Problems for Chapter 5

 

Analytical Techniques and Solutions for Linear Elastic Solids

 

 

 

5.7.  Energy Methods

 

 

5.7.1.      A shaft with length L and square cross section is fixed at one end, and subjected to a twisting moment T at the other.  The shaft is made from a linear elastic solid with Young’s modulus E and Poisson’s ratio ν MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabe27aUbaa@37F1@ .  The torque causes the top end of the shaft to rotate through an angle ϕ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabew9aMbaa@3801@ .

5.7.1.1.            Consider the following displacement field

v 1 = ϕ L x 1 x 3 v 2 = ϕ L x 2 x 3 v 3 =0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWG2bWaaSbaaSqaaiaaigdaaeqaaO Gaeyypa0JaeyOeI0YaaSaaaeaacqaHvpGzaeaacaWGmbaaaiaadIha daWgaaWcbaGaaGymaaqabaGccaWG4bWaaSbaaSqaaiaaiodaaeqaaO GaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaadAhadaWgaaWcbaGaaG OmaaqabaGccqGH9aqpdaWcaaqaaiabew9aMbqaaiaadYeaaaGaamiE amaaBaaaleaacaaIYaaabeaakiaadIhadaWgaaWcbaGaaG4maaqaba GccaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPa VlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaadAhadaWgaa WcbaGaaG4maaqabaGccqGH9aqpcaaIWaGaaGPaVdaa@7588@

Show that this is a kinematically admissible displacement field for the twisted shaft.

5.7.1.2.            Calculate the strains associated with this kinematically admissible displacement field

5.7.1.3.            Hence, show that the potential energy of the shaft is

You may assume that the potential energy of the torsional load is Tϕ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqGHsislcaWGubGaeqy1dyMaaGPaVd aa@37DE@

5.7.1.4.            Find the value of ϕ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaHvpGzcaaMc8oaaa@3618@  that minimizes the potential energy, and hence estimate the torsional stiffness of the shaft.

 

 

 

 

5.7.2.      In this problem you will use the principle of minimum potential energy to find an approximate solution to the displacement in a pressurized cylinder.  Assume that the cylinder is an isotropic, linear elastic solid with Young’s modulus E MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGfbaaaa@3390@  and Poisson’s ratio ν MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaH9oGBaaa@347E@ , and subjected to internal pressure p at r=a.

5.7.2.1.            Approximate the radial displacement field as u r = C 1 + C 2 r MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWG1bWaaSbaaSqaaiaadkhaaeqaaO Gaeyypa0Jaam4qamaaBaaaleaacaaIXaaabeaakiabgUcaRiaadoea daWgaaWcbaGaaGOmaaqabaGccaWGYbaaaa@3B3E@ , where C 1 , C 2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGdbWaaSbaaSqaaiaaigdaaeqaaO GaaiilaiaadoeadaWgaaWcbaGaaGOmaaqabaaaaa@36DF@  are constants to be determined.  Assume all other components of displacement are zero.  Calculate the strains in the solid ε rr , ε θθ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaH1oqzdaWgaaWcbaGaamOCaiaadk haaeqaaOGaaiilaiabew7aLnaaBaaaleaacqaH4oqCcqaH4oqCaeqa aaaa@3C80@

5.7.2.2.            Find an expression for the total strain energy of the cylinder per unit length, in terms of C 1 , C 2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGdbWaaSbaaSqaaiaaigdaaeqaaO GaaiilaiaadoeadaWgaaWcbaGaaGOmaaqabaaaaa@36DF@  and relevant geometric and material parameters

5.7.2.3.            Hence, write down the potential energy (per unit length) of the cylinder.

5.7.2.4.            Find the values of C 1 , C 2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGdbWaaSbaaSqaaiaaigdaaeqaaO GaaiilaiaadoeadaWgaaWcbaGaaGOmaaqabaaaaa@36DF@  that minimize the potential energy

