y}=\frac{128.4\,\text{kPa}}{4.5\,×\,{10}^{7}\text{kPa}}=2.85\,×\,{10}^{-6}.\]

Significance

Observe that the normal force acting on the cross-sectional area of the pillar is not constant along its length, but varies from its smallest value at the tiptop to its largest value at the lesser of the pillar. Thus, if the pillar has a uniform cross-sectional area forth its length, the stress is largest at its base.

Check Your Agreement

Find the compressive stress and strain at the base of Nelson's cavalcade.

[reveal-answer q="fs-id1163709836524″]Show Solution[/reveal-answer]

[hidden-answer a="fs-id1163709836524″]

\[206.8\,\text{kPa};\,4.6\,×\,{10}^{-5}\]

[/subconscious-answer]

Case

Stretching a Rod

A 2.0-k-long steel rod has a cross-sectional expanse of

\[0.30\,{\text{cm}}^{2}.\]

The rod is a part of a vertical support that holds a heavy 550-kg platform that hangs attached to the rod'south lower end. Ignoring the weight of the rod, what is the tensile stress in the rod and the elongation of the rod nether the stress?

Strategy

First nosotros compute the tensile stress in the rod under the weight of the platform in accordance with (Figure). Then we invert (Figure) to detect the rod'south elongation, using

\[{L}_{0}=2.0\,\text{m}.\]

From (Figure), Young's modulus for steel is

\[Y=\,2.0\,×\,{10}^{11}\text{Pa}.\]

Solution

Substituting numerical values into the equations gives the states

\[\begin{array}{ccc}\hfill \frac{{F}_{\perp }}{A}& =\hfill & \frac{(550\,\text{kg})(9.8\,{\text{m/s}}^{2})}{3.0\,×\,{10}^{-5}\,{\text{m}}^{2}}=1.8\,×\,{10}^{8}\text{Pa}\hfill \\ \hfill \text{Δ}L& =\hfill & \frac{{F}_{\perp }}{A}\,\frac{{L}_{0}}{Y}=(1.8\,×\,{10}^{8}\text{Pa})\,\frac{2.0\,\text{m}}{2.0\,×\,{10}^{11}\text{Pa}}=1.8\,×\,{10}^{-3}\text{m}=\,1.8\,\text{mm.}\hfill \end{array}\]

Significance

Similarly as in the example with the column, the tensile stress in this instance is not uniform forth the length of the rod. Unlike in the previous example, withal, if the weight of the rod is taken into consideration, the stress in the rod is largest at the top and smallest at the bottom of the rod where the equipment is fastened.

Check Your Understanding

A 2.0-k-long wire stretches 1.0 mm when subjected to a load. What is the tensile strain in the wire?

[reveal-respond q="fs-id1163713286423″]Testify Solution[/reveal-reply]

[subconscious-reply a="fs-id1163713286423″]

\[5.0\,×\,{10}^{-4}\]

[/hidden-answer]

Objects tin can ofttimes experience both compressive stress and tensile stress simultaneously (Figure). One instance is a long shelf loaded with heavy books that sags between the terminate supports under the weight of the books. The meridian surface of the shelf is in compressive stress and the bottom surface of the shelf is in tensile stress. Similarly, long and heavy beams sag nether their own weight. In modernistic building construction, such bending strains can be almost eliminated with the employ of I-beams (Figure).

Figure A is a schematic drawing of forces experienced by the object during bending downward. It experiences tensile stress (stretching) in the upper section and compressive stress (compressing) in the lower section. Figure B shows a photograph of weightlifter during the lifting. The iron bar that he is holding is bent.
Figure 12.twenty (a) An object angle downwardly experiences tensile stress (stretching) in the upper department and compressive stress (compressing) in the lower section. (b) Elite weightlifters often curve iron bars temporarily during lifting, as in the 2012 Olympics competition. (credit b: modification of work past Oleksandr Kocherzhenko)
Figure is a photograph of steel I-beams are used in construction.
Figure 12.21 Steel I-beams are used in construction to reduce bending strains. (credit: modification of work by "US Ground forces Corps of Engineers Europe District"/Flickr)

A heavy box rests on a table supported by three columns. View this demonstration to motility the box to meet how the compression (or tension) in the columns is affected when the box changes its position.

