ME 323 Examination #2 April 11, 2018

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1 ME 2 Eamination #2 April, 2 PROBLEM NO. 25 points ma. A thin-walled pressure vessel is fabricated b welding together two, open-ended stainless-steel vessels along a 6 weld line. The welded vessel has an internal radius of R = m and a thickness t =. m. The gas pressure inside the vessel is p. a) Determine the aial and hoop stresses, σ! and σ!, in terms of the internal pressure p. Draw the state of stress at point A and its Mohr s circle. b) Use the Mohr s circle to determine the components of stress, σ!, σ! and τ!", in terms of the internal pressure p, along the weld line. c) Use the Mohr s circle to determine the maimum gas pressure p, such that BOTH of the following conditions are met: The maimum shear stress of the vessel does not eceed the ield strength of stainless-steel, σ! = MPa, AND: The maimum normal stress along the weld line does not eceed the welding strength in tension, σ wt = MPa.

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3 ME 2 Eamination #2 April, 2 PROBLEM NO points ma. The beam AB is supported b a pin at A, a roller at B and a post having a 2 cm diameter at C. The post and the beam have a Young s modulus of E = 2 GPa, and the second area moment of the beam is I = 2 6 mm 4. In our analsis, neglect the contribution of shear strain to the total strain energ. a) Using Castigliano s second theorem: i. Determine the reactions on the beam at A, B and C. ii. Determine the slope of the beam deflection at B. b) Under what assumptions can ou neglect the contribution of shear strain to the total strain energ? If the dimensions of the beam cross section are much smaller than the length of the beam. A FBD p C ( ) ( 2) A C B B A A p M H V C FBD B FBD 2 C p FBD 4 ( ) A C K B C A C V 2 From FBD ( ) + B ( ) p ( 4) = B = C + 4 p M A = C : F = A + B C p = A = 5 C + 4 p The problem is INDETERMINATE: will choose C as the redundant load. From FBDs and 4: M h = A + p 2 + M = M ( ) = 4 p + 5 C 2 p 2 M K = A + p 2 + C ( ) + = ( ) = C + 4 p + C 2 p 2 METHOD # consider the strain energ from onl the beam ( C is an eternal applied force) U = 2 ( )d ( )d

4 Using Castigliano s theorem, along with rod elongation equation: v C = C / : v C = U C = = + C = M M d + C 5 4 p + 5 C 2 p 2 d p C p d M C d 9C C +2 p C p p d = 5 2 p C 5 64 p 4 + 9C C +2 p C p + 64 p 4 = 5 2 p C C where: Solving gives C = 49. kn ( ) 5 = = 64 p ( ) ( ) C +2 p ( 6 mm 4 ) m / mm I 2 A = 64 9 ( ) + ( ) 4 ( ) 2 =.2 m 2 π π.2 / 2 B = C + 4 p = 2.5 kn A = 5 C + 4 p = 9. kn + C + 4 p + C 2 p 2 d + C = = 9 64 C p m2 + C ( ) + + C 64 p ( 496 ) + I A C METHOD #2 consider the strain energ from both the beam and rod together ( C is a reaction force) U = 2 ( )d + M ( )d + C 2 2 Using Castigliano s theorem: = U C = M M d + C M M 2 d C ( ) + C This gives the same equation for C as does METHOD #.

5 ME 2 Eamination #2 April, 2 PROBLEM NO. 25 points ma. Beam BD is supported b a fied wall at end B, and b a roller support at C. A concentrated couple M acts at D. The beam is made up of a material with a Young s modulus E and has a constant cross section with a second area moment of I over the full length of the beam. Using either the second-order or fourthorder integration integration methods: a) determine the reaction force acting on the beam at C. b) determine the slope of the deflection θ D of the beam at D Leave our answers in terms of, at most: M, E, I and a. M V H 2a a M B D M C Section BC M H B C D = M() + C 2a θ ( ) = θ ( ) + v( ) = v( ) + Since v 2a ( ) C ( ) M = M() = 2aC M ( ) C ( ) =, we have: = 2 2aC M Also, θ C = θ ( 2a) = 2aC M d = + 2aC ( M ) 2 C 2 ( 2aC M ) 2 C 2 d = + ( 2 2aC M ) 2 6 C ( ) 2a ( ) 2 6 C ( 2a ) 2a M 4 a M 2a = C a 2M a 2 C = 4 ( ) 2 Section CD M K = M M() = M() = M Therefore, θ ( a) = θ ( 2a) + a M d = 2 2a 4 M a M M a 2a ( ) 2 M a = am 2 am = 2 am M V B C D M ( a 2a) = M 2 a = θ D K