5.7.2.5.            Plot a graph showing the normalized radial displacement field E( b 2 a 2 ) u r (r)/(pa b 2 ) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGfbWaaeWaaeaacaWGIbWaaWbaaS qabeaacaaIYaaaaOGaeyOeI0IaamyyamaaCaaaleqabaGaaGOmaaaa aOGaayjkaiaawMcaaiaadwhadaWgaaWcbaGaamOCaaqabaGccaGGOa GaamOCaiaacMcacaGGVaGaaiikaiaadchacaWGHbGaamOyamaaCaaa leqabaGaaGOmaaaakiaacMcaaaa@43F1@  as a function of the normalized position (ra)/(ba) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaGGOaGaamOCaiabgkHiTiaadggaca GGPaGaai4laiaacIcacaWGIbGaeyOeI0IaamyyaiaacMcaaaa@3BAF@  in the cylinder, for ν=0.3 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaH9oGBcqGH9aqpcaaIWaGaaiOlai aaiodaaaa@37AD@ , and b/a=1.1 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGIbGaai4laiaadggacqGH9aqpca aIXaGaaiOlaiaaigdaaaa@3874@  and b/a=2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGIbGaai4laiaadggacqGH9aqpca aIYaaaaa@3708@ .   On the same graph, plot the exact solution, given in Section 4.1.9.

 

 

 

 

5.7.3.      A bi-metallic strip is made by welding together two materials with identical Young’s modulus and Poisson’s ratio E,ν MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGfbGaaiilaiabe27aUbaa@35F8@ , but with different thermal expansion coefficients α 1 , α 2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaHXoqydaWgaaWcbaGaaGymaaqaba GccaGGSaGaeqySde2aaSbaaSqaaiaaikdaaeqaaaaa@388D@ , as shown in the picture.  At some arbitrary temperature T 0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGubWaaSbaaSqaaiaaicdaaeqaaa aa@3485@  the strip is straight and free of stress.   The temperature is then increased to a new value T MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGubaaaa@339F@ , causing the strip to bend.  Assume that, after heating, the displacement field in the strip can be approximated as u 1 =λ x 1 + x 1 x 2 /R u 2 = x 1 2 /(2R)β x 2 u 3 =0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWG1bWaaSbaaSqaaiaaigdaaeqaaO Gaeyypa0Jaeq4UdWMaamiEamaaBaaaleaacaaIXaaabeaakiabgUca RiaadIhadaWgaaWcbaGaaGymaaqabaGccaWG4bWaaSbaaSqaaiaaik daaeqaaOGaai4laiaadkfacaaMc8UaaGPaVlaaykW7caaMc8UaaGPa VlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaamyDamaaBaaaleaaca aIYaaabeaakiabg2da9iabgkHiTiaadIhadaqhaaWcbaGaaGymaaqa aiaaikdaaaGccaGGVaGaaiikaiaaikdacaWGsbGaaiykaiabgkHiTi abek7aIjaadIhadaWgaaWcbaGaaGOmaaqabaGccaaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaadwhadaWgaaWcba GaaG4maaqabaGccqGH9aqpcaaIWaaaaa@6DBB@ , where λ,R,β MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaH7oaBcaGGSaGaamOuaiaacYcacq aHYoGyaaa@3852@  are constants to be determined.

5.7.3.1.            Briefly describe the physical significance of the shape changes associated with λ,R,β MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaH7oaBcaGGSaGaamOuaiaacYcacq aHYoGyaaa@3852@ .

5.7.3.2.            Calculate the distribution of (infinitesimal) strain associated with the kinematically admissible displacement field

5.7.3.3.            Hence, calculate the strain energy density distribution in the solid.  Don’t forget to account for the effects of thermal expansion

5.7.3.4.            Minimize the potential energy to determine values for λ,R,β MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaH7oaBcaGGSaGaamOuaiaacYcacq aHYoGyaaa@3852@  in terms of relevant geometric and material parameters.