Bulk Stress, Strain, and Modulus

When you dive into h2o, you feel a forcefulness pressing on every part of your trunk from all directions. What yous are experiencing and then is bulk stress, or in other words, pressure. Majority stress always tends to decrease the volume enclosed by the surface of a submerged object. The forces of this "squeezing" are ever perpendicular to the submerged surface (Figure). The effect of these forces is to decrease the book of the submerged object by an amount

\[\text{Δ}V\]

compared with the volume

\[{V}_{0}\]

of the object in the absence of bulk stress. This kind of deformation is called bulk strain and is described by a change in volume relative to the original volume:

\[\text{bulk strain}=\frac{\text{Δ}V}{{V}_{0}}.\]

Figure is a schematic drawing of forces experienced by an object under the bulk stress. Equal forces perpendicular to the surface act from all directions and reduce the volume by the amount delta V compared to the original volume, V0.
Figure 12.22 An object under increasing bulk stress ever undergoes a decrease in its volume. Equal forces perpendicular to the surface act from all directions. The effect of these forces is to decrease the volume by the corporeality

\[\text{Δ}V\]

compared to the original book,

\[{V}_{0}.\]

The bulk strain results from the majority stress, which is a force

\[{F}_{\perp }\]

normal to a surface that presses on the unit surface surface area A of a submerged object. This kind of physical quantity, or pressure p, is defined as

\[\text{pressure}=p\equiv \frac{{F}_{\perp }}{A}.\]

We will report pressure level in fluids in greater detail in Fluid Mechanics. An of import characteristic of pressure is that it is a scalar quantity and does not have any particular direction; that is, pressure level acts equally in all possible directions. When you submerge your hand in water, you sense the same amount of force per unit area interim on the top surface of your hand every bit on the bottom surface, or on the side surface, or on the surface of the pare between your fingers. What you are perceiving in this case is an increase in pressure level

\[\text{Δ}p\]

over what you lot are used to feeling when your hand is non submerged in water. What you feel when your mitt is not submerged in the water is the normal pressure level

\[{p}_{0}\]

of i atmosphere, which serves as a reference point. The bulk stress is this increase in pressure, or

\[\text{Δ}p,\]

over the normal level,

\[{p}_{0}.\]

When the bulk stress increases, the bulk strain increases in response, in accordance with (Effigy). The proportionality abiding in this relation is called the bulk modulus, B, or

\[B=\frac{\text{bulk stress}}{\text{bulk strain}}=-\frac{\text{Δ}p}{\text{Δ}V\,\text{/}\,{V}_{0}}=\text{−}\text{Δ}p\,\frac{{V}_{0}}{\text{Δ}V}.\]

The minus sign that appears in (Effigy) is for consistency, to ensure that B is a positive quantity. Note that the minus sign

\[(-)\]

is necessary considering an increase

\[\text{Δ}p\]

in pressure (a positive quantity) always causes a decrease

\[\text{Δ}V\]

in volume, and decrease in book is a negative quantity. The reciprocal of the bulk modulus is called compressibility

\[k,\]

or

\[k=\frac{1}{B}=-\frac{\text{Δ}V\,\text{/}\,{V}_{0}}{\text{Δ}p}.\]

The term 'compressibility' is used in relation to fluids (gases and liquids). Compressibility describes the change in the volume of a fluid per unit increase in force per unit area. Fluids characterized by a large compressibility are relatively easy to compress. For example, the compressibility of water is

\[4.64\,×\,{10}^{-5}\text{/atm}\]

and the compressibility of acetone is

\[1.45\,×\,{10}^{-4}\text{/atm}.\]

This means that under a i.0-atm increase in pressure, the relative decrease in volume is approximately three times as large for acetone every bit information technology is for water.

