Exercise F-1.1.3 TechMech 2 Mechanics of Materials

Practice Exercise F-1.1.3

Uniaxial Stress State: Loaded Homogeneous Bar with Constant Thickness and Linearly Varying Width

Problem Statement

A homogeneous bar with constant thickness $d=20~\mathrm{mm}$ and linearly varying width is subjected to a compressive force $F=1~\mathrm{kN}$.

The rod stands on an indicated ground. Its lower base is 2b x d, its upper base is b x d. Its height is dimensioned as h. From the top, it is subjected to a force F, represented by a downward arrow, along its axis of symmetry. The running coordinate x extends from the bottom to the top. The gravitational acceleration g is indicated with an arrow pointing downward. — Fig. 1: Homogeneous bar with constant thickness and linearly varying width

Given: $b=15~\mathrm{mm}$, $h=80~\mathrm{mm}$, $\varrho = 7,85~\mathrm{g/cm^3}$

What is the formula for calculating the cross-sectional area $A(x)$ at any location x in a section perpendicular to the axis of the rod?
What is the formula for calculating the normal force $N(x)$ at any location x in a section perpendicular to the axis of the rod?
What is the magnitude of the normal stress $\sigma$ at the location $x=30~\mathrm{mm}$ in a section perpendicular to the axis of the rod?

Short Solution

What is the formula for calculating the cross-sectional area $A(x)$ at any location x in a section perpendicular to the axis of the rod?

$$ A(x) =b\cdot d \cdot \Bigl(2-\dfrac{x}{h}\Bigr) $$

What is the formula for calculating the normal force $N(x)$ at any location x in a section perpendicular to the axis of the rod?

$$ N(x) = \dfrac{1}{2} \varrho \cdot g \cdot b \cdot d \Bigl[x\Bigl(4 - \dfrac{x}{h}\Bigr) - 3h\Bigr] - F $$

What is the magnitude of the normal stress $\sigma$ at the location $x=30~\mathrm{mm}$ in a section perpendicular to the axis of the rod?

$$ \sigma(x=30~\mathrm{mm}) = -2,054 ~\frac{\mathrm{N}}{\mathrm{mm^2}} $$

Comprehensive Solution

Preliminary Considerations

So, you want to crack the stress state in the rod? Cool stuff! But no worries, it's a piece of cake.

First things first, chill out! The rod stands firm, not toppling over anytime soon. The compressive force is pushing down, including the weight force of the rod. But the support reactions hold their ground. That creates a uniaxial stress state – all easy, right?

Ultimately, in task c), you're on the lookout for the normal stress. For that, you need this formula:

$$ \begin{aligned} \sigma_x = \dfrac{N_x}{A} \end{aligned} $$

(1.3)

Okay, okay, looks complicated, but it's not that bad.

$N_x$ is the normal force in the cut. It depends on the weight force of the cut-off section of the rod. And thus, on the position $x$ in the rod: The higher you cut, the less rod mass presses downward, the smaller $N_x$.

$A$ is the cross-sectional area of the rod at point $x$. The higher you cut, the smaller the area, the smaller $A$.

You see, both $N_x$ and $A$ change with $x$. That's why you have to use formula 1.3b to calculate stress at every point $x$.

$$ \begin{aligned} \sigma(x) = \dfrac{N(x)}{A(x)} \end{aligned} $$

(1.3b)

But no panic! With a bit of math and patience, you've got this covered.

Tip: Imagine slicing the rod into many thin slices. Calculate the stress for each slice. That way, you get a feel for how stress is distributed in the rod.

Good luck! And if you're ever stuck, just peek into the complete solution below.

P.S.: Don't forget that the rod is made of a specific material. The material influences the stress. So, you still need to consider the material properties.

P.P.S.: Oh, and the units! Be careful to use the right units. Otherwise, the result won't be right in the end.

What is the formula for calculating the cross-sectional area $A(x)$ at any location x in a section perpendicular to the axis of the rod?

According to the problem statement, the rod has a constant thickness $d$ and a linearly varying width $B(x)$:

This illustration shows that the cross-sectional area of the rod changes with the position x. — Fig. 2: Cross-sectional area of the rod

The cross-sectional area is rectangular at each position $x$. Therefore, the formula is:

$$ \begin{align} \tag{1} A(x)=B(x) \cdot d \end{align} $$

We need to express the factor $B(x)$ as a function of $x$, while the factor $d$ is not dependent on $x$ and remains constant.

Due to the symmetry of the width view and the constant thickness $d$, we can simplify our considerations by only focusing on the range from the centerline to the outer edge.

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