. The strength of materials; a text-book for engineers and architects. at the centre o an angle 0 called the angle of torsion. This angle of torsion may be regarded as a torsional deflection. If we imagine the shaft divided up into a number of very small equal slices it follows that since each slice is exactly like every other slice the angle of twist in each sHce wiU be equal, and if we regard each slice for convenience as of unit length we have 0 Angle of twist per unit length of shaft = I Now consider the slice contained between sections x x andY Y, Fig. 142, the thickness x being regarded
. The strength of materials; a text-book for engineers and architects. at the centre o an angle 0 called the angle of torsion. This angle of torsion may be regarded as a torsional deflection. If we imagine the shaft divided up into a number of very small equal slices it follows that since each slice is exactly like every other slice the angle of twist in each sHce wiU be equal, and if we regard each slice for convenience as of unit length we have 0 Angle of twist per unit length of shaft = I Now consider the slice contained between sections x x andY Y, Fig. 142, the thickness x being regarded as very small,and consider a square abed of side x at distance r from the 318 THE STRENGTH OF :\L\TERIALS centre. In our figure 6 c is appreciably curved, but that isonlv because we cannot draw the fio^ure clearly without makingX of appreciable size. The result of the twisting action is to make abed take upthe form a h c d, this being the typical form (cf. Fig. 1) indi-cating pure shear strain, and the initial shear strain /? is givenby the relation X Y -^ JU^ X -^. Y Fig. U2.—Torsion of Shafts. Shear stress — shear modulus x unital shear strain= y3G X G (1) Xow the angle hob — angle of twist in length x = angle of twist per unit length x x_6x_I and bb = ar c = radius x angle = —^ rOx .G xlrOGI . •. In (1) shear stress (2) TORSION AND TWISTING OF SHAFTS 319 This gives the important result that: The shear stress atany point in a shaft is proportional to the distance of that pointfrom the centre. The shear stress is therefore the same at all points on circlesconcentric with the shaft and the variation of shear stressis indicated by the triangle o eg, the stress at the extremefibre being s and that at any other radius r being equal to , _s r^ ^ R Now consider a very small element of area a at a point p in the section at distance r from the centre, the area being so small that the stress over it is constant. r OGThen from (2) stress on element = —v- •. Force on elemen
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