ME 323 Examination #2


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1 ME 33 Eamination # SOUTION Novemer 14, 17 ROEM NO. 1 3 points ma. The cantilever eam D of the ending stiffness is sujected to a concentrated moment M at C. The eam is also supported y a roller at. Using Castigliano s theorem: a) Determine the reaction force at the roller. ) Determine the rotation angle of the eam aout z ais at the end. Ignore the shear energy due to ending. Epress your answers in terms of M, E, and I. M d M M D SOUTION Eternal reactions Using FD of entire eam: M D F y = y + D y = = M y ( ) + M d + M D = rolem is INDETERMINTE. Will choose y as the redundant reaction: M D = y M D y = y (1) () C (3) D M d M d H y D y M 1 V 1 K M Strain energy From FD with cut through section (1): M H = M 1 + M d = M 1 = From FD with cut through section (): M H = M + M d y ( ) = ( ) M = + y From FD with cut through section (3): M H = M 3 + M d y M 3 = + y ( ) M ( ) + M = M d y V C y M Q M 3 V 3
2 From this, we have: U = U 1 +U +U 3 = 1 d y ( ) d y ( ) M d age of 1 Castigliano s theorem Since y is our redundant reaction, we can write: = U y = + 1 = y = y Md = + y ( ) ( + )d + 1 Md = 3 ( )d y ( + )d M 1 ( 3 )3 3 ( ) + = y M = 8 y 3 3 M 3 Therefore: y = 9 16 M 3 3 ( ) M ( )d ( ) + 1 ( 3 3 ) 3 ( ) 3 M Md = ( ) ( ) ( ) ( ) ( )d + ( 3 ) 3
3 lso: U θ = M d = 1 = y M d = M d d + 1 Md = ( )d y y ( ) Md ( 1 )d + 1 = ( ) M d = y 1 ( ) ( ) y 1 3 = y = y = 9 16 = 1 8 M 3 1 y + M M + M M ( ) ( ) 3 + y ( 3 ) + M age 3 of 1 ( ) M Md ( 1 )d = ( 3 )
4 ME 33 Eamination # age 4 of 1 SOUTION Novemer 14, 17 ROEM NO. 5 points ma. t a point aove the neutral ais of the eam shown in the figure, the state of plane stress can e descried y the insert on the righthand side of the figure. The maimum inplane shear stress at this point is τ!"# = 13Ma, the normal stress in the direction is σ! = Ma, and normal stress in the ydirection is σ! = Ma. a) Determine the magnitude of the shear stress, τ!", on the and y faces. ) Determine the sign of the shear stress τ!". HINT: Determine first the direction of the shear force acting on the crosssection with normal at point. c) Draw Mohr s circle corresponding to the state of stress at point. Clearly indicate the location of the center of the circle, the radius of the circle and point X (which represents the stress state on the face) in this drawing. d) Determine the two inplane principal stresses at this point. Determine the rotation angle of the stress element for each principal stress. e) Show the locations of the principal stresses and of the inplane maimum shear stress on your Mohr s circle in c) aove. y y + State of stress at point SOUTION Internal resultants = +V = V = F y M V Stress transformation results: σ ave = σ +σ y = = 1 Ma R = σ σ y + τ y = + τ = 1 + τ
5 Since: we can write: τ ma,in plane = R = τ = 13 τ = ± 13 1 = ± 69 Choose the  sign ased on the direction of the applied force. age 5 of 1 Mohr s circle and principal stresses The Mohr s circle is centered at σ ave, ( ) = ( 1,) Ma and has a radius of R = 13 Ma, as shown. From this, the principal components of stress are: σ 1 = σ ave + R = = 3 Ma σ = σ ave R = 1 13 = 3 Ma The location X of the ais on Mohr s circle is: (, 69) Ma. From the figure, we see that the rotation angles from the ais to the aove principal components of stress are: θ = 18 tan = θ = 7.1 θ θ 1 = θ +18 θ 1 = θ + 9 = 16.1 θ S = θ + 9 θ S = θ + 45 = σ = 3 τ X R = 13 σ = ais τ ma = 13 τ y = 69 σ θ 1 τ ma = 13 θ S σ ave = 1 σ 1 = 3
6 ME 33 Eamination # age 6 of 1 SOUTION Novemer 14, 17 ROEM NO. 3 5 points ma. The propped cantilever in the figure is simply supported at end and fied at end. It supports a linearly distriuted load of maimum intensity w! on the span. a) Draw a free ody diagram of the structure. ssume the reactions forces act in the direction of positive and y aes, and the reaction moments act counterclockwise. ) State the equations of equilirium of the structure and indicate whether it is statically determinate or indeterminate. c) Indicate all the oundary conditions that correspond to this prolem. d) Use the secondorder integration method (or the fourthorder integration method) to determine an epression for the reaction(s) at the support. Epress the result as a sole function of, w! and. e) Determine an epression for the deflection of the eam as a sole function of, w! and. f) Sketch the deflection curve. w / w M y y 1 w SOUTION Equilirium From FD of entire eam: M = 1 w 3 + y + M = y y + M = 1 3 w C V M F y = y + y 1 w = y + y = 1 w From FD with cut through eam at location : M C = y + 1 w M = M() = y 1 w 3 6