Example

Hydraulic Printing

In a hydraulic press (Effigy), a 250-liter book of oil is subjected to a 2300-psi pressure increment. If the compressibility of oil is

\[2.0\,×\,{10}^{-5}\,\text{/}\,\,\text{atm},\]

discover the bulk strain and the absolute decrease in the volume of oil when the printing is operating.

Figure is a schematic drawing of a hydraulic press. A small piston is displaced downward and causes the large piston holding object to move upward.
Effigy 12.23 In a hydraulic press, when a small piston is displaced downwards, the pressure level in the oil is transmitted throughout the oil to the large piston, causing the large piston to motion upwards. A pocket-size force practical to a small-scale piston causes a large pressing strength, which the large piston exerts on an object that is either lifted or squeezed. The device acts every bit a mechanical lever.

Strategy

We must invert (Figure) to find the bulk strain. Start, we catechumen the pressure increase from psi to atm,

\[\text{Δ}p=2300\,\text{psi}=2300\,\text{/}\,14.7\,\text{atm}\approx \,160\,\text{atm},\]

and place

\[{V}_{0}=\,250\,\text{L}.\]

Solution

Substituting values into the equation, we have

\[\begin{array}{cc} \text{bulk strain}=\frac{\text{Δ}V}{{V}_{0}}=\frac{\text{Δ}p}{B}=k\text{Δ}p=(2.0\,×\,{10}^{-5}\text{/atm})(160\,\text{atm})=0.0032\hfill \\ \text{answer:}\enspace\text{Δ}V=0.0032\,{V}_{0}=0.0032(250\,\text{L})=0.78\,\text{L.}\hfill \end{array}\]

Significance

Observe that since the compressibility of water is two.32 times larger than that of oil, if the working substance in the hydraulic press of this problem were changed to water, the bulk strain as well equally the volume change would be two.32 times larger.

Check Your Agreement

If the normal force acting on each face of a cubical

\[1{\text{.0-m}}^{3}\]

piece of steel is inverse by

\[1.0\,×\,{10}^{7}\text{N},\]

find the resulting modify in the book of the piece of steel.

[reveal-answer q="fs-id1163713080974″]Bear witness Solution[/reveal-answer]

[hidden-reply a="fs-id1163713080974″]

63 mL

[/hidden-answer]

Shear Stress, Strain, and Modulus

The concepts of shear stress and strain concern simply solid objects or materials. Buildings and tectonic plates are examples of objects that may be subjected to shear stresses. In general, these concepts do non utilize to fluids.

Shear deformation occurs when ii antiparallel forces of equal magnitude are applied tangentially to opposite surfaces of a solid object, causing no deformation in the transverse direction to the line of forcefulness, as in the typical example of shear stress illustrated in (Figure). Shear deformation is characterized by a gradual shift

\[\text{Δ}x\]

of layers in the direction tangent to the acting forces. This gradation in

\[\text{Δ}x\]

occurs in the transverse direction along some altitude

\[{L}_{0}.\]

Shear strain is defined past the ratio of the largest deportation

\[\text{Δ}x\]

to the transverse distance

\[{L}_{0}\]

\[\text{shear strain}=\frac{\text{Δ}x}{{L}_{0}}.\]

Shear strain is caused by shear stress. Shear stress is due to forces that act parallel to the surface. We use the symbol

\[{F}_{\parallel }\]

for such forces. The magnitude

\[{F}_{\parallel }\]

per surface expanse A where shearing forcefulness is practical is the measure out of shear stress

\[\text{shear stress}=\frac{{F}_{\parallel }}{A}.\]

The shear modulus is the proportionality constant in (Effigy) and is defined past the ratio of stress to strain. Shear modulus is unremarkably denoted by S:

\[S=\frac{\text{shear stress}}{\text{shear strain}}=\frac{{F}_{\parallel }\text{/}\,A}{\text{Δ}x\,\text{/}\,{L}_{0}}=\frac{{F}_{\parallel }}{A}\,\frac{{L}_{0}}{\text{Δ}x}.\]

