Problems_Solution - ∑ = ⎯→ ⎯ + x F-800N+N sin60+ V...

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Problem 5 -Solution. Part a. The bar is sectioned (Fig b) and the internal resultant loading consists only of an axial force for which P=800 N. Average stress. The average normal stress is determined by: () () KPa m m N A P 500 04 . 0 04 . 0 800 = = = σ No shear stress exists on the section since the shear force at the section is zero 0 = avg τ Part b- if the bar is sectioned along b-b, the free body diagram of the left segment is shown in the figure below. Here both the normal force N and the shear force V act on the sanctioned are
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Unformatted text preview: ∑ = ⎯→ ⎯ + x F-800N+N sin60+ V cos60=0 ∑ = y F V sin 60 – N cos 60 = 0 Solving for the 2 equations: N =692.8 N V= 400 N Average stresses: In this case, the sectioned area has a thickness and depth of 40 mm and 40 mm/sin 60= 46.19 mm respectively. Thus, the average normal stress is: ( ) ( ) KPa m m N A N 375 04619 . 04 . 8 . 692 = = = σ And the average shear stress is: ( ) ( ) KPa m m N A V avg 217 04619 . 04 . 400 = = = τ Problem 6....
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Problems_Solution - ∑ = ⎯→ ⎯ + x F-800N+N sin60+ V...

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