 

 

 

 

5.7.4.      By guessing the deflected shape, estimate the stiffness of a clamped MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFuacaaa@37E7@ clamped beam subjected to a point force at mid-span.   Note that your guess for the deflected shape must satisfy w( x 1 )= dw d x 1 =0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFH e9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaa caqabeaacmqaamaaaOqaaiaadEhacaGGOaGaamiEamaaBaaaleaaca aIXaaabeaakiaacMcacqGH9aqpdaWcaaqaaiaadsgacaWG3baabaGa amizaiaadIhadaWgaaWcbaGaaGymaaqabaaaaOGaeyypa0JaaGimaa aa@3C3C@ , so you can’t assume that it bends into a circular shape as done in class.  Instead, try a deflection of the form w( x 1 )=1cos( 2π x 1 /L ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFH e9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaa caqabeaacmqaamaaaOqaaiaadEhacaGGOaGaamiEamaaBaaaleaaca aIXaaabeaakiaacMcacqGH9aqpcaaIXaGaeyOeI0Iaci4yaiaac+ga caGGZbWaaeWaaeaacaaIYaGaeqiWdaNaamiEamaaBaaaleaacaaIXa aabeaakiaac+cacaWGmbaacaGLOaGaayzkaaaaaa@419F@ , or a similar function of your choice (you could try a suitable polynomial, for example).  If you try more than one guess and want to know which one gives the best result, remember that energy minimization always overestimates stiffness.  The best guess is the one that gives the lowest stiffness.

 

 

 

 

5.7.5.      A slender rod with length L and cross sectional area A is subjected to an axial body force b=b( x 2 ) e 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFH e9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaa caqabeaacmqaamaaaOqaaiaahkgacqGH9aqpcaWGIbGaaiikaiaadI hadaWgaaWcbaGaaGOmaaqabaGccaGGPaGaaCyzamaaBaaaleaacaaI Yaaabeaaaaa@385D@ .  Our objective is to determine an approximate solution to the displacement field in the rod.

5.7.5.1.            Assume that the displacement field has the form

u 2 =w( x 2 ) u 1 = u 3 =0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFH e9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaa caqabeaacmqaamaaaOqaaiaadwhadaWgaaWcbaGaaGOmaaqabaGccq GH9aqpcaWG3bGaaiikaiaadIhadaWgaaWcbaGaaGOmaaqabaGccaGG PaGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caWG1bWaaSbaaSqaai aaigdaaeqaaOGaeyypa0JaamyDamaaBaaaleaacaaIZaaabeaakiab g2da9iaaicdaaaa@45F2@

where the function w is to be determined.  Find an expression for the strains in terms of w and hence deduce the strain energy density.

 

5.7.5.2.            Show that the potential energy of the rod is

V(w)=EA 1ν ( 12ν )( 1+ν ) 0 L 1 2 { dw d x 2 } 2 d x 2 A 0 L b( x 2 )w( x 2 )d x 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFH e9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaa caqabeaacmqaamaaaOqaaiaadAfacaGGOaGaam4DaiaacMcacqGH9a qpcaWGfbGaamyqamaalaaabaGaaGymaiabgkHiTiabe27aUbqaamaa bmaabaGaaGymaiabgkHiTiaaikdacqaH9oGBaiaawIcacaGLPaaada qadaqaaiaaigdacqGHRaWkcqaH9oGBaiaawIcacaGLPaaaaaWaa8qC aeaadaWcaaqaaiaaigdaaeaacaaIYaaaamaacmaabaWaaSaaaeaaca WGKbGaam4DaaqaaiaadsgacaWG4bWaa0baaSqaaiaaikdaaeaaaaaa aaGccaGL7bGaayzFaaWaaWbaaSqabeaacaaIYaaaaaqaaiaaicdaae aacaWGmbaaniabgUIiYdGccaWGKbGaamiEamaaBaaaleaacaaIYaaa beaakiabgkHiTiaadgeadaWdXbqaaiaadkgacaGGOaGaamiEamaaBa aaleaacaaIYaaabeaakiaacMcacaWG3bGaaiikaiaadIhadaWgaaWc baGaaGOmaaqabaGccaGGPaGaamizaiaadIhadaWgaaWcbaGaaGOmaa qabaaabaGaaGimaaqaaiaadYeaa0Gaey4kIipaaaa@6537@

