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Unformatted text preview: the plane of the joint are P.cos( θ ) and P.sin( θ ) respectively. Hence, ' P.cos ' A q s = ' P.sin ' A q t = Substituting the value of A’, we get ' cos P. ' 2 A q s = MPa allowable 20 = s Q MPa A 20 ' cos P. 2 = ∴ q ∴ P.cos 2 ( θ ) = 20 x1300……….(1) And, ' sin . cos P. ' A q q t = MPa allowable 10 = t Q MPa A 20 ' sin . cos P. = ∴ q q ∴ P.sin( θ ) .Cos( θ ) = 10 x1300……….(2) Dividing (2) by (1), we have, Tan( θ ) = 0.5 Which gives θ = 26.56 0 Now, We put back the value of θ in (1) to get ) 56 . 26 ( cos 1300 20 2 × = ∴ P ∴ P = 32500 N ∴ P = 32.5 KN Now, we check whether the stresses induced in the bar are within safe limits 1300 32500 = s ∴σ = 25 Mpa < 56 Mpa, hence, safe 1300 2 32500 × = t ∴τ = 12.5 Map < 28 Mpa, hence, safe. Final Answer The maximum allowable axial load that the bar can carry is 32.5 KN and the corresponding value of the angle is 26.56 degrees....
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This note was uploaded on 11/22/2011 for the course EML 6653 taught by Professor Law during the Fall '09 term at University of South Florida  Tampa.
 Fall '09
 Law
 Shear, Stress

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