Strain: 2.5 Statically Indeterminate Problems Illustrative Example 2.6 The Concrete Post in Fig. (A) Is Reinforced

Chapter II  Use Hooke’s law to express the deformations
STRAIN (strains) in the compatibility equations in terms

of forces (or stresses).
2.5 Statically Indeterminate Problems  Solve the equilibrium and compatibility
equations for the unknown forces.
If the equilibrium equations are su‰cient
to calculate all the forces (including support ILLUSTRATIVE EXAMPLE
reactions) that act on a body, these forces are 2.6 The concrete post in Fig. (a) is reinforced
said to be statically determinate.
axially with four symmetrically placed steel bars,
In statically determinate problems, the
each of cross-sectional area 900 mm2. Compute
number of unknown forces is always equal to the
number of independent equilibrium equations. the stress in each material when the 1000-kN
If the number of unknown forces exceeds axial load is applied. The moduli of elasticity are
the number of independent equilibrium 200 GPa for steel and 14 GPa for concrete.
equations, the problem is said to be statically
indeterminate.
Static indeterminacy does not imply that the

problem cannot be solved; it simply means that the
solution cannot be obtained from the equilibrium
equations alone.
A statically indeterminate problem

always has geometric restrictions imposed on its
deformation.
The mathematical expressions of these
restrictions, known as the compatibility
equations, provide us with the additional
equations needed to solve the problem (the
term compatibility refers to the geometric
compatibility between deformation and the
imposed constraints).
Because the source of the compatibility
equations is deformation, these equations
contain as unknowns either strains or
elongations.
We can, however, use Hooke’s law to
express the deformation measures in terms of Solution:
stresses or forces. The equations of equilibrium Equilibrium
and compatibility can then be solved for the The FBD in Fig. (b) was drawn by isolating the
unknown forces. portion of the post above section a-a, where Pco
is the force in concrete and Pst denotes the total
Procedure for Solving Statically force carried by the steel rods. For equilibrium,
Indeterminate Problems we must have
In summary, the solution of a statically ∑ 𝐹𝑦 = 0 +↑

indeterminate problem involves the following
steps: 𝑃𝑠𝑡 + 𝑃𝑐𝑜 − 1.0 × 106 = 0
 Draw the required free-body diagrams and
which, written in terms of stresses, becomes
derive the equations of equilibrium.
 Derive the compatibility equations. To 𝜎𝑠𝑡 𝐴𝑠𝑡 + 𝜎𝑐𝑜 𝐴𝑐𝑜 = 1.0 × 106 𝑒𝑞. (𝑎)
visualize the restrictions on deformation, it is
often helpful to draw a sketch that Equation (a) is the only independent equation
exaggerates the magnitudes of the of equilibrium that is available in this problem.
deformations. Because there are two unknown stresses, we
conclude that the problem is statically
indeterminate.
Compatibility
For the deformations to be compatible, the
changes in lengths of the steel rods and the
concrete must be equal; that is, 𝛿𝑠𝑡 = 𝛿𝑐𝑜.
Because the lengths of steel and concrete are
identical, the compatibility equation, written in
terms of strains, is
𝜖𝑠𝑡 = 𝜖𝑐𝑜 𝑒𝑞. (𝑏)
Hooke’s Law
For the deformations to be compatible, the
changes in
𝜎𝑠𝑡 𝜎𝑐𝑜
= 𝑒𝑞. (𝑐)
𝐸𝑠𝑡 𝐸𝑐𝑜
Equations (a) and (c) can now be solved for the

stresses. From Eq. (c) we obtain
𝐸𝑠𝑡 𝜎𝑐𝑜
𝜎𝑠𝑡 = = 14.286𝜎𝑐𝑜 𝑒𝑞. (𝑑)
𝐸𝑐𝑜
Substituting the cross-sectional areas

