Four 1,9-m pipes, having outside diameters of 60 mm and inside diameters of
52 mm each, are used as the legs of a stand.
The maximum compressive stress in the pipe material must not exceed
65 MPa . Young’s modulus for the pipe material is 210 GPa .
6.1.1 Calculate the maximum mass that can be placed on the stand.

Ask by Harris Turner. in South Africa

Mar 30,2025

Question

Four 1,9-m pipes, having outside diameters of 60 mm and inside diameters of
52 mm each, are used as the legs of a stand.
The maximum compressive stress in the pipe material must not exceed
65 MPa . Young’s modulus for the pipe material is 210 GPa .
6.1.1 Calculate the maximum mass that can be placed on the stand.

Ask by Harris Turner. in South Africa

Mar 30,2025

UpStudy AI · Accepted Answer

**Step 1. Compute the cross‐sectional area of one pipe**

The pipe is a hollow circular tube. Its cross‐sectional (wall) area is given by

$$
A=\frac{\pi}{4}\left(D_o^2-D_i^2\right)
$$

where 
- $$D_o=60$$ mm is the outside diameter,
- $$D_i=52$$ mm is the inside diameter.

Calculate the squares:

$$
60^2=3600,\quad 52^2=2704
$$

Thus,

$$
A=\frac{\pi}{4}(3600-2704)=\frac{\pi}{4}(896)
$$

Since $$896\div4=224$$, we can write

$$
A=224\pi\quad \text{(mm}^2\text{)}
$$

---

**Step 2. Maximum compressive load on one pipe**

The maximum compressive stress allowed is

$$
\sigma_{\max}=65\; \text{MPa}=65\; \text{N/mm}^2
$$

For one pipe the maximum force is:

$$
P_{\text{pipe}}=\sigma_{\max}A=65\; \text{N/mm}^2 \times 224\pi\; \text{mm}^2=65\times224\pi\; \text{N}
$$

Calculating the numerical product:

$$
65\times224=14560
$$

Thus,

$$
P_{\text{pipe}}=14560\pi\; \text{N}\approx14560\times3.1416\approx45754\; \text{N}
$$

Since there are 4 pipes in parallel, the total allowable load is

$$
P_{\text{total}}=4\times P_{\text{pipe}}=4\times45754\; \text{N}\approx183016\; \text{N}
$$

---

**Step 3. Maximum mass that can be placed on the stand**

Using the relation between force and weight, where $$g=9.81\; \text{m/s}^2$$:

$$
P_{\text{total}}=Mg
$$

Thus the maximum mass is:

$$
M=\frac{P_{\text{total}}}{g}=\frac{183016\; \text{N}}{9.81\; \text{m/s}^2}\approx18655\; \text{kg}
$$

---

**Step 4. Shortening of the pipes under the maximum load**

Each pipe acts as a compressive member. The shortening (axial deformation) is given by

$$
\delta=\frac{PL}{EA}
$$

For one pipe:
- $$P=P_{\text{pipe}}=45754\; \text{N}$$
- $$L=1.9\; \text{m}$$ (pipe length)
- $$E=210\; \text{GPa}=210\times10^9\; \text{N/m}^2$$
- The cross‐sectional area $$A$$ must be in $$\text{m}^2$$. Since
  $$
  A=224\pi\; \text{mm}^2=224\pi\times10^{-6}\; \text{m}^2,
  $$
  
Substitute these into the deformation formula:

$$
\delta=\frac{45754\; \text{N}\times1.9\; \text{m}}{210\times10^9\; \text{N/m}^2 \times 224\pi\times10^{-6}\; \text{m}^2}
$$

Compute the numerator:

$$
45754\times1.9\approx86933\; \text{N}\cdot\text{m}
$$

Now the denominator:

$$
210\times10^9\times224\pi\times10^{-6}=210\times224\times\pi\times10^{9-6}=210\times224\times\pi\times10^3
$$

Calculate $$210\times224$$:

$$
210\times224=47040
$$

Thus the denominator is:

$$
47040\pi\times10^3\; \text{N} \approx47040\times3.1416\times10^3\; \text{N}\approx147823200\; \text{N}
$$

Now, the shortening is

$$
\delta\approx \frac{86933}{147823200}\; \text{m}\approx0.000588\; \text{m}
$$

Converting to millimeters:

$$
0.000588\; \text{m}\times1000\approx0.588\; \text{mm}
$$

---

**Final Answers**

1. The maximum mass that can be placed on the stand is approximately

$$
M\approx1.87\times10^4\; \text{kg}
$$

2. Under this load, each pipe will shorten by approximately

$$
\delta\approx0.59\; \text{mm}
$$
The maximum mass that can be placed on the stand is approximately 18,655 kg. Each pipe will shorten by about 0.59 mm under this load.

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