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1) Air at 15.6°C and 1 atm abs flows across a bank tube consisting of total 10 tubes of 30-mm outside diameter in a staggered
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Eq.6.12 gives (3) ...... 47.01 4(1.92) 0.87(415) 4 s φ φ τ φ σ == = bd Ld Eq.6.13 gives d f Lo V M L d )( 1 when s σ = +≤ To find M1, we need xu mm 157.46 (300) 0.36(20) (942) 0.87(415) 0.36 0.87 = = = bf Af x ck sty u xu,max = 0.48(400) = 192 mm Since xu < xu,max ; M1 = 0.87 fy Ast (d – 0.42 xu) = 0.87(415) (942) {400 – 0.42(157.46)} = 113.55 kNm ….. (4) From Eqs. (3) and (4), we have (5) ...... )( 01.47 1 L

At the point of inflection Lo is the maximum of d (= 400 mm) or φ = mm). 240 ( 12 With Lo = 400 mm, we can write from Eq.(5) 400 } )10(200 )10(55.113 { 01.47 3 6 φ ≤ + which gives φ ≤ 20.58 mm. Thus, use of φ = mm 20 satisfies the bond requirement.

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