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

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.

Initial and 30^o Rotated Stress Element

The stresses are requested in both the +30^o and -30^o. For the +30^o, 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 30^o in the positive direction. Using the stress transformation equations, the stresses in the new x'-y' coordinate system are

Simplifying gives

     σ_x´ = 37.5 - 12.5(0.5) - 25(0.8660) = 9.600 ksi

     σ_y´ = 37.5 + 12.5(0.5) + 25(0.8660) = 65.40 ksi

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

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

Simplifying gives

     σ_x´ = 37.5 - 12.5(0.5) - 25(-0.8660) = 52.90 ksi

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

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

Stress vs. Rotation 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 180^o.