Consider a test in which loading has progressed along the elastic portion of the force–displacement curve until F s ϭ F (Fig. 9.7a). For the
next small increment of force, ⌬F, the displacement increases by ⌬u and the increment of work
done by the force on the specimen spring is:
(9.3)
⌬W is called the virtual work and ⌬u is called the
virtual displacement. We use (9.3) to calculate the
virtual work associated with the specimen and
testing machine in the post-peak stress regime. In
that regime, suppose the actuator applies forces
F s ϭ P and F m ϭϪP so the system is in equilibrium.
Imagine replacing the actuator with a perfectly
rigid bar that transmits this force and couples the
specimen and machine springs so they must displace by the same amount (Fig. 9.6c). If a downward virtual displacement, ⌬u, is imposed on the
upper horizontal bar (Fig. 9.6d), the forces within
the springs change as:
(9.4)
post-peak stress
Using (9.3) the virtual work for the machine and
specimen are:
(9.5)
Note that ⌬u is small (less than unity), so (⌬u)
2 is
very small. Thus, the leading terms in (9.5) are
larger than the second terms, so ⌬W m is a negative
quantity and ⌬W s is positive. That is, work is done
by the machine on the specimen.
The next step in the mechanical analysis is to
determine if the system is stable by asking: what
are the conditions under which the system will
spontaneously evolve to a different state? Such a
spontaneous response could represent catastrophic failure of the specimen. To address this
question we apply the principle of virtual work: for
any possible virtual displacement of an elastic
body in equilibrium, the total virtual work done
by the internal forces, the body forces, and the
surface forces must vanish (Venkatraman and
Patel, 1970, p. 120). Possible virtual displacements
⌬W s ϭ P⌬u Ϫ
1
2 | fЈ(u s )|(⌬u) 2
⌬W m ϭ ϪP⌬u ϩ
1
2 C m (⌬u) 2
⌬F s ϭ f Ј(u s )⌬u ϭ Ϫ| fЈ(u s ) |⌬u,
⌬F m ϭ C m ⌬u
⌬W ϭ F⌬u ϩ
1
2 ⌬F⌬u
are those that are not prevented by the boundary
conditions, such as the fixed lower bar. Applying
this principle to the system illustrated in Fig. 9.7
we add the two equations in (9.5) and set the
results equal to zero:
(9.6)
The system is in static equilibrium only if the
machine stiffness, C m , and the magnitude of the
specimen stiffness, |f Ј(u s )|, are identical.
To understand the implications of (9.6) we
examine the two cases where equilibrium is not
attained. First, consider the machine to be stiffer
than the specimen, so C m Ͼ |f Ј(u s )| and area 2 is
greater than area 3 (Fig. 9.7a, b). For this condition, the work that the machine spring can do on
the rest of the system is not sufficient to meet the
requirements of the work done on the specimen
spring. To achieve the downward displacement,
⌬u, work must be done by some external force, for
example the hydraulic pump supplies more fluid
to the actuator, thereby increasing the applied
force. These testing conditions are described as
stable because the system will remain in static
equilibrium unless some external action is taken.
If this external action is carefully controlled, the
test can proceed under stable conditions and the
complete stress–strain curve will be captured.
Now suppose the machine is softer than the
specimen spring, so C m Ͻ |f Ј(u s )| and area 2 is less
than area 4 (Fig. 9.7a, c). The machine spring can
do more work than the specimen spring requires;
so if the distance between the upper bar and the
specimen spring is not changed, the machine will
drive the top of the specimen downward at an
accelerating rate. For an actual testing machine,
this could lead to a catastrophic result, and the
stress–strain information for the remainder of the
test would be lost. These testing conditions are
described as unstable because the system selfdestructs unless some external action is taken, for
example the hydraulic pump drains fluid out of
the actuator, thus decreasing the applied force.
