The steel rod A B C is attached to rigid supports and is...

The steel rod $A B C$ is attached to rigid supports and is unstressed at a temperature of $25^{\circ} \mathrm{C}$. The stecl is assumed elastoplastic with $E=200$ GPa and $a_{y}=250$ MPa. The temperature of both portions of the rod is then raised to $150^{\circ} \mathrm{C}$. Knowing that $\alpha=11.7 \times 10^{-6} /^{\circ} \mathrm{C}$ determine $(a)$ the stress in both portions of the rod, $(b)$ the deflection of point $C$.

Key Concepts

Thermal Expansion

Thermal expansion is the tendency of a material to change in length upon a change in temperature, characterized by the coefficient of thermal expansion (?). When a material’s temperature increases, it tends to expand, and the change in length can be calculated using the relation ?L = ?·?T·L, where ?T is the temperature change and L is the original length. This concept is fundamental in understanding how temperature variations affect structural components.

Thermal Stress in Constrained Structures

When thermal expansion is restricted by supports or boundary conditions, the inability of the material to freely expand or contract induces thermal stresses. In a constrained member, the thermal strain that would have occurred in a free expansion is counteracted by mechanical strain, leading to internal stresses. The induced stress can be estimated in the elastic range by ? = E·?·?T, but if this value exceeds the material’s yield strength, plastic deformation may occur.

Elastic Modulus and Yield Strength

The elastic modulus (E) is a material property that measures the stiffness of the material, defining the linear relationship between stress and strain in the elastic region. The yield strength (or yield stress, a_y) is the stress level at which a material begins to deform plastically and will not return to its original shape when the load is removed. Both parameters are critical in determining whether a structure remains in the elastic range or experiences permanent (plastic) deformations under applied loads, including thermal stresses.

Elastoplastic Material Behavior

Elastoplastic behavior describes how materials respond to stress beyond the elastic limit. Initially, the material behaves elastically according to Hooke’s Law until the stress reaches the yield point. Beyond this point, the material deforms plastically, meaning that additional deformations occur at a nearly constant stress level. This concept is essential when analyzing structures subjected to thermal loads, as the induced stresses might lead to yielding and permanent deformations if the temperature change is significant.

Transcript

00:02 Okay, so the steel rot abc is attached to rigid supports and is unstressed at 25 degrees c.

00:10 Steel is assumed elastoplastic and our energy is 200 gigapascals, and we've got a sigma that is 250 megapascals.

00:26 The temperature in both portions of the rod is raised to 150, knowing that we've got, let's see here.

00:44 Our stress, we've got a constant 11 .7 times 10 to the minus 6 per degree c.

00:52 We're going to find the stress in both portions of the rod and the deflection of c.

00:57 Okay.

00:59 So let's figure out what we're doing.

01:02 We'll have to get some data changes here with our area of a, c will equal, that was 500.

01:17 So 500 times 10 to the minus 6 meters squared.

01:24 Our area of cb will be 300 times 10 to the minus 6 meters squared.

01:36 Our l of ac and our l of cb will be 0 .150 meters and 0 .250 meters.

01:48 Okay.

01:51 And then we're going to have our deformation due to change in temperature and produce force should be zero.

02:05 Okay.

02:05 And our required temperature to cause yielding.

02:21 Okay.

02:21 This is going to take moment.

02:37 L -a -b.

02:42 Okay.

02:42 So let's get this delta t out of there.

03:03 Okay.

03:26 Still here.

03:30 Okay.

03:30 Now let's substitute some values in.

03:35 Move this over.

03:38 That'll be 300.

03:40 Times 10 to the minus 6 times 2 .5 times 10 to the 6 over 0 .40 times 10 to the 9th.

04:01 And i believe that these are pascal's and 11 .7 times 10 to the minus 6 for c degrees.

04:19 We'll take that times 0 .15 over 500.

04:27 Times 10 to the minus 6 plus 0 .250 over 300 times 10 to the minus 6.

04:40 Okay, so we can solve for this.

04:43 We'll get a delta t of 90 .812 degrees c...

Please give Ace some feedback