7 Integrations Will need to enforce the following oundary conditions: v() = v() = θ ( ) =. θ() = θ() + 1 M ( )d = θ + 1 = θ y 1 w 4 4 y 1 w 3 d 6 v() = v() + θ ( )d = + θ y 1 w 4 4 d = θ y 3 1 w 5 1 age 7 of 1 Enforcing the oundary conditions at : = θ() = θ y 1 4 w 3 θ = 1 1 y w 3 = v() = θ y w 4 θ = y w 3 Equating the aove two epressions for θ : 1 1 y w 3 = y w 3 1 y w = 1 6 y w y = w y = w 1 and: θ = 1 1 y w 3 = 1 1 w w 3 = 1 w 3 1 Therefore: v() = 1 w w w 4 = w eam deflec)on w
8 ME 33 Eamination # age 8 of 1 SOUTION Novemer 14, 17 ROEM NO. 4  RT 4 points ma. y y h / 4 h z h / 4 h / 4 / / h / 4 C D eam cross sec*on at C Consider the cantilevered eam aove with the concentrated load at end D. Determine the shear stress on the neutral surface of the eam at location C along the eam. SOUTION Internal resultant shear force F y = +V = V = M V Shear stress with: Therefore, τ = V* y * It * y * = h h h 4 I = 1 1 h3 1 h 1 t = 4 = τ = 7h / 64 15h 3 / 19 = 14 / 5 3 3h 8 = h3 h + 4 h h 4 = 7 64 h
9 ME 33 Eamination # age 9 of 1 SOUTION Novemer 14, 17 ROEM NO. 4  RT 3 points ma. y a / / C D eam cross sec*on at C Consider the cantilevered eam aove with the concentrated load at end D. Consider the aial components of stress at points a and (σ a and σ, respectively) at location C along the eam. Circle the response elow that most accurately descries the relative sizes of the magnitudes of these two stresses: a) σ a > σ ) σ a = σ y c) σ a < σ SOUTION et O e the centroid of the cross section. Therefore, σ a σ = Mh / I Md / I = h d > 1 σ a > σ z h d a O
10 ME 33 Eamination # age 1 of 1 SOUTION Novemer 14, 17 ROEM NO. 4  RT C 4 points ma. t r g C W Consider the thinwalled pressure vessel aove that contains a gas under a pressure of p. The vessel is attached to a fied support at and has a plate of weight W attached to it at end C. Ignore the weight of the vessel. Determine the weight W of the plate for which the maimum inplane shear stress in the vessel at point is zero. SOUTION σ h = pr t σ a = pr t + W πrt For zero maimum inplane shear stress: R = = pr t σ σ y = pr t + W πrt + τ y = W = π pr σ h σ a + = σ h σ a σ h = σ a
11 ME 33 Eamination # age 11 of 1 SOUTION Novemer 14, 17 ROEM NO. 4  RT D 4 points ma. y σ y =? σ =? σ 4 ksi τ ksi 1 ksi Consider the state of plane stress shown aove left where the two normal components of stress, σ and σ y, are unknown. The Mohr s circle for this state of stress is provided in the figure aove right. Determine numerical values for the two normal components of stress σ and σ y. There may e more than one set of answers; you need only find one set. SOUTION From the aove Mohr s circle, we see that: σ ave = 1 + = 7 ksi R = 1 = 5 ksi Therefore: σ ave = σ +σ y R = σ σ y σ σ y = ± σ +σ y = σ ave = 14 ksi + τ y σ σ y + τ y = R R τ y = ± 5 ( 4) = ± 6ksi Choosing the + sign and solving gives: σ = 1 ksi and σ y = 4 ksi. Choosing the  sign and solving gives: σ = 4 ksi and σ y = 1 ksi.
12 ME 33 Eamination # age 1 of 1 SOUTION Novemer 14, 17 ROEM NO. 4  RT E 5 points ma. d d d C D H rod is made up of three circular crosssection components: C,CD and DH. The material for all components have a Young s modulus of E. Suppose you are to develop a finite element model for the rod using one element for each component. If the equilirium equations after the enforcement of oundary conditions are to e written as: K { u} = { F} determine the stiffness matri Spring stiffnesses: ( ) k 1 = Eπ d / ( ) ( ) Eπ d / k = k 3 = Eπ d / = π 4 = π = π 8 K and the load vector { F}. Ed Ed Ed F C D H Stiffness and forcing Therefore the gloal stiffness matri and forcing vector efore enforcing Cs are: F K 6 4 = π Ed and { F} = F H Eliminating the first and last row and column of [K] and the first and last row of {F}: K 6 4 = π Ed and { F} = F H
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