Stress analysis of a stepped bar Problem Find the stresses induced in the axially loaded stepped bar shown in Figure. The bar has cross-sectional areas of A ) and A ) over the lengths l ) and l ), respectively. Assume the following data: A ) = cm, A ) = cm, l ) = l ) = cm, E ) = E ) = E = 7 N/cm, P 3 = N. Solution Idealization Let the bar be considered as an assemblage of two elements. By assuming the bar to be a one-dimensional structure, we have only axial displacement at any point in the element. As there are three nodes, the axial displacements of the nodes Φ, Φ, and Φ 3 will be taken as unknowns. Displacement model In each of the elements, we assume a linear variation of axial displacement φ φx) = a + bx, ) Figure : A stepped bar under axial load
where a and b are constants. If we consider the end displacements φ at x = l e) ) as unknowns, we obtain φ at x = ) and a =, b = Φe) )/l e), ) where the superscript e denotes the element number. Thus, Element stiffness matrix φx) = + ) x. 3) l e) The element stiffness matrices can be derived from the principle of minimum potential energy. For this, we write the potential energy of the bar I under axial deformation as I = strain energy work done by external forces = π ) + π ) W p, 4) where π e) represents the strain energy of element e, W p denotes the work done by external forces, and l e) π e) = A e) σe) ε e) dx = Ae) E e) l e) ε e)) dx, 5) where A e) is the cross-sectional area of element e, l e) is the length of element, σ e) is the stress in element e, ε e) is the strain in element e, and E e) is the Young s modulus of elelemnt e. From the expression of φx), we can write and hence pi e) = Ae) E e) = Ae) E e) l e) ε e) = φ x = Φe) l e), 6) l e) ) + ) + l e) ) ) Φ e) ) Φ e) Φe) Φe) ) dx = 7) This expression for π e) can be written in matrix form as π e) = T K e), 8) where or = Φ ) = Φ Φ is the vector of nodal displacements of element e, 9) for e =, Φ ) = Φ Φ 3 for e =, )
3 and K e) = Ae) E e) l e) is called the stiffness matrix of element e. ) Since there are only concentrated loads acting at the nodes of the bar no distributed load), the work done by external forces can be expressed as W p = Φ P + Φ P + Φ 3 P 3 = Φ T P, ) where P i denotes the force applied in the direction of the displacement Φ i i =,, 3). In this example, P = reaction at fixed node, P =, and P 3 =.. If the bar as a whole is in equilibrium under the loads P = P, P, P 3 ) T, the principle of minimum potential energy gives Φ i =, i =,, 3. 3) This equation can be rewritten as ) = π e) W p =, Φ i Φ i i =,, 3, 4) e= where the summation sign indicates the addition of the strain energies scalars) of the elements. Equation 4) can be written as e= K e) P e)) =, 5) where the summation sign indicates the assembly of vectors not the addition of vectors) in which only the elements corresponding to a particular degree of freedom in different vectors are added. Assembly of element stiffness matrices and element load vectors This step includes the assembly of element stiffness matrices K e) and element load vectors P e) to obtain the overall or global equilibrium equations. Eq. 5) can be rewritten as K Φ P = 6) where K is the assembled or global stiffness matrix, and Φ is the vector of global displacements. For the given data, the element matrices would be K ) = A) E ) l ) K ) = A) E ) l ) = 6 = 6 4 4 4 4, 7). 8)
4 The overall stiffness matrix of the bar can be obtained by assembling the two element stiffness matrices. Since there are three nodal displacement unknowns, the global stiffness matrix K will be of order three. To obtain K, the elements of K ) and K ) corresponding to the unknowns Φ, Φ, and Φ 3 are added as shown below 4 4 K = 6 4 4 + = 6 3. 9) The global load vector can be written as P = P P P 3 = P, ) where P denotes the reaction at node. Thus, the overall equilibrium equations become Φ P 6 3 Φ = ) Φ 3 Eq. ) can be derived in a much simpler way as follows. stepped bar can be expressed as Then, I = π ) + π ) W p = A ) E ) Φ l ) + Φ ) Φ Φ + + A ) E ) Φ l ) + Φ ) 3 Φ Φ 3 P Φ P Φ P 3 Φ 3 Solution for displacements The potential energy of the ) = A) E ) Φ l ) Φ Φ ) P = 3) = A) E ) Φ l ) Φ Φ ) + A) E ) l ) Φ Φ 3 ) P = 4) = A) E ) Φ 3 l ) Φ 3 Φ ) P 3 = 5) If we try to solve Eq. ) for the unknowns Φ, Φ, and Φ 3, we will not be able to do it since the matrix K is singular. This is because we have not incorporated the known geometric boundary condition Φ =. The final equilibrium equations can be written as The solution of Eq. 6) gives 6 3 Φ Φ 3 = 6) Φ =.5 6 cm and Φ 3 =.75 6 cm 7)
5 Element strains and stresses Once the displacements are computed, the strains in the elements can be found as ε ) = φ x = Φ) Φ ) l ) = Φ Φ l ) =.5 7, 8) ε ) = φ x = Φ) Φ ) l ) = Φ 3 Φ l ) =.5 7. 9) The stresses in the elements are given by σ ) = E ) ε ) = 7 ).5 7 ) =.5 N/cm, 3) σ ) = E ) ε ) = 7 ).5 7 ) =. N/cm. 3)