 

5.7.5.3.            To minimize the potential energy, suppose that w is perturbed from the value the minimizes V  to a value w+δw MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFH e9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaa caqabeaacmqaamaaaOqaaiaadEhacqGHRaWkcqaH0oazcaWG3baaaa@34E6@ .  Assume that δw MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFH e9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaa caqabeaacmqaamaaaOqaaiabes7aKjaadEhaaaa@3308@  is kinematically admissible, which requires that δw=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFH e9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaa caqabeaacmqaamaaaOqaaiabes7aKjaadEhacqGH9aqpcaaIWaaaaa@34C8@  at any point on the bar where the value of w is prescribed.  Calculate the potential energy V(w+δw) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFH e9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaa caqabeaacmqaamaaaOqaaiaadAfacaGGOaGaam4DaiabgUcaRiabes 7aKjaadEhacaGGPaaaaa@371A@  and show that it can be expressed in the form

V(w+δw)=V(w)+δV+ 1 2 δ 2 V MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFH e9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaa caqabeaacmqaamaaaOqaaiaadAfacaGGOaGaam4DaiabgUcaRiabes 7aKjaadEhacaGGPaGaeyypa0JaamOvaiaacIcacaWG3bGaaiykaiab gUcaRiabes7aKjaadAfacqGHRaWkdaWcaaqaaiaaigdaaeaacaaIYa aaaiabes7aKnaaCaaaleqabaGaaGOmaaaakiaadAfaaaa@448E@

where V MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFH e9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaa caqabeaacmqaamaaaOqaaiaadAfaaaa@3142@  is a function of w only, δV MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFH e9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaa caqabeaacmqaamaaaOqaaiabes7aKjaadAfaaaa@32E7@  is a function of w and δw MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFH e9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaa caqabeaacmqaamaaaOqaaiabes7aKjaadEhaaaa@3308@ , and δ 2 V MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFH e9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaa caqabeaacmqaamaaaOqaaiabes7aKnaaCaaaleqabaGaaGOmaaaaki aadAfaaaa@33DA@  is a function of δw MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFH e9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaa caqabeaacmqaamaaaOqaaiabes7aKjaadEhaaaa@3308@  only.

 

5.7.5.4.            As discussed in Section 8 of the online notes (or in class), if V MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFH e9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaa caqabeaacmqaamaaaOqaaiaadAfaaaa@3142@  is stationary at δw=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFH e9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaa caqabeaacmqaamaaaOqaaiabes7aKjaadEhacqGH9aqpcaaIWaaaaa@34C8@ , then δV=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFH e9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaa caqabeaacmqaamaaaOqaaiabes7aKjaadAfacqGH9aqpcaaIWaaaaa@34A7@ .  Show that, to satisfy δV=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFH e9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaa caqabeaacmqaamaaaOqaaiabes7aKjaadAfacqGH9aqpcaaIWaaaaa@34A7@ , we must choose w to satisfy

