3. Figure shows the singly symmetrical cross-section of a thin-walled open section beam of constant wall thickness t, which has a narrow longitudinal slit at the corner 15. Calculate and sketch the distribution of shear flow due to a vertical shear force S acting through the shear centre, S and note the principal values .Also determine the position of shear center. (10)4. Draw the shear flow distribution for the cross-section shown in figure. 1. Determine the Moment of inertia for the singly-symmetrical cross-section.

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2021-09-29T04:37:20-0400 The answer to your question is provided in the image:solution:Write the expression for shear flow:-q_(s)=(sy)/(I_(xx))int_(0)^(s)tydsfrom the figure find the moment of inertia{:[I_(xx)=2[int_(0)^(b)t((2h)/(b)s)^(2)+int_(0)^(a)t((h)/(2a)s)^(2)ds]],[I_(xx)=(th^(2)(b+a))/(6)]:}on surface (1-2) :{:[q_(1-2)=-(s_(y))/(I_(ax))int_(0)^(s_(1))t(-h//2s_(1))ds_(1)],[=(s_(yth))/((s_(1)^(2))/(2))],[=(s_(x)xth)/(2b((t^(2)(b+a))/(6)(s_(1)^(2))/(2))]:}=(3sy)/(2bh(b+a))s1^(2)^(2)q_(2)=(3sy)/(2bh(b+a))(b)^(2)v_(2)=(3sy(b))/(2h(b+a))q_(2)=(3bsy)/(2b(b+a))q_(23)=-(sy)/(I_(xx))int_(0)^(s_(1))t(-b//2a(a-s_(2))ds_(2)+q_(2) ... See the full answer