# chapter 10 answers - Problem 10.1 A prismatic bar with...

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Problem 10.1 A prismatic bar with cross-sectional area A =0 . 1m 2 is loaded at the ends in two ways: (a) by 100-Pa uniform normal tractions; (b) by 10-N axial forces acting at the centroid of the bar’s cross section. What are the normal and shear stress distributions at the plane P in the two cases? Diagram: Solution: (a) Because the applied loads are in the axial direction and the plane is normal to the axis of the bar, there is no shear stress. In the Frst case, the applied load in a uniform normal traction, so the normal stress MUST be: ANS: σ = 100 Pa As stated above, since the applied load is in the axial direction and the plane is normal to the axis of the bar: ANS: τ (b) In the second case, the normal stress distribution is: σ = P/A = (10 N) / (0 . 2 ) ANS: σ = 100 Pa As stated above, since the applied load is in the axial direction: ANS: τ Problem 10.2 A prismatic bar with cross-sectional area A =4in 2 is subjected to tensile axial loads P .I t consists of a material that will safely support a tensile normal stress of 60 ksi . Based on this criterion, what is the largest safe value of P ? Diagram: Solution: Using the deFnition of normal stress: P ALLOW = σ ALLOW A = (60 , 000 lb / in 2 )(4 in 2 ) ANS: P ALLOW = 240 , 000 lb = 240 kip

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Problem 10.3 A prismatic bar has a solid circular cross section with 20-mm diameter. It consists of a ma- terial that will safely support a tensile normal stress of 300 MPa . Based on this criterion, what is the largest tensile load P to which the bar can be subjected? Free Body Diagram: Solution: The cross-sectional area of the bar is: A = πr 2 = π (0 . 01 m) 2 =3 . 142 × 10 4 m 2 For a maximum normal stress of 300 × 10 6 Pa : 300 × 10 6 N / m 2 = P MAX 3 . 142 × 10 4 m 2 ANS: P MAX =94 . 2kN Problem 10.4 The cross-sectional area of bar AB is 0 . 5in 2 . If the force F = 3 kips , what is the normal stress on a plane perpendicular to the axis of bar AB ? Free Body Diagrams: Solution: Summing moments about point C : Σ M C = 0 = (3000 lb)(6 ft) F AB (4 ft) F AB = 4500 lb (T) The normal stress in member AB is: σ AB = F AB /A AB = (4500 lb) / (0 . 2 ) ANS: σ AB = 9000 psi = 9 ksi
Problem 10.5 Bar AB of the frame in Problem 10.4 consists of material that will safely support a tensile nor- mal stress of 20 ksi . If you want to design the frame to support forces F as large as 8 kip , what is the minimum required cross-sectional area of bar AB ? Free Body Diagram: Solution: The maximum safe load which can be supported by member AB is: [1] ( F AB ) MAX = σ ALLOW A AB = (20 , 000 lb / in 2 )( A AB ) Summing moments about point C : Σ M C =0= F (6 ft) ( F AB ) MAX (4 ft) = (8000 lb)(6 ft) ( F AB ) MAX (4 ft) ( F AB ) MAX =12 , 000 lb Using Equation [1] to Fnd the cross-sectional area of member AB : 12 , 000 lb = (20 , 000 lb / in 2 )( A AB ) ANS: A AB =0 . 6in 2 Problem 10.6 The mass of the suspended box is 800 kg . The mass of the crane’s arm (not including the hydraulic actuator BC )is 200 kg , and its center of mass is2mtothe right of A . The cross-sectional area of the upper part of the hydraulic actuator is 0 . 004 m 2 . What is the normal stress on a plane perpendicular to the axis of the upper part of the actuator?

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## This note was uploaded on 04/03/2008 for the course MAE 101 taught by Professor Orient during the Winter '08 term at UCLA.

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chapter 10 answers - Problem 10.1 A prismatic bar with...

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