Introduction. Solid bodies are generally considered rigid and not deformable in shape and volume. But it is possible to deform them either temporarily or permanently by applying stress forces. There are three kinds of stress forces: tensions, compressions, and shears.
(a) A tensile stress is applied to a solid body when two forces, equal in magnitude but opposite in direction along the same line of action, are exerted on the opposite ends of the body, thereby increasing its length.How a body responds to any of the above stresses depends upon its composition, shape, temperature, etc. But in general the amount of deformation is directly proportional to the applied stress, provided that the force does not exceed a certain limit. For example, in the case of tension when a 20 lb weight is exerted on a certain wire it is stretched 0.1 in. Doubling the weight to 40 lb, then the wire will be stretched to 0.2 in, tripling the weight to 60 lb, then it will be stretched to 0.3 in, and so forth. When the weights are removed, the wire returns to its original length. This direct proportionality is called Hooke's law, and is written(b) A compression stress is applied to a solid body when two forces, equal in magnitude but toward each other along the same line of action, are exerted on the opposite ends of the body, thereby decreasing its length.
(c) A shearing stress is applied to a solid body when two forces, equal in magnitude but opposite in direction, are exerted on the body not along the same line of action, thereby changing its shape without changing its volume. To put it in simpler terms, tension stretches a body, compression shrinks a body, and shear twists a body upon which they act.
Solid bodies obey Hooke's law up to a limit, called the elastic limit, which refers to the maximum deformation a body will undergo without being permanently deformed. That is, when the elastic limit is exceeded, the body does not return to its original size or shape, and may be near a rupture of the body. Brittle materials like glass or cast iron break at or near their elastic limits. As an example, a glass rod with a cross section of 1 in2 follows Hooke's law as long as the tension or compression on it are less than about 10,000 lb; above that value it will break. Most metals may be deformed greatly beyond their elastic limits, a property that is known as ductility. Copper, for example, is a very ductile metal; a copper bar with a cross section of 1 in2 has a elastic limit of about 6000 lb and will not rupture until the stress has reached 33,000 lb. A graph of elongation of the bar versus tensional stress is a straight line, corresponding to Hooke's Law, until it the elastic limit is reached, then the curve becomes horizontal, corresponding to the shift from a reversible to an irreversible change in length. The curve ends at the point where the bar ruptures.
Elastic Moduli. Hooke's Law can be expressed for
each kind of stress in such a way that a single constant is need
to be known for a particular material in order to relate the stress
applied to a body of that material to the resulting deformation.
This constant is called elastic modulus and in general
is defined
Modulus =
stress/strain, (2)
where the strain is the deformation produced by the
stress.
Young's Modulus. It has been found experimentally
that the change in size or shape of a body under stress is proportional
to the ratio of the applied force to the cross-sectional area
of the body. Thus a thick rod will be stretched less than a thin
rod with a given force. Tensile stress is defined as the
ratio of the tensile force to cross-sectional area; that is,
tensile stress =
(tensile force)/(cross-sectional area) =
F/A.
Tensile strain is defined as ratio of the elongation of
the body to the original unstressed length; that is,
tensile strain =
(elongation)/(unstressed length) =
ΔL/L0.
In the case of a rod of initial length L0 and a
cross-sectional area A, a tensional or compressional force F
produces a change of ΔL in the length of the rod; this
relationship is expressed by
ΔL/L0 =
(1/Y)(F/A). (3)
The rod increases in length by ΔL if a tension stress is
applied, and decreases by that amount if the same force is applied
as a compression stress. The value of the constant Y,
called the Young's modulus, depends upon the material of
which the rod is composed; Y is customarily expressed in
lb/in2 in the British System of Units and it holds only below
the elastic limit. When the applied stress is tensile or compression
and the strain produced is the elongation or shorten of its length,
Young's Modulus is defined by
Y =
(tensile stress)/(tensile strain) =
(F/A)/(ΔL/L0). (4)
Bulk Modulus. When inward compressional forces
acts over the entire surface of a body, its volume decreases by
some amount ΔV from its original volume V0.
Only force components, that are perpendicular or normal to the body's
surface where they act, are effective in compressing it, since
the tangential or parallel components produce shear stress.
If the compressional force per unit area F/A is the
same over the entire surface of the body, then the relation between
the stress and the strain is given by
ΔV/V0 =
-(1/B)(F/A). (5)
The minus sign indicates that an increase in force leads to decrease
in volume. The quantity B is called the bulk modulus.
The bulk or volume stress is defined by the ratio
F/A, where F is force perpendicular or normal
to the surfaces, and it is commonly called pressure.
P = F/A. (6)
Here A is the area over which the force F is applied.
The SI unit of pressure is the pascal (Pa), which is equal to
1 nt/m2. Under uniform pressure the body will contract, and its
fractional change in volume , is called bulk or volume
strain. Thus the bulk modulus is defined by
B =
(volume stress)/(volume strain) =
-[(F/A)/(ΔV/V0)] =
-[F/(ΔV/V0)]. (7)
The negative sign is used to this defining equation so that the
bulk modulus will be positive quantity; an increase in pressure
results in a decrease in volume, that is, ΔV is negative
when P is positive.
Shear Modulus. Shear stress changes the shape of
the body upon which they act. Consider a block of thickness d
whose lower face is fixed in place and the upon whose upper face
the force F acts. A measure of relative distortion of
the block caused by the shear stress is the ratio
Δx/d between the displacement Δx
of the block's faces and the distance d between them.
The greater the area A of these faces, the
less they will be displaced by the shear force F. The
relationship between the deformation and the shear stress is expressed by
Δx/d = (1/S)(F/A). (8)
Here the applied forces are parallel or tangential to the faces
upon which they act and not perpendicular or normal as in the
case of tension and compression. The quantity S is called
shear modulus, and the value of S for any material
is usually is less than its Young's modulus. This means that
it is easier to slide the atoms of a solid body past one another
than it is to pull them apart or squeeze them together. The
shear stress is defined by the ratio F/A,
where the shear force F is acting over the area A.
Note that in contrast to tensile and compressional stresses,
a shear is produced by a force acting in the plane of the area
A; that is, the force is tangential to the area.
The shear strain is defined by the ratio Δx/d,
where d is the separation between the two areas over which the two
opposing tangential forces are acting. When the applied stress is shear and
strain produced is slippage, the shear modulus is defined by
S = (shear stress)/(shear strain) =
(F/A)/(Δx/d). (9)
| Material | Young's Modulus 1010 nt/m2 | Bulk Modulus 1010 nt/m2 | Shear Modulus 1010 nt/m2 |
|---|---|---|---|
| Aluminum | 7.0 | 7.5 | 2.7 |
| Gold | 8.0 | 16.5 | 2.8 |
| Brass | 9.1 | 6.1 | 3.6 |
| Iron | 9.1 | 10 | 7.0 |
| Copper | 12.0 | 14.0 | 4.0 |
| Steel | 20 | 17 | 8 |
| Nickel | 21 | 26 | 7.7 |
| Tungsten | 35 | 20 | 14 |
| Glass | 6.5-7.8 | 5.0-5.5 | 2.6-3.2 |
| Quartz | 5.6 | 2.7 | 2.6 |
| Lead | 1.6 | 0.77 | 0.56 |