Solution:
𝐴𝑠𝑡 = 4(900 × 10−6 ) = 3.6 × 10−3 𝑚2
Equilibrium
𝐴𝑐𝑜 = 0.32 − 3.6 × 10−3 = 86.4 × 10−3 𝑚2 We assume that the rod deforms enough so that
the bearing plate makes contact with the tube,
and Eq. (d) into Eq. (a) yields as indicated in the FBD in Fig. (b). From this FBD
(14.286𝜎𝑐𝑜 )(3.6 × 10−3 ) + 𝜎𝑐𝑜 (3.6 × 10−3 ) = 1.0 × 106 we get
Solving for the stress in concrete, we get ∑ 𝐹𝑦 = 0 +↑
𝜎𝑐𝑜 = 7.55 × 106 𝑃𝑎 = 𝟕. 𝟐𝟓𝟓 𝑴𝑷𝒂 𝑃𝑐𝑢 + 𝑃𝑎𝑙 − 𝑃 = 0 𝑒𝑞. (𝑎)
From Eq. (d), the stress in steel is Because no other equations of equilibrium are
available, the forces Pcu and Pal are statically
𝜎𝑠𝑡 = 14.286(7.255) = 𝟏𝟎𝟑. 𝟔 𝑴𝑷𝒂
indeterminate.
Compatibility
2.7 Figure (a) shows a copper rod that is placed
Figure (c) shows the changes in the lengths of
in an aluminum tube. The rod is 0.005 in. longer
the two materials (the deformations have been
than the tube. Find the maximum safe load P
greatly exaggerated). We see that the
that can be applied to the bearing plate, using compatibility equation is
the following data:
𝛿𝑐𝑢 = 𝛿𝑎𝑙 + 0.005 𝑒𝑞. (𝑏)
Copper
Hooke’s Law
Area = 2 in.2
Substituting 𝛿 = 𝜎𝐿/𝐸 into Eq. (b), we get
E = 17x106 psi
Alow. Stress = 20 ksi 𝜎𝐿 𝜎𝐿
( ) = ( ) + 0.005
𝐸 𝑐𝑢 𝐸 𝑎𝑙
Aluminum 𝜎𝑐𝑢 (10.005) 𝜎𝑎𝑙 (10)
Area = 3 in.2 17 × 10 6
=
10 × 106
+ 0.005
E = 10x106 psi
Alow. Stress = 10 ksi which reduces to
𝜎𝑐𝑢 = 1.6992𝜎𝑎𝑙 + 8496 𝑒𝑞. (𝑐)
From Eq. (c) we find that if 𝜎𝑎𝑙 = 10000 𝑝𝑠𝑖, the Solution:
copper will be overstressed to 25500 psi. Equilibrium
Therefore, the allowable stress in the copper The free-body diagram of the bar, shown in Fig.
(20000 psi) is the limiting condition. The (b), contains four unknown forces. Since there
corresponding stress in the aluminum is found are only three independent equilibrium
from Eq. (c): equations, these forces are statically
20000 = 1.6992𝜎𝑎𝑙 + 8496 indeterminate. The equilibrium equation that
𝜎𝑎𝑙 = 6770 𝑝𝑠𝑖 does not involve the pin reactions at A is
∑ 𝑀𝐴 = 0 +↺
From Eq. (a), the safe load is
0.6𝑃𝑠𝑡 + 1.6𝑃𝑏𝑟 − 2.4(50 × 103 ) = 0 𝑒𝑞. (𝑎)
𝑃 = 𝑃𝑐𝑢 + 𝑃𝑎𝑙 = 𝜎𝑐𝑢 𝐴𝑐𝑢 + 𝜎𝑎𝑙 𝐴𝑎𝑙
𝑃 = 20000(2) + 6770(3) Compatibility
The displacement of the bar, consisting of a
𝑷 = 𝟔𝟎. 𝟑 𝒌𝒊𝒑𝒔 rigid-body rotation about A, is shown greatly
exaggerated in Fig. (c). From similar triangles, we
2.8 Figure (a) shows a rigid bar that is supported see that the elongations of the supporting rods
must satisfy the compatibility condition
by a pin at A and two rods, one made of steel
and the other of bronze. Neglecting the weight 𝛿𝑠𝑡 𝛿𝑏𝑟
= 𝑒𝑞. (𝑏)
of the bar, compute the stress in each rod 0.6 1.6
caused by the 50-kN load, using the following
data:
Hooke’s Law
When we substitute 𝛿 = 𝜎𝐿/𝐸 into Eq. (b), the
compatibility equation becomes
Steel
Area = 600 mm2 1 𝑃𝐿 1 𝑃𝐿
( ) = ( )
E = 200 GPa 0.6 𝐴𝐸 𝑠𝑡 1.6 𝐴𝐸 𝑏𝑟
1 𝑃𝑠𝑡 (1.0) 1 𝑃𝑏𝑟 (2.0)
Bronze =
0.6 (200)(600) 1.6 (83)(300)
Area = 300 mm2
E = 83 Gpa which simplifies to
𝑃𝑠𝑡 = 3.614𝑃𝑏𝑟 𝑒𝑞. (𝑐)
Note that we did not convert the areas from mm2

to m2 , and we omitted the factor 109 from the moduli of
elasticity. Since these conversion factors appear on both
sides of the equation, they would cancel out.
Solving Eqs. (a) and (c), we obtain

𝑃𝑠𝑡 = 115.08 × 103 𝑁
𝑃𝑏𝑟 = 31.84 × 103 𝑁
The stresses are

𝑃𝑠𝑡 115.08 × 103
𝜎𝑠𝑡 = =
𝐴𝑠𝑡 600 × 10−6
𝝈𝒔𝒕 = 𝟏𝟗𝟏. 𝟖 𝑴𝑷𝒂
𝑃𝑏𝑟 106.1 × 103

𝜎𝑏𝑟 = =
𝐴𝑏𝑟 300 × 10−6
𝝈𝒃𝒓 = 𝟏𝟎𝟔. 𝟏 𝑴𝑷𝒂

Strain: 2.5 Statically Indeterminate Problems Illustrative Example 2.6 The Concrete Post in Fig. (A) Is Reinforced

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Chapter II  Use Hooke’s law to express the deformations

STRAIN (strains) in the compatibility equations in terms

Static indeterminacy does not imply that the

A statically indeterminate problem

In summary, the solution of a statically ∑ 𝐹𝑦 = 0 +↑

Equations (a) and (c) can now be solved for the

Substituting the cross-sectional areas

Solving for the stress in concrete, we get ∑ 𝐹𝑦 = 0 +↑

𝜎𝑐𝑜 = 7.55 × 106 𝑃𝑎 = 𝟕. 𝟐𝟓𝟓 𝑴𝑷𝒂 𝑃𝑐𝑢 + 𝑃𝑎𝑙 − 𝑃 = 0 𝑒𝑞. (𝑎)

Note that we did not convert the areas from mm2

Solving Eqs. (a) and (c), we obtain

The stresses are

𝑃𝑏𝑟 106.1 × 103

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