Science, asked by ciayahh, 6 months ago

calculate the strain when a body of length 10cm elongates to 12cm on the application of an external load.​

Answers

Answered by Ameya09
0

Answer:

ne pascal

=

1.0

Pa

=

1.0

N

1.0

m

2

.

In the British system of units, the unit of stress is ‘psi,’ which stands for ‘pound per square inch’  

(

lb/in

2

)

.

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

1

psi

=

6895

Pa

and

1

Pa

=

1.450

×

10

4

psi

1

atm

=

1.013

×

10

5

Pa

=

14.7

psi.

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

The greater the stress, the greater the strain; however, the relation between strain and stress does not need to be linear. Only 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

stress

=

(elastic modulus)

×

strain.

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

We can also 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 small elastic modulus means that stress produces large strain and noticeable deformation. For example, a stress on a rubber band produces larger strain (deformation) than the same stress on a steel band of the same dimensions because the elastic modulus for rubber is two orders of magnitude smaller than the elastic modulus for steel.

The elastic modulus for tensile stress is called Young’s 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 conditions, such as varying temperature, and collected in engineering data tables for reference ((Figure)). These tables are valuable references for industry and for anyone involved in engineering or construction. In the next section, we discuss strain-stress relations beyond the linear limit represented by (Figure), in the full range of stress values up to a fracture point. In the remainder of this section, we study the linear limit expressed by (Figure).

Approximate Elastic Moduli for Selected Materials

Material Young’s modulus

×

10

10

Pa

Bulk modulus

×

10

10

Pa

Shear modulus

×

10

10

Pa

Aluminum 7.0 7.5 2.5

Bone (tension) 1.6 0.8 8.0

Bone (compression) 0.9  

Brass 9.0 6.0 3.5

Brick 1.5  

Concrete 2.0  

Copper 11.0 14.0 4.4

Crown glass 6.0 5.0 2.5

Granite 4.5 4.5 2.0

Hair (human) 1.0  

Hardwood 1.5  1.0

Iron 21.0 16.0 7.7

Lead 1.6 4.1 0.6

Marble 6.0 7.0 2.0

Nickel 21.0 17.0 7.8

Polystyrene 3.0  

Silk 6.0  

Spider thread 3.0  

Steel 20.0 16.0 7.5

Acetone  0.07  

Ethanol  0.09  

Glycerin  0.45  

Mercury  2.5  

Water  0.22  

Tensile or Compressive Stress, Strain, and Young’s Modulus

Tension or compression occurs when two antiparallel forces of equal magnitude act on an object along only one of its dimensions, in such a way that the object does not move. One way to envision such a situation is illustrated in (Figure). A rod segment is either stretched or squeezed by a pair of forces acting along its length and perpendicular to its cross-section. The net effect 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 L that it has under the action of the forces. This change in length  

Δ

L

=

L

L

0

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

L

0

)

or contraction (when L is smaller than the original length  

L

0

)

.

s

Explanation:

Similar questions