Figure is a schematic drawing of an object under shear stress: Two antiparallel forces of equal magnitude are applied tangentially to opposite parallel surfaces of the object. As the result, the object is transformed from the rectangle to the parallelogram, shape. While the height of the object remains the same, top corners move to the right by the Delta X.
Figure 12.24 An object nether shear stress: Two antiparallel forces of equal magnitude are applied tangentially to opposite parallel surfaces of the object. The dashed-line contour depicts the resulting deformation. In that location is no change in the direction transverse to the acting forces and the transverse length

\[{L}_{0}\]

is unaffected. Shear deformation is characterized by a gradual shift

\[\text{Δ}x\]

of layers in the direction tangent to the forces.

Example

An Quondam Bookshelf

A cleaning person tries to move a heavy, former bookcase on a carpeted flooring by pushing tangentially on the surface of the very top shelf. However, the only noticeable effect of this effort is like to that seen in (Figure), and it disappears when the person stops pushing. The bookcase is 180.0 cm tall and 90.0 cm broad with four 30.0-cm-deep shelves, all partially loaded with books. The total weight of the bookcase and books is 600.0 N. If the person gives the top shelf a 50.0-Due north push button that displaces the top shelf horizontally by 15.0 cm relative to the motionless bottom shelf, find the shear modulus of the bookcase.

Strategy

The only pieces of relevant information are the concrete dimensions of the bookcase, the value of the tangential force, and the deportation this force causes. We identify

\[{F}_{\parallel }=50.0\,\text{N},\,\text{Δ}x=15.0\,\text{cm},\]

\[{L}_{0}=180.0\,\text{cm},\]

and

\[A=\text{(30.0 cm)}\,\text{(90.0 cm)}=2700.0\,{\text{cm}}^{2},\]

and nosotros use (Figure) to compute the shear modulus.

Solution

Substituting numbers into the equations, we obtain for the shear modulus

\[S=\frac{{F}_{\parallel }}{A}\,\frac{{L}_{0}}{\text{Δ}x}=\frac{50.0\,\text{N}}{2700.0\,{\text{cm}}^{2}}\,\frac{180.0\,\text{cm}\text{.}}{15.0\,\text{cm}\text{.}}=\frac{2}{9}\,\frac{\text{N}}{{\text{cm}}^{2}}=\frac{2}{9}\,×\,{10}^{4}\frac{\text{N}}{{\text{m}}^{2}}=\frac{20}{9}\,×\,{10}^{3}\text{Pa}=\text{2.222 kPa.}\]

We can also detect shear stress and strain, respectively:

\[\begin{array}{c}\frac{{F}_{\parallel }}{A}=\frac{50.0\,\text{N}}{2700.0\,{\text{cm}}^{2}}=\frac{5}{27}\,\text{kPa}=\text{185.2 Pa}\hfill \\ \frac{\text{Δ}x}{{L}_{0}}=\frac{15.0\,\text{cm}}{180.0\,\text{cm}}=\frac{1}{12}=0.083.\hfill \end{array}\]

Significance

If the person in this example gave the shelf a good for you push, it might happen that the induced shear would plummet it to a pile of rubbish. Much the aforementioned shear mechanism is responsible for failures of earth-filled dams and levees; and, in general, for landslides.

Cheque Your Understanding

Explain why the concepts of Young's modulus and shear modulus do not apply to fluids.

[reveal-answer q="fs-id1163713558669″]Show Solution[/reveal-answer]
[subconscious-answer a="fs-id1163713558669″]

Fluids have unlike mechanical properties than those of solids; fluids menstruum.

