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Mechanics of Materials – MAE 243 (Section 002) Spring 2008. Dr. Konstantinos A. Sierros. Problem 3.3-3

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Mechanics of Materials – MAE 243 (Section 002)

Spring 2008

Dr. Konstantinos A. Sierros


Problem 3.3-3

While removing a wheel to change a tire, a driver applies forces P = 25 lb at the ends of two of the arms of a lug wrench (see figure). The wrench is made of steel with shear modulus of elasticity G = 11.4 x (10^6) psi. Each arm of the wrench is 9.0 in. long and has a solid circular cross section of diameter d = 0.5 in.

(a) Determine the maximum shear stress in the arm that is turning the lug nut (arm A).

(b) Determine the angle of twist (in degrees) of this same arm.


Problem 5.5-4

A simply supported wood beam AB with span length L = 3.5 m carries a uniform load of intensity q = 6.4 kN/m (see figure).

Calculate the maximum bending stress max due to the load q if the

beam has a rectangular cross section with width b = 140 mm and height

h = 240 mm.


5.8: Shear stresses in beams of rectangular cross-section

  • Vertical and horizontal shear stresses.
  • We can isolate a small element mn of the beam. There are horizontal shear stresses acting between horizontal layers of the beam as well as vertical shear stresses acting on the cross-sections
  • At any point in the beam, these complementary shear stresses are equal in magnitude
  • τ = 0 where y = ±h/2

5.8: Shear formula

  • A formula for the shear stress τ in a rectangular beam can be derived

Where V is the shear force, I is the moment of inertia and b is the width of the beam. Q is the first moment of the cross-sectional area above the level at which the shear stress τ is being evaluated.

The shear formula can be used to determine the shear stress τ at any point in the cross-section of a rectangular beam


5.8:Distribution of shear stresses in a rectangular beam

  • We can determine the distribution of the shear forces in a beam of rectangular cross-section
  • The distribution of shear stresses over the height of the beam is parabolic. Note that τ = 0 where y = ±h/2
  • The maximum value of shear stress occurs at the neutral axis (y1 = 0) where the first moment Q has its maximum value.

Where A = bh is the cross-sectional area


5.8: Limitations

  • A common error is to apply the shear formula to cross-sectional shapes for
  • which it is not applicable. It is not applicable to triangular or semicircular
  • cross-sections
  • The formula should be applied when:
  • The edges of the cross section are parallel to the y-axis
  • The shear stress is uniform across the width of the cross section
  • The beam is prismatic