EA(1ν) (12ν)(1+ν) 0 L dw d x 2 dδw d x 2 d x 2 A 0 L bδw d x 2 =0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFH e9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaa caqabeaacmqaamaaaOqaamaalaaabaGaamyraiaadgeacaGGOaGaaG ymaiabgkHiTiabe27aUjaacMcaaeaacaGGOaGaaGymaiabgkHiTiaa ikdacqaH9oGBcaGGPaGaaiikaiaaigdacqGHRaWkcqaH9oGBcaGGPa aaamaapehabaWaaSaaaeaacaWGKbGaam4DaaqaaiaadsgacaWG4bWa aSbaaSqaaiaaikdaaeqaaaaakmaalaaabaGaamizaiabes7aKjaadE haaeaacaWGKbGaamiEamaaBaaaleaacaaIYaaabeaaaaGccaWGKbGa amiEamaaBaaaleaacaaIYaaabeaakiabgkHiTiaadgeadaWdXbqaai aadkgacqaH0oazcaWG3baaleaacaaIWaaabaGaamitaaqdcqGHRiI8 aaWcbaGaaGimaaqaaiaadYeaa0Gaey4kIipakiaadsgacaWG4bWaaS baaSqaaiaaikdaaeqaaOGaeyypa0JaaGimaaaa@60BF@

 

5.7.5.5.            Integrate the first term by parts to deduce that, to minimize, V, w must satisfy

d 2 w d x 2 2 +b( x 2 )=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFH e9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaa caqabeaacmqaamaaaOqaamaalaaabaGaamizamaaCaaaleqabaGaaG OmaaaakiaadEhaaeaacaWGKbGaamiEamaaDaaaleaacaaIYaaabaGa aGOmaaaaaaGccqGHRaWkcaWGIbGaaiikaiaadIhadaWgaaWcbaGaaG OmaaqabaGccaGGPaGaeyypa0JaaGimaaaa@3DB5@

Show that this is equivalent to the equilibrium condition

d σ 22 d x 2 + b 2 =0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFH e9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaa caqabeaacmqaamaaaOqaamaalaaabaGaamizaiabeo8aZnaaBaaale aacaaIYaGaaGOmaaqabaaakeaacaWGKbGaamiEamaaBaaaleaacaaI YaaabeaaaaGccqGHRaWkcaWGIbWaaSbaaSqaaiaaikdaaeqaaOGaey ypa0JaaGimaaaa@3C24@

Furthermore, deduce that if w MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFH e9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaa caqabeaacmqaamaaaOqaaiaadEhaaaa@3163@  is not prescribed at either x 2 =0, x 2 =L MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFH e9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaa caqabeaacmqaamaaaOqaaiaadIhadaWgaaWcbaGaaGOmaaqabaGccq GH9aqpcaaIWaGaaiilaiaaykW7caaMc8UaaGPaVlaaykW7caWG4bWa aSbaaSqaaiaaikdaaeqaaOGaeyypa0Jaamitaaaa@3EB8@  or both, then the boundary conditions on the end(s) of the rod must be

dw d x 2 =0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFH e9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaa caqabeaacmqaamaaaOqaamaalaaabaGaamizaiaadEhaaeaacaWGKb GaamiEamaaBaaaleaacaaIYaaabeaaaaGccqGH9aqpcaaIWaaaaa@36F4@

Show that this corresponds to the condition that σ 22 =0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFH e9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaa caqabeaacmqaamaaaOqaaiabeo8aZnaaBaaaleaacaaIYaGaaGOmaa qabaGccqGH9aqpcaaIWaaaaa@3598@  at a free end.

 

5.7.5.6.            Use your results in (2.5) to estimate the displacement field in a bar with mass density ρ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFH e9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaa caqabeaacmqaamaaaOqaaiabeg8aYbaa@3227@ , which is  attached to a rigid wall at x 2 =L MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFH e9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaa caqabeaacmqaamaaaOqaaiaadIhadaWgaaWcbaGaaGOmaaqabaGccq GH9aqpcaWGmbaaaa@342D@ , is free at x 2 =0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFH e9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaa caqabeaacmqaamaaaOqaaiaadIhadaWgaaWcbaGaaGOmaaqabaGccq GH9aqpcaaIWaaaaa@3416@ , and subjected to the force of gravity (acting vertically downwards…)