Chapter 4_Torsion

# Chapter 4_Torsion - CHAPTER 4 TORSION OF THIN-WALLED...

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1 CHAPTER 4 TORSION OF THIN-WALLED STRUCTURES Closed sections under torque Single-cell closed sections Thin-walled open sections Multi-cell closed sections Consider a slender body such as a wing or a fuselage subjected to torsional moment or torque. ) ( x α : angle of twist measured from a reference plane dx d x θ = ) ( : angle of twist per unit length Then one can show that ) ( ) ( ) ( x x K x T T = ( 1 ) where T : torsional moment acting over a cross-section K T : torsional rigidity Torsional rigidity K T is dependent on material property (shear modulus) and the geometry of the cross-section over which T is acting. For a section of single material, K T can be expressed as K GJ T = ( 2 ) where G : shear modulus J : torsion constant that depends entirely on the cross-sectional geometry Then T GJ = ( 3 ) Thin-walled closed sections are very effective in resisting torque. Accordingly, we will now focus our attention on them. In this section we will learn how to determine J for a given cross-section. For this purpose, it is adequate to work with cross-sections that are uniform along the axis and are under a constant torque.

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2 4.1 Closed Sections under Torque Consider a closed thin-walled section as shown in the sketch. s : coordinate along the mid contour t ( s ) : wall thickness τ () s : shear stress in the s direction Note: (a) The sidewalls are stress free. Accordingly, 0 = = xn nx . Also note xs = . (b) For closed thin walls, one can assume that is constant through thickness. 1) For convenience, introduce shear flow q defined as ) ( ) ( s t s q = Then it can be shown that q is constant along the contour . 2) For a closed section, it can be shown that A q T 2 = where A : area inside the mid-contour of the cross-section. Note : A is NOT the area of the cross-section. A S q t(s) z y s= 0 s n x nx xn xs
3 1) Shear flow q , defined as ) ( ) ( s t s q τ = , remains constant along the contour. That is, shear stress and thickness can be functions of s but their product is constant. To prove this, consider the equilibrium of a free body created by introducing (imaginary) cuts at x, x-dx, A and B as shown below. For the equilibrium of the forces in the x direction, 0 = dx t dx t A A B B 0 ) ( = dx t t A A B B 0 = A A B B t t A B q q = Cuts at A and B are arbitrarily located along the contour. Accordingly, q is constant along the contour.

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Chapter 4_Torsion - CHAPTER 4 TORSION OF THIN-WALLED...

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