Stress-strain relations

Transcription

1 SICLLY INDRMIN SRSS SYSMS staticall determinate stress sstem simple eample of this is a bar loaded b a weight, hanging in tension. he solution for the stress is simpl W/ where is the cross sectional area. Note that we did not need to know anthing about the mechanical properties of the material of the bar to arrive at this solution. For this tpe of problem, where the forces acting upon the bod are specified, the stress is independent of the mechanical behaviour. If we had defined the problem b specifing the displacement imposed on the end rather than the force, then we would need to know the material s modulus to evaluate the stress. W Structure of a Solid Mechanics problem In general, we need to satisf three conditions to obtain the solution to a Solid Mechanics problem. hese are shown in the diagram. he above problem was a special case for which onl the quilibrium stress) equilibrium condition was required. geometrical condition ma need to be satisfied also, and this will take the form of an equation involving strains. Finall, there must be a condition involving both stress and strain that relates the two in our case, this is Hooke s Law. hree-bar eample. Stress-strain relations Geometrical conditions strain) his is a problem in which the geometrical condition is eplicit. sstem consists of two rigid end-plates, tied together b three horiontal bars as shown. hrough a fabrication error, the central bar is.5l too short. ll bars are of the same cross section and of steel with GPa. Find the stress in each bar after the sstem has been mechanicall pulled together so that the gap is closed. ns. -5 MPa 7 MPa.

2 Solution Let the length changes in the bars be δ in the bars and δ in the bar. Note that δ is a compression, and so negative, and δ is a stretch and positive. he geometrical condition can be written as: -δ δ ). Now we use Hooke s Law for the bars and the bar : δ L δ L δ L δ L We now use the above two equations to substitute for δ and δ in ): L ) ). So far we have the above equation containing two unknowns. We need another equation, which comes from the equilibrium condition. Make a vertical cut in the sstem to create a free bod diagram: Horiontal equilibrium gives c Where is the cross section of each bar. Hence, - ). quations ) and ) form a sstem of two equations in two unknowns. o solve, substitute for from ) into ): L ).

3 From this we obtain.5 Mpa L which is 5 Mpa. quation ) now gives 7 Mpa. wo-dimensional eample B m.5m C D mm thick sheet is held within a rigid frame, which prevents an movement in the direction. It is strained b % along the direction. It has a Young s modulus of GPa.and a Poisson s ratio of.4. What is the sum of the forces acting in the members C and BD? ns. 47.6kN hermal strains Recall Hooke s Law week 4) for the case of varing temperature: ν )) ν )) ) ) ν )) ).

4 When there is no stress along or, it reduces to ) 4) where is the coefficient of epansion and the temperature rise. Bimetallic strip eample. bimetallic strip is heated up and not allowed to bend. What stresses are set up as a result? Solution Let the cross-sectional areas, moduli and coefficients of epansion be as indicated. Suppose the bimetallic strip is subject to a temperature rise. hen the stresses and strains in each strip are related b Hooke s Law 4) above: 5). We now appl the geometrical condition the strips are bonded together and there is no bending. his means that there is a constant strain in each strip, and the strains in the two strips are equal: 6). he third condition see page!) is equilibrium. he total horiontal force on the bimetallic strip is ero: 7). quations 5), 6) and 7) are four in number, so should be soluble for the unknowns,, and. We are onl interested in the stresses so proceed b eliminating the strains. Using 6) in 5) gives 4

5 5 8). Now from 7),. Substituting in 8) gives. his now needs to be rearranged to give. Put the two terms together: ) o give the stress ) which looks a bit nicer if we multipl top and bottom b : ). Notice that when the epansion coefficients are equal, there is no stress as ou would epect. Finall, we get using the epression obtained above,. his gives us ).