This must happen at a sufficient rate for the system
to remain in equilibrium. Another solution is to
control the test using the displacement of the
specimen, u s , instead of the force from the actuator. Modern testing machines use servo-control
⌬W m ϩ ⌬W s ϭ
1
2 ΄C m Ϫ | fЈ(u s )|΅(⌬u) 2 ϭ 0
9.2 STRENGTH OF LABORATORY SAMPLES
341
next small increment of force, ⌬F, the displacement increases by ⌬u and the increment of work
done by the force on the specimen spring is:
(9.3)
⌬W is called the virtual work and ⌬u is called the
virtual displacement. We use (9.3) to calculate the
virtual work associated with the specimen and
testing machine in the post-peak stress regime. In
that regime, suppose the actuator applies forces
F s ϭ P and F m ϭϪP so the system is in equilibrium.
Imagine replacing the actuator with a perfectly
rigid bar that transmits this force and couples the
specimen and machine springs so they must displace by the same amount (Fig. 9.6c). If a downward virtual displacement, ⌬u, is imposed on the
upper horizontal bar (Fig. 9.6d), the forces within
the springs change as:
(9.4)
post-peak stress
Using (9.3) the virtual work for the machine and
specimen are:
(9.5)
Note that ⌬u is small (less than unity), so (⌬u)
2 is
very small. Thus, the leading terms in (9.5) are
larger than the second terms, so ⌬W m is a negative
quantity and ⌬W s is positive. That is, work is done
by the machine on the specimen.
The next step in the mechanical analysis is to
determine if the system is stable by asking: what
are the conditions under which the system will
spontaneously evolve to a different state? Such a
spontaneous response could represent catastrophic failure of the specimen. To address this
question we apply the principle of virtual work: for
any possible virtual displacement of an elastic
body in equilibrium, the total virtual work done
by the internal forces, the body forces, and the
surface forces must vanish (Venkatraman and
Patel, 1970, p. 120). Possible virtual displacements
⌬W s ϭ P⌬u Ϫ
1
2 | fЈ(u s )|(⌬u) 2
⌬W m ϭ ϪP⌬u ϩ
1
2 C m (⌬u) 2
⌬F s ϭ f Ј(u s )⌬u ϭ Ϫ| fЈ(u s ) |⌬u,
⌬F m ϭ C m ⌬u
⌬W ϭ F⌬u ϩ
1
2 ⌬F⌬u
are those that are not prevented by the boundary
conditions, such as the fixed lower bar. Applying
this principle to the system illustrated in Fig. 9.7
we add the two equations in (9.5) and set the
results equal to zero:
(9.6)
The system is in static equilibrium only if the
machine stiffness, C m , and the magnitude of the
specimen stiffness, |f Ј(u s )|, are identical.
To understand the implications of (9.6) we
examine the two cases where equilibrium is not
attained. First, consider the machine to be stiffer
than the specimen, so C m Ͼ |f Ј(u s )| and area 2 is
greater than area 3 (Fig. 9.7a, b). For this condition, the work that the machine spring can do on
the rest of the system is not sufficient to meet the
requirements of the work done on the specimen
spring. To achieve the downward displacement,
⌬u, work must be done by some external force, for
example the hydraulic pump supplies more fluid
to the actuator, thereby increasing the applied
force. These testing conditions are described as
stable because the system will remain in static
equilibrium unless some external action is taken.
If this external action is carefully controlled, the
test can proceed under stable conditions and the
complete stress–strain curve will be captured.
Now suppose the machine is softer than the
specimen spring, so C m Ͻ |f Ј(u s )| and area 2 is less
than area 4 (Fig. 9.7a, c). The machine spring can
do more work than the specimen spring requires;
so if the distance between the upper bar and the
specimen spring is not changed, the machine will
drive the top of the specimen downward at an
accelerating rate. For an actual testing machine,
this could lead to a catastrophic result, and the
stress–strain information for the remainder of the
test would be lost. These testing conditions are
described as unstable because the system selfdestructs unless some external action is taken, for
example the hydraulic pump drains fluid out of
the actuator, thus decreasing the applied force.
This must happen at a sufficient rate for the system
to remain in equilibrium. Another solution is to
control the test using the displacement of the
specimen, u s , instead of the force from the actuator. Modern testing machines use servo-control
⌬W m ϩ ⌬W s ϭ
1
2 ΄C m Ϫ | fЈ(u s )|΅(⌬u) 2 ϭ 0
9.2 STRENGTH OF LABORATORY SAMPLES
341