[/hidden-reply]

Summary

  • External forces on an object (or medium) crusade its deformation, which is a change in its size and shape. The strength of the forces that cause deformation is expressed past stress, which in SI units is measured in the unit of force per unit area (pascal). The extent of deformation under stress is expressed past strain, which is dimensionless.
  • For a small stress, the relation between stress and strain is linear. The elastic modulus is the proportionality constant in this linear relation.
  • Tensile (or compressive) strain is the response of an object or medium to tensile (or compressive) stress. Here, the elastic modulus is chosen Young's modulus. Tensile (or compressive) stress causes elongation (or shortening) of the object or medium and is due to an external forces acting along only one direction perpendicular to the cross-section.
  • Bulk strain is the response of an object or medium to bulk stress. Hither, the elastic modulus is chosen the bulk modulus. Bulk stress causes a change in the book of the object or medium and is caused by forces acting on the body from all directions, perpendicular to its surface. Compressibility of an object or medium is the reciprocal of its bulk modulus.
  • Shear strain is the deformation of an object or medium under shear stress. The shear modulus is the elastic modulus in this case. Shear stress is caused past forces acting along the object's 2 parallel surfaces.

Conceptual Questions

Note: Unless stated otherwise, the weights of the wires, rods, and other elements are causeless to be negligible. Rubberband moduli of selected materials are given in (Effigy).

Why can a squirrel leap from a tree co-operative to the ground and run away undamaged, while a human could break a bone in such a fall?

[reveal-respond q="fs-id1163713160498″]Prove Solution[/reveal-answer]

[hidden-answer a="fs-id1163713160498″]

In contact with the basis, stress in squirrel'due south limbs is smaller than stress in man's limbs.

[/subconscious-respond]

When a glass bottle total of vinegar warms up, both the vinegar and the drinking glass expand, merely the vinegar expands significantly more with temperature than does the glass. The canteen will break if information technology is filled upwardly to its very tight cap. Explicate why and how a pocket of air to a higher place the vinegar prevents the canteen from breaking.

A thin wire strung between ii nails in the wall is used to support a large movie. Is the wire likely to snap if it is strung tightly or if it is strung so that it sags considerably?

[reveal-answer q="fs-id1163713201312″]Testify Solution[/reveal-answer]

[hidden-answer a="fs-id1163713201312″]

tightly

[/hidden-respond]

Review the relationship between stress and strain. Can you detect any similarities between the two quantities?

What type of stress are yous applying when you press on the ends of a wooden rod? When you pull on its ends?

[reveal-answer q="fs-id1163713296478″]Testify Solution[/reveal-reply]

[hidden-respond a="fs-id1163713296478″]

compressive; tensile

[/hidden-answer]

Can compressive stress be applied to a prophylactic band?

Can Young's modulus have a negative value? What about the bulk modulus?

[reveal-answer q="fs-id1163713263717″]Testify Solution[/reveal-answer]

[subconscious-answer a="fs-id1163713263717″]

no

[/hidden-reply]

If a hypothetical fabric has a negative majority modulus, what happens when you squeeze a slice of information technology?

Discuss how you might measure the majority modulus of a liquid.

Problems

The "lead" in pencils is a graphite composition with a Immature's modulus of approximately

\[1.0\,×\,{10}^{9}\text{N}\,\text{/}\,{\text{m}}^{2}.\]

Summate the change in length of the lead in an automatic pencil if yous tap it directly into the pencil with a forcefulness of iv.0 N. The lead is 0.fifty mm in diameter and 60 mm long.

[reveal-answer q="fs-id1163709736795″]Testify Solution[/reveal-answer]

[hidden-reply a="fs-id1163709736795″]

0.three mm

[/hidden-answer]

Television broadcast antennas are the tallest bogus structures on Globe. In 1987, a 72.0-kg physicist placed himself and 400 kg of equipment at the top of a 610-m-high antenna to perform gravity experiments. Past how much was the antenna compressed, if we consider it to be equivalent to a steel cylinder 0.150 m in radius?

Past how much does a 65.0-kg mountain climber stretch her 0.800-cm bore nylon rope when she hangs 35.0 m below a rock outcropping? (For nylon,

\[Y=1.35\,×\,{10}^{9}\text{Pa}\text{.)}\]

[reveal-answer q="fs-id1163709830348″]Show Solution[/reveal-reply]

[subconscious-reply a="fs-id1163709830348″]

9.0 cm

[/subconscious-answer]

When water freezes, its book increases by 9.05%. What force per unit area is water capable of exerting on a container when it freezes?

