. The strength of materials; a text-book for engineers and architects. v//f///////// / Fig. 228. to consider the shear stress in the concrete itself, w^hich willbe constant from the reinforcement to the and willthen vary towards the top as indicated in Fig. 228. The difference of pulls T — T is distributed over a hori-zontal rectangular area one side of which is x and the otherside of which is b, the breadth of the beam. •. If 5 is the shear stress we shall have s xb X X = T ~-Tbut T - T == ^^ (from (1))(B - B) s = X . a .b (4) II 482 THE STRENGTH OF IVIATERIALS But as before we may put —
. The strength of materials; a text-book for engineers and architects. v//f///////// / Fig. 228. to consider the shear stress in the concrete itself, w^hich willbe constant from the reinforcement to the and willthen vary towards the top as indicated in Fig. 228. The difference of pulls T — T is distributed over a hori-zontal rectangular area one side of which is x and the otherside of which is b, the breadth of the beam. •. If 5 is the shear stress we shall have s xb X X = T ~-Tbut T - T == ^^ (from (1))(B - B) s = X . a .b (4) II 482 THE STRENGTH OF IVIATERIALS But as before we may put —-— = shearing force S X S a 0and for reactangular beams as above S ,9 = b d n (5)(6) GRAPHICAL TREATMENT FOR FINDING DISTRIBUTIONOF SHEAR STRESS ON A CROSS SECTION Consider the section, composed of joists and plates, shownin Fig. 229. The first step is to mass the section up about. p \Q ft,-—; 1^ ^L^ .- ir i _j /I ?1 1 A IS B \ -^ c- 111 JC «- ^ 11 1 ^ ^ r Fig. 229. a vertical centre line : this is done by drawing horizontal linesacross the joists, and adding on each side of the centre joist thecorresponding horizontal ordinate of the outside joists. Thisgives the section shown in the figure { d = ah -]-hc -{- c d).Consider any line p p. We have shown that the mean shear stress along p p = <s, m Ic^h Now a . y = first moment of area above p p about neutralaxis X X. Draw the first moment curve of the section above x xabout the line x x, as explained on p. 176. As the section issymmetrical about a vertical axis, we need draw the curvefor one half of the area only, x q c being such curve. DISTRIBUTION OF SHEAR STRESSES 483 Then a x y = 2 area J x Q n x h, .*. Mean shear along p Pi = ^„-. • = Now find the sum-curve j r s of the first moment curve,taking the polar distance p = , . Then n e. x ^^ = area of first moment curve above p p N R X A^2 h = ar
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