6 ME 2 Eamination #2 April, 2 PROBLEM NO. 4 - PART A 7 points ma. A beam is made up a material with a Young s modulus of E and has a constant cross section with a second area moment of I. A downward, constant line load p (force/length) acts along the full length of the beam. The beam has roller supports at B and C, along with a pin joint support at end D. Using the superposition approach, determine the reaction force acting on the beam at the roller support C. p 2a B C D a v( ) ( ) v ( 2 ) C v B C D B D B C D

7 ME 2 Eamination #2 April, 2 PROBLEM NO. 4 - PART B points ma. Cantilevered beams A, B and C shown below are acted upon b a point load P acting at the free end of the beam. The cross sections and lengths of each beam are the same. Beams A and B are made up of steel and aluminum, respectivel, whereas Beam C is made up of both steel and aluminum components over its length. Let σ A ma, σ B ma and σ C ma responses below: be the maimum normal stress in Beams A, B and C, respectivel. Circle the correction TRUE or FALSE: σ A ma = σ B ma TRUE or FALSE: σ A ma = σ C ma All structures are DETERMINATE. Stresses do not depend on materials. TRUE or FALSE: σ B ma = σ C ma P P P steel aluminum steel aluminum L L L Beam A Beam B Beam C PROBLEM NO. 4 - PART C 2 points ma. The proppped-cantilevered beams A, B and C shown below are acted upon b a point load P acting at the free end of the beam. The cross sections and lengths of each beam are the same. Beams A and B are made up of steel and aluminum, respectivel, whereas Beam C is made up of both steel and aluminum components over its length. Let σ A ma, σ B ma and σ C be the maimum normal stress in Beams A, B and C, respectivel. Circle the ma correction responses below: TRUE or FALSE: σ A ma = σ C ma TRUE or FALSE: σ B ma = σ C ma All structures are INDETERMINATE. Stresses depend on deflections, which are dependent on material properties. P P P steel aluminum steel aluminum L L L Beam A Beam B Beam C

8 ME 2 Eamination #2 April, 2 PROBLEM NO. 4 - PART D 7 points ma. A rod is made up of solid, circular cross-section segments (), (2) and (), where Segment () has a length of 2L and diameter 2d, Segment (2) has a length of L and diameter d, and Segment () has a length of L and diameter 2d. All segments are made up of a material having a Young s modulus of E. Loads of P and 2P act on the rigid connectors, as shown below. You are asked to set up a three element, finite element model for displacement analsis of this rod, using one element for each of the rod segments. To this end, write down the stiffness matri K and load vector F { } for the model after the boundar conditions have been enforced on the model. () ( 2) ( ) 2d P d 2P 2d 2L L L k = L = Eπ ( 2d / 2) 2 = π 2L 2 ( ) 2 k 2 = 2 = Eπ d / 2 L 2 L = π 4 Ed 2 L = 2 π Ed 2 4L Ed 2 L = π Ed 2 4L F () P ( 2) ( ) 2P F k = L = Eπ ( 2d / 2) 2 = π Ed 2 L L 4 π Ed 2 4L Therefore, before enforcing BCs: K = π Ed 2 4L ; { F} = F P 2P F After enforcing BCs (removing st and 4 th rows and columns of [K] and the st and 4 th row of {F}: K = π Ed 2 P ; { F} = 4L 5 2P

9 ME 2 Eamination #2 April, 2 PROBLEM NO. 4 - PART E 6 points ma. A state of stress is characterized b its unknown - components on the stress element shown below. When the stress element is rotated through an angle of 2, the state of stress has the n-t components shown below right. a) Draw the Mohr s circle for this state of stress on the aes provided below. Carefull label the center of the circle as well as its radius. Show the -ais on the Mohr s circle. b) What is the maimum in-plane shear stress? c) What are the σ, σ and τ components of this state of stress? σ ave = σ P +σ P = = 7 ksi 2 2 R = σ P σ P = = 5 ksi 2 2 τ = R = 5 ksi ma,in plane,t σ =?? τ =??,n σ =?? t n 2 22 ksi 2 ksi From Mohr s circle shown: σ = σ ave Rcos6 = 7 ( 5)cos6 = 4.5 ksi σ = σ ave + Rcos6 = 7 + ( 5)cos6 = 9.5 ksi τ = R sin6 = ( 5)sin6 = 4. ksi ais 6 X τ σ 2θ P = 24 ais τ σ σ σ P2 = 2 ksi σ P = 22 ksi σ ave = 7 ksi

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