A farmer making grape juice fills a glass bottle to the skirt and caps it tightly. The juice expands more than the glass when information technology warms up, in such a fashion that the book increases past 0.2%. Calculate the force exerted by the juice per square centimeter if its bulk modulus is

\[1.8\,×\,1{0}^{9}N\,\text{/}\,{m}^{2},\]

bold the bottle does not interruption.

[reveal-answer q="fs-id1163709680492″]Show Solution[/reveal-answer]

[hidden-reply a="fs-id1163709680492″]

\[4.0\,×\,{10}^{2}\,{\text{N/cm}}^{2}\]

[/hidden-respond]

A disk between vertebrae in the spine is subjected to a shearing force of 600.0 Northward. Discover its shear deformation, using the shear modulus of

\[1.0\,×\,1{0}^{9}\,{\text{N/m}}^{2}.\]

The disk is equivalent to a solid cylinder 0.700 cm high and four.00 cm in bore.

A vertebra is subjected to a shearing strength of 500.0 N. Find the shear deformation, taking the vertebra to exist a cylinder 3.00 cm loftier and 4.00 cm in diameter. How does your effect compare with the upshot obtained in the preceding problem? Are spinal bug more common in disks than in vertebrae?

[reveal-answer q="fs-id1163713274089″]Testify Solution[/reveal-answer]

[hidden-answer a="fs-id1163713274089″]

\[0.149\,\text{μm}\]

[/hidden-reply]

Summate the forcefulness a piano tuner applies to stretch a steel piano wire by 8.00 mm, if the wire is originally one.35 thou long and its diameter is 0.850 mm.

A 20.0-m-tall hollow aluminum flagpole is equivalent in force to a solid cylinder four.00 cm in diameter. A strong wind bends the pole as much as a horizontal 900.0-N force on the top would do. How far to the side does the elevation of the pole flex?

[reveal-reply q="fs-id1163709733857″]Show Solution[/reveal-answer]

[hidden-respond a="fs-id1163709733857″]

0.57 mm

[/hidden-respond]

A copper wire of diameter 1.0 cm stretches 1.0% when information technology is used to lift a load upward with an acceleration of

\[2.0\,{\text{m/s}}^{2}.\]

What is the weight of the load?

As an oil well is drilled, each new section of drill piping supports its own weight and the weight of the piping and the drill chip below it. Calculate the stretch in a new 6.00-m-long steel pipe that supports a 100-kg drill bit and a 3.00-km length of pipe with a linear mass density of 20.0 kg/m. Treat the pipe as a solid cylinder with a 5.00-cm diameter.

[reveal-reply q="fs-id1163713356867″]Evidence Solution[/reveal-answer]

[hidden-respond a="fs-id1163713356867″]

eight.59 mm

[/subconscious-answer]

A large compatible cylindrical steel rod of density

\[\rho =7.8\,{\text{g/cm}}^{3}\]

is 2.0 m long and has a bore of 5.0 cm. The rod is fastened to a concrete flooring with its long axis vertical. What is the normal stress in the rod at the cross-department located at (a) 1.0 m from its lower end? (b) 1.5 m from the lower terminate?

A 90-kg mountain climber hangs from a nylon rope and stretches it past 25.0 cm. If the rope was originally 30.0 m long and its diameter is 1.0 cm, what is Immature'due south modulus for the nylon?

[reveal-answer q="fs-id1163709646172″]Show Solution[/reveal-respond]

[hidden-answer a="fs-id1163709646172″]

\[1.35\,×\,{10}^{9}\text{Pa}\]

[/hidden-respond]

A suspender rod of a suspension span is 25.0 one thousand long. If the rod is made of steel, what must its diameter exist so that it does non stretch more than 1.0 cm when a

\[2.5\,×\,{10}^{4}\text{-kg}\]

truck passes by it? Assume that the rod supports all of the weight of the truck.

A copper wire is 1.0 m long and its diameter is i.0 mm. If the wire hangs vertically, how much weight must be added to its costless end in order to stretch it 3.0 mm?

