Determine the thermal stresses that develop when the temperature of one member is increased.
A quasi-rigid link pivots about a hole at its right end. The link rests atop a cylindrical brass rod. A cylindrical steel rod anchors down the left end of the link. The temperature of the brass rod is increased from 20° C (the stress-free temperature of the assembly) to 50° C. The resulting thermal expansion induces a tensile stress in the steel hold-down rod. Since the brass rod is not free to expand unimpeded, an axial compressive stress develops in it. We will compare this axial stress to the theoretical solution.
Rigid elements are not available in the subject analysis type. Therefore, quasi-rigid behavior is achieved by making the link relatively large and setting the material stiffness to the maximum supported value. In the diagram that follows, all dimensions are in millimeters.
Property | Link | Brass Rod | Steel Rod |
---|---|---|---|
Modulus of Elasticity (MPa) | 1.3 x 106 | 105,000 | 200,000 |
Poisson's Ratio | 0.3 | 0.31 | 0.3 |
Thermal Expansion Coefficient (/ °C) | 1.2 x 10-5 | 1.88 x 10-5 | 1.2 x 10-5 |
Holes in the link for both rods are sized for clearance. Therefore, contact only occurs in two locations:
The settings for both contact sets are as follows:
Considering the preceding diagram, we have the following relationship from the Beer and Johnston reference:
ΣME = 0 RA * (0.75 m) - RB * (0.3 m) = 0 RA = 0.4 RB
The method of superposition is used. With RB removed, the temperature rise of the cylinder causes point B to move down a distance δT. The reaction RB causes a displacement δ1 of the same magnitude as δT so that the final deflection of point B is zero.
Due to a temperature rise of 30° C (50° C - 20° C), the length of the Brass Rod, when unconstrained, increases by δT according to the following equation:
δT = L(ΔT)α = (0.3 m)(30° C)(1.88 x 10-5/° C) = 0.0001692 m
We note that δD = 0.4 δC and that δ1 = δD + δB-D, where δB-D is the change in length of the Brass Rod.
δC = RAL / (AE)
Given that, for the Steel Rod, L = 0.9 m, A = πD2/4, D = 0.03 m, and E = 105 x 109 Pa:
δC = 4 RA(0.9 m) / [π(0.03 m)2(105 x 109 Pa)] = 11.83797097 x 10-9 RA δD = 0.4 δC = 0.4 (11.83797097 x 10-9 RA) = 4.73518839 x 10-9 RA δB-D = RBL / (AE)
Given that, for the Brass Rod, L = 0.3 m, A = πD2/4, D = 0.022 m, and E = 105 x 109 Pa:
δB-D = 4 RB(0.9 m) / [π(0.03 m)2(105 x 109 Pa)] = 4.04203030 x 10-9 RB
Given that RA = 0.4 RB:
δ1 = δD + δB-D = [4.73518839 x 10-9 (0.4 RB) + 4.04203030 x 10-9 RB] = 5.936105657 x 10-9 RB
But, δT = δ1:
0.0001692 m = 5.936105657 x 10-9 RB RB = 28,503.536 N
σB = RB/A = 4 (28,503.536 N) / [π (0.03 m)2] = 40,324,254 Pa
Therefore, σB = 40.324254 MPa
The stress in the Brass Rod is essentially axial. However, the solid representation of the rod, and the way that it is constrained, make it possible for bending deflection and bending stresses to occur. This behavior is not experienced in a true truss, which is the basis of the theoretical solution. To ensure that we do not include bending effects in our comparison, we select a point along the neutral axis of the rod (where the bending stress component is zero). In addition, we do not want to check stresses close to a constraint or contact area, where the results may be skewed by local effects. Therefore, we compare the theoretical axial stress of the truss (Brass Rod) to the Fusion Normal YY Stress result at the centroid of the Brass Rod (-300, -150, 0). The minus sign of the result indicates that the stress is compressive, as expected (whereas positive stress components are tensile):
Location | Theoretical Result (Beer and Johnston) | Fusion Result | % Difference |
---|---|---|---|
Centroid of Brass Rod | -40.324254 MPa | -39.805 MPa | -1.288 % |
Despite these known differences between the theoretical and solid finite element models, the results are still quite close.
Beer, Ferdinand P. and Johnston, Jr., E. Russell, Mechanics of Materials , McGraw-Hill, Inc., 1981, sample problem 2.4, page 58.