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Sideways stress: shear modulus

[link] illustrates what is meant by a sideways stress or a shearing force . Here the deformation is called Δ x size 12{Δx} {} and it is perpendicular to L 0 size 12{L rSub { size 8{0} } } {} , rather than parallel as with tension and compression. Shear deformation behaves similarly to tension and compression and can be described with similar equations. The expression for shear deformation    is

Δ x = 1 S F A L 0 , size 12{Δx= { {1} over {S} } { {F} over {A} } L rSub { size 8{0} } } {}

where S size 12{F} {} is the shear modulus (see [link] ) and F size 12{F} {} is the force applied perpendicular to L 0 size 12{L rSub { size 8{0} } } {} and parallel to the cross-sectional area A size 12{A} {} . Again, to keep the object from accelerating, there are actually two equal and opposite forces F size 12{F} {} applied across opposite faces, as illustrated in [link] . The equation is logical—for example, it is easier to bend a long thin pencil (small A size 12{A} {} ) than a short thick one, and both are more easily bent than similar steel rods (large S size 12{S} {} ).

Shear deformation

Δ x = 1 S F A L 0 , size 12{Δx= { {1} over {S} } { {F} over {A} } L rSub { size 8{0} } } {}

where S size 12{S} {} is the shear modulus and F size 12{F} {} is the force applied perpendicular to L 0 size 12{L rSub { size 8{0} } } {} and parallel to the cross-sectional area A size 12{A} {} .

Bookcase sheared by a force applied at the bottom right toward the bottom left, and at the top left toward the top right.
Shearing forces are applied perpendicular to the length L 0 and parallel to the area A , producing a deformation Δx . Vertical forces are not shown, but it should be kept in mind that in addition to the two shearing forces, F size 12{F} {} , there must be supporting forces to keep the object from rotating. The distorting effects of these supporting forces are ignored in this treatment. The weight of the object also is not shown, since it is usually negligible compared with forces large enough to cause significant deformations.

Examination of the shear moduli in [link] reveals some telling patterns. For example, shear moduli are less than Young’s moduli for most materials. Bone is a remarkable exception. Its shear modulus is not only greater than its Young’s modulus, but it is as large as that of steel. This is why bones are so rigid.

The spinal column (consisting of 26 vertebral segments separated by discs) provides the main support for the head and upper part of the body. The spinal column has normal curvature for stability, but this curvature can be increased, leading to increased shearing forces on the lower vertebrae. Discs are better at withstanding compressional forces than shear forces. Because the spine is not vertical, the weight of the upper body exerts some of both. Pregnant women and people that are overweight (with large abdomens) need to move their shoulders back to maintain balance, thereby increasing the curvature in their spine and so increasing the shear component of the stress. An increased angle due to more curvature increases the shear forces along the plane. These higher shear forces increase the risk of back injury through ruptured discs. The lumbosacral disc (the wedge shaped disc below the last vertebrae) is particularly at risk because of its location.

The shear moduli for concrete and brick are very small; they are too highly variable to be listed. Concrete used in buildings can withstand compression, as in pillars and arches, but is very poor against shear, as might be encountered in heavily loaded floors or during earthquakes. Modern structures were made possible by the use of steel and steel-reinforced concrete. Almost by definition, liquids and gases have shear moduli near zero, because they flow in response to shearing forces.

Questions & Answers

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Practice Key Terms 6

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