[reveal-reply q="fs-id1163713527113″]Testify Solution[/reveal-reply]

[hidden-answer a="fs-id1163713527113″]

259.0 N

[/hidden-answer]

A 100-N weight is attached to a gratuitous cease of a metallic wire that hangs from the ceiling. When a second 100-N weight is added to the wire, it stretches iii.0 mm. The diameter and the length of the wire are 1.0 mm and ii.0 m, respectively. What is Young'south modulus of the metal used to manufacture the wire?

The bulk modulus of a cloth is

\[1.0\,×\,{10}^{11}\,{\text{N/m}}^{2}.\]

What fractional change in volume does a piece of this material undergo when it is subjected to a bulk stress increment of

\[{10}^{7}\,{\text{N/m}}^{2}\text{?}\]

Assume that the force is practical uniformly over the surface.

[reveal-answer q="fs-id1163713247853″]Show Solution[/reveal-reply]

[subconscious-reply a="fs-id1163713247853″]

0.01%

[/hidden-answer]

Normal forces of magnitude

\[1.0\,×\,{10}^{6}\text{N}\]

are applied uniformly to a spherical surface enclosing a volume of a liquid. This causes the radius of the surface to decrease from l.000 cm to 49.995 cm. What is the bulk modulus of the liquid?

During a walk on a rope, a tightrope walker creates a tension of

\[3.94\,×\,1{0}^{3}N\]

in a wire that is stretched betwixt two supporting poles that are 15.0 m autonomously. The wire has a diameter of 0.fifty cm when it is non stretched. When the walker is on the wire in the eye betwixt the poles the wire makes an bending of

\[5.0\text{°}\]

below the horizontal. How much does this tension stretch the steel wire when the walker is this position?

[reveal-answer q="fs-id1163709766552″]Show Solution[/reveal-respond]

[hidden-reply a="fs-id1163709766552″]

1.44 cm

[/hidden-answer]

When using a pencil eraser, y'all exert a vertical force of 6.00 Northward at a distance of 2.00 cm from the hardwood-eraser joint. The pencil is 6.00 mm in bore and is held at an angle of

\[20.0\text{°}\]

to the horizontal. (a) By how much does the wood flex perpendicular to its length? (b) How much is it compressed lengthwise?

Normal forces are applied uniformly over the surface of a spherical volume of water whose radius is 20.0 cm. If the pressure on the surface is increased by 200 MPa, by how much does the radius of the sphere decrease?

[reveal-respond q="fs-id1163713365459″]Bear witness Solution[/reveal-answer]

[subconscious-answer a="fs-id1163713365459″]

0.63 cm

[/hidden-reply]

Glossary

bulk modulus
rubberband modulus for the bulk stress
bulk strain
(or volume strain) strain under the bulk stress, given every bit fractional modify in volume
majority stress
(or volume stress) stress caused by compressive forces, in all directions
compressibility
reciprocal of the majority modulus
compressive strain
strain that occurs when forces are contracting an object, causing its shortening
compressive stress
stress caused by compressive forces, but in one direction
elastic modulus
proportionality constant in linear relation between stress and strain, in SI pascals
normal pressure
pressure of one atmosphere, serves as a reference level for pressure level
pascal (Pa)
SI unit of stress, SI unit of measurement of pressure
pressure
force pressing in normal management on a surface per the surface area, the majority stress in fluids
shear modulus
rubberband modulus for shear stress
shear strain
strain caused by shear stress
shear stress
stress caused past shearing forces
strain
dimensionless quantity that gives the amount of deformation of an object or medium nether stress
stress
quantity that contains information about the magnitude of force causing deformation, defined as force per unit of measurement area
tensile strain
strain under tensile stress, given every bit fractional alter in length, which occurs when forces are stretching an object, causing its elongation
tensile stress
stress caused by tensile forces, only in one direction, which occurs when forces are stretching an object, causing its elongation
Young's modulus
elastic modulus for tensile or compressive stress

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Source: https://opentextbc.ca/universityphysicsv1openstax/chapter/12-3-stress-strain-and-elastic-modulus/

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