Stress Element from Plane Fuselage
(Fibers oriented in 0o, 90o, 30o, and -30o)


A plane fuselage undergoes both a pressure and twist load. This causes a tension stress in both the longitudinal and circumferential directions and a twisting load or a shear stress.

If a stress element is cut from the fuselage, the induced stresses can be shown in a common x-y coordinate system. This longitudinal, or x-direction would be 25 ksi. The circumferential or y-direction would be 50 ksi. The shear stress would be 25 ksi. These stresses are shown in the diagram at the left. The shear arrows in the diagram are in the negative direction, and thus the shear is a negative shear stress.

  Stresses in the +30o Fiber Direction


Initial and 30o Rotated Stress Element


The stresses are requested in both the +30o and -30o. For the +30o, the initial stress element is shown at the left with the positive directions and thus the shear stress is negative.

The stress element needs to be rotated 30o in the positive direction. Using the stress transformation equations, the stresses in the new x'-y' coordinate system are



Simplifying gives

     σ = 37.5 - 12.5(0.5) - 25(0.8660) = 9.600 ksi

     σ = 37.5 + 12.5(0.5) + 25(0.8660) = 65.40 ksi

     τx´y´ = 12.5(0.8660) - 25(0.5) = -1.675 ksi



  Stresses in the -30o Fiber Direction

Initial and -30o Rotated Stress Element

The method to find the stresses in the -30o is the same as for the +30o. Starting with the basic stress transformation equations, gives


Simplifying gives

     σ = 37.5 - 12.5(0.5) - 25(-0.8660) = 52.90 ksi

     σ = 37.5 + 12.5(0.5) + 25(-0.8660) = 22.10 ksi

     τx´y´ = 12.5(-0.8660) - 25(0.5) = -23.33 ksi

Stress vs. Rotation Angle

  Any Angle


It is interesting to plot the changing stresses as a function of angle. As expected, the stresses vary in a periodic cycle. Due to the double angle trigonometry terms in all three equations, the period is 180o.

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