# Stress at the base of a chimney

• dvep
In summary, a steel chimney is 28 m high, 1.4 m external diameter, and 20 mm thick. It is rigidly fixed at the base and is acted upon by a horizontal wind pressure of intensity 1.1 kN/m2 on the projected area. The chimney weighs 74 kN per metre of height. Calculate the maximum stress in the chimney at the base.

## Homework Statement

A steel chimney is 28 m high, 1.4 m external diameter, and 20 mm thick. It is rigidly fixed at the base and is acted upon by a horizontal wind pressure of intensity 1.1 kN/m2 on the projected area. 2
The chimney weighs 74 kN per metre of height. Calculate the maximum stress in the chimney at the base.

## The Attempt at a Solution

total chimney weight = 74kN x 28 = 2072kN

bending moment due to wind pressure = 1/2 x 28^2 x 1100 = 341.2kN/m^2

How am I meant to add the forces when they are incompatible units?

Also do I use the formula, stress = Force/area or stress = My/I?

Hi dvep, welcome to PF!

The formula $\sigma=F/A$ is for axial loads; the formula $\sigma=-My/I$ is for bending moments. Does this help?

Mapes said:
Hi dvep, welcome to PF!

The formula $\sigma=F/A$ is for axial loads; the formula $\sigma=-My/I$ is for bending moments. Does this help?

Thank you for your reply. Can I ask why you have -My/I ? When I see it elsewhere as being positive.

So is this correct then:

stress due to weight = (weight x Area x height)/Area
= 2072000 x 28
=58016kN/m

I = 0.020645
ymax = 0.7 m

bending moment stress = My/I = (341200 x 0.7)/0.020645 = 11568.9kN/m

Maximum stress at base of chimney = stress due to weight + bending moment stress
= 58016000 + 11568.9
= 58027.5689kN/m

If not, I would be grateful for some help.

many thanks

Last edited:
dvep said:
stress due to weight = (weight x Area x height)/Area
= 2072000 x 28
=58016kN/m
?? As Mapes pointed out, the axial stress at the base is F/A, where F is the weight (which you calculated correctly the first time) and A is the cross sectional area at the base.
Bending Moment Stress = My/I = (341200 x 0.7)/0.020645 = 11568.9kN/m
Your moment M is incorrect. The wind pressure acts on the projected surface area of the part of the chimney facing the wind.
You must watch your units. Stress has units of force/area (N/m^2)

PhanthomJay said:
?? As Mapes pointed out, the axial stress at the base is F/A, where F is the weight (which you calculated correctly the first time) and A is the cross sectional area at the base. Your moment M is incorrect. The wind pressure acts on the projected surface area of the part of the chimney facing the wind.
You must watch your units. Stress has units of force/area (N/m^2)

So,

stress due to weight = 2072000/[(Pi*0.7^2) - (Pi*0.68^2) = 2072000/0.0867
= 23896.31kN/m^2

Bending Moment = 1100 x 28 x 14
=431.2kN/m^2

bending moment stress = My/I = (431200*0.7)/0.020645
=14620.49kN/m^2

Total stress = 14620.49kN/m^2 + 23896.31kN/m^2

Is that correct?

dvep said:

So,

stress due to weight = 2072000/[(Pi*0.7^2) - (Pi*0.68^2) = 2072000/0.0867
= 23896.31kN/m^2

Bending Moment = 1100 x 28 x 14
=431.2kN/m^2
No. Bending moment has units of force times length (Newton-meters). The uniformly distributed wind force on the pole is 1100 N/m^2 times 1.4 m = 1540 N/m, which acts along the full 28 m chimney height. So that's 1540 N/m times 28 m = 43120 N, or 43.12 kN total wind force. Now apply that force at the cg of the chimney, and calculate the moment at the base.

PhanthomJay said:
No. Bending moment has units of force times length (Newton-meters). The uniformly distributed wind force on the pole is 1100 N/m^2 times 1.4 m = 1540 N/m, which acts along the full 28 m chimney height. So that's 1540 N/m times 28 m = 43120 N, or 43.12 kN total wind force. Now apply that force at the cg of the chimney, and calculate the moment at the base.

ok so,

Bending moment at base = Total wind force x cg = 43.12 * 14 = 603.68kN/m

stress due to bending moment = My/I = (603680*0.7)/0.020645 = 20468.685kN/m^2

Total stress at base = 20468.684 + 23896.31
= 44364.994kN/m^2

?

dvep said:
ok so,

Bending moment at base = Total wind force x cg = 43.12 * 14 = 603.68kN/m

stress due to bending moment = My/I = (603680*0.7)/0.020645 = 20468.685kN/m^2

Total stress at base = 20468.684 + 23896.31
= 44364.994kN/m^2

?
Yes, that looks right; although i didn't check your math all that carefully, the method is good. This is the maximum compressive stress at the base, which occuirs at the far edge of the base. Note that kN/m^2 = kPa. By the way, I was in error earlier, the wind resultant force is applied at the cg of the distributed load, not the cg of the chimney (which happens to be the same in this example, so the answer is still correct). An alternate way to find the moment at the base of a cantilever member, under a uniformly distributed load, is to use the cookbook formula M = wL^2/2, where w has units of force per unit length. You've got to be careful with units and be sure to use the proper value for the distributed load. And check all zeroes before and after the decimal point...the SI system has many. Also, your answer should be rounded to 2 significant figures...max compressive stress = 44 000 kPa. You can't get a much better result than that.

dvep: I checked the math, and your answer is correct. By the way, please note the following international standard for writing units.

1. Always leave a space between a numeric value and its following unit symbol. E.g., 2072 kN, not 2072kN. See the international standard for writing units (ISO 31-0).

2. For long numbers having five or more digits, the international standard says you can write the digits in groups of three, separated by spaces. E.g., 0.020 645 m^4, instead of 0.020645 m^4.

3. As PhanthomJay pointed out, kN/m^2 is called kPa. And N/mm^2 is called MPa. Always use the correct, special name for a unit. E.g., 44 365 kPa, not 44 365 kN/m^2. Or even better, you could write this as 44.365 MPa.

dvep said:
Can I ask why you have -My/I ? When I see it elsewhere as being positive.

Here I'm using the convention that a positive bending moment bends a horizontal beam into a "smile" shape (vs. a "frown" shape), that positive y is up, and that tensile stress is positive. Then the minus sign indicates correctly that the stress is compressive in the upper half of the beam and tensile in the lower half of the beam.

The smile shape works for horizontal members such as the usual beams, but, with vertical members, you have to turn you head to one side. Which way? It doesn't matter that much. What matters is that you recognise that the axial stress is combined with the bending stress in two different ways. you take the extreme fibre stresses to be F/A + or - My/I. You can work out which stress is on the windward side, and which on the leeward side, by thinking of how the chimney deflects.

Lots of thanks guys, that was very helpful.