. The strength of materials; a text-book for engineers and architects. as the previous case, but we can find the shear stress at the simply as follows— In this case we have 7rD2 2D 6 =D ttD^ 2D 8 • Stt . •. 5„ . = m D2164 w .D= 1-33 m So that the mean shear stress along the is 1J times themean shear stress over the whole section. In this case it is interesting to note that the maximum shearstress along the is 1*45 m. (3) Pipe Section.—Let a thin pipe be of mean diameterD and thickness t (Fig. 226). Then a = —^ D y ^^ IT D2 ^ ~ 8 6=2^ ttD^ D . •. 5v A = W* X -:^ = 2 m ^ X 2* 476
. The strength of materials; a text-book for engineers and architects. as the previous case, but we can find the shear stress at the simply as follows— In this case we have 7rD2 2D 6 =D ttD^ 2D 8 • Stt . •. 5„ . = m D2164 w .D= 1-33 m So that the mean shear stress along the is 1J times themean shear stress over the whole section. In this case it is interesting to note that the maximum shearstress along the is 1*45 m. (3) Pipe Section.—Let a thin pipe be of mean diameterD and thickness t (Fig. 226). Then a = —^ D y ^^ IT D2 ^ ~ 8 6=2^ ttD^ D . •. 5v A = W* X -:^ = 2 m ^ X 2* 476 THE STRENGTH OF MATERIALS So that the mean stress shear along the is twice themean shear stress over the whole section. (4) I Section.—To calculate the proportion of the shearingforce carried by the flanges and web, respectively. Take a beam of I section of breadth h and height h, andlet the thickness of the flanges and the web be t and w,respectively. First consider a horizontal line p p in the flange at distanceX from the top edge (Fig. 226).. i^ Pi P-h7--f —?4^ uy T -?, B \K Ml Fig. 226. Then mean shear along p p = m . ^^g ; s,. = m .b X {h — x) 6 F 2 — 9 y^ 2 V ^ ^ ) (1) This depends on x^, so that the curve showing the variationof stress is a parabola. When X = t, i. e. at the junction of web and flange, St = 2^2 (^ ^ - ^) (2) Now consider a horizontal line Pj^ p^ in the web at distancex^ from the top. Then mean shear along pj p^ — - — DISTRIBUTION OF SHEAR STRESSES 477 In this case ay = first moment of area above p^p^ about _b t {h — t) ^ ^,_ ^^ \h /^ ^ , ^1 — ^ + w{x,-t)\^^-[t+^-^) _b t {h — t) w {Xi — t) (h — x^ — t) - 2 + 2~ also b = w in general expression for shear stress. n 9 1,2 m fbt{h — t) (% — t) (h — x^ — t) 2 k^ I w w _ m , mt (h — t) {b — w) .ox - 2 p (^ ^1 - ^-1 ) + -^2k^w « ^^^ The second term of this expression is constant for all valuesof x-^^ and the first term is the shear stress which would oc
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