Download Chapter 3-6 Lecture Notes

Chapter 3-5-3-6 Lecture Notes
The Chapter 3 Objectives are to:
1. Explain and apply stresses and strains in circular bar principles under uniform and non-uniform
conditions,
2. Explain and apply statically indeterminate principles,
3. Explain and apply stresses on inclined plane principles,
4. Explain and apply transmission of power principles,
5. Demonstrate positive teamwork, and
6. Demonstrate problem solving ability.
Chapter 3-5 Lecture Notes
Stresses in Pure Shear
We build on the shear stress and strain principles developed earlier. Recall the
sign convention from Section 1-7 is slightly complex.
First, there is the notion of positive and negative faces. If the face of an
element has an outward normal in the direction of the positive direction
of the face axes, the face is positive. Conversely, if the outward normal
is in the direction of the negative direction of the face axes, the face is
negative.
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Second, shear stress acting on a positive face is positive if it’s acting in the positive direction of
the axis and negative if it acts in the negative direction of the axis.
Third, shear stress acting on a negative face is positive if it’s acting the direction of a negative
axis and negative if it’s acting in the positive direction of the axis.
Fourth, shear strain (angular deformation) is positive when the resulting strain causes the angle
between two positive faces (or two negative faces) to reduce. Shear strain (angular deformation)
is negative when the resulting strain causes the angle between two positive faces (or two negative
faces) to increase.
It may be helpful to think of the shear stresses as two opposing equal couples (moments) where
counter clockwise moments are positive and clockwise moments are negative.
The stresses on inclined planes are more easily understood with a
trigonometric analysis of a stress element. We find the lengths of a
rectangular stress element in terms of one side, a. Using the
trigonometric relationships shown, we can find the lengths of sides b
and c in terms of the length of side a and the angle θ. The area of
each face is the length of each side times the width.
In the force balance to the left:
V = τθA0secθ
N = σθA0secθ
V1 = τθA0
V2 = τθA0tanθ
Using the above and left relationships, we can express the stresses
and forces on the stress element shown in Figure 3-25.
After some algebraic efforts on the above expressions, we get:
σθ = τsin2θ
and
τθ = τcos2θ
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Figure 3-26 reveals the behavior of the two stresses and Figure 3-37 illustrates the planes for minimum
and maximum stresses. One of the important takeaways is that torsion causes both shear and normal
stresses. The material will fail in the plane with the lowest modulus of elasticity. It explains why chalk
fails in tension on a 45 degree plane as illustrated below in Figure 3-28. Table 1 summarizes the failure
modes.
τ>G
σ<E
Shear
Failure Mode
τ<G
No Failure
Table 1 Failure Mode Summary
σ>E
Shear or Normal Failure Mode
(which mode depends on relative magnitudes
of the stresses and moduli of elasticity)
Normal Failure Mode
Strains in Pure Shear
Examining Figure 3-29a, we see that the effect of the shear stress is to deform the stress element. The
sides don’t elongate, but the right angles decrease by the angle, γ, as described earlier. Consequently, the
diagonal dimension increases and the cross-diagonal dimension decreases. Examining Figure 3-29b, we
see that the effect of the normal stresses is to lengthen the dimension that is in tension and shorten the
dimension that is in compression. Because of the Poisson effect, additional strains are caused by both the
tension and compression stresses.
The strain in the -135 deg direction caused by σmax = τ is -ντ/E and the strain in the -135 deg
direction caused by σmin =-τ is -τ/E so the total strain in the -135 deg direction is the sum of the
strains: εmax = -τ(1+ν)/E in the 135 deg direction and the dimension is shortened.
The same discussion holds for strain in the 45 deg direction except that the strains are positive so
we get: εmax = τ(1+ν)/E in the 45 deg direction and the dimension is lengthened.
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Chapter 3-6 Lecture Notes
Relationship Between the Modulii of Elasticity and Shear
Examining Figure 3-32, we find the relationships that lead to two important principles:
εmax = γ/2 and G = E(1+ν)/2
First, from Figure 3-32a, the un-deformed length of the diagonal db is Ldb = 21/2 h and the elongation, δ,
caused by the stresses, τ, is εLdb, so the length, Ldb, caused by the stresses is Ldb = 21/2h(1+ε). Next, from
Figure 3-3b, from the angle dab perspective and the law of cosines, we get:
Ldb2 = h2 +h 2 - 2h2cos(π/2 + γ) and from Figure 3-32a result
Ldb2 = (21/2(1+ν)2
After plowing through the algebraic manipulations, we get
1 + 2εmax + εmax2 = 1 + sinγ and for small ε and small γ, εmax2 ≈ 0 and sinγ = γ
So the final result is εmax = γ/2.
Now to develop the final result, we have
εmax = γ/2 = τ/(G2) from the shear stress-strain relationship and
εmax = τ(1+ν) from the normal stress-strain relationship and the two strains must be equal.
Therefore, we have the final result
G = E/(2(1+ν))
See Ex 3-6 to illustrate the principles.
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