How to Calculate Deflection

Sounds like homework.

yet 2 guru's are fighting

Weld a plate over the hole position either side before drilling it, and then ignore the hole.
Del

In this case I can't because the support pin needs to be removable, otherwise it's a good idea.

The reduced moment of inertia at the center of hole is:

I_r = I_tube - I_h

where I_h = 2*t*h³/12 = moment of inertia of the hole through both walls.

t = wall thickness

h = height of hole

The moment of inertia is variable across the width of pin because the height of hole varies from 1" to 0". This variation makes deflection calculations more difficult, so you may want to make some reasonable assumption as to the variation of 'I' over that short distance. In practice, it would probably be ignored and the moment of inertia would be taken as constant over the entire length of beam.

The loss of strength at the middle of the pin is the plastic moment capacity of the removed material or 2*F_y*t*h²/4 where F_y is the yield strength of the material.

WRONG you can sum or deduct ONLY moments of inertia but not the W (resistance )! The loss in this case is less than what you wrote. Rethink it.

Sorry nickname, but you are the one who is wrong! You rethink it!

Fred,

The loss in strength is quite small. The strength or plastic moment of a rectangular tube may be expressed as Z*F_y where Z is the plastic modulus of the tube and F_y is the yield strength of the material.

Approximating the shape to a rectangular ring, i.e. ignoring the rounded corners, Z = 1/4(2*3² - 1.75*2.75²) = 1.191 in³.

The reduction in plastic modulus for a 1" dia. hole is 1/4(2*1/8*1²) = 0.0625 in³. This reduces Z to 1.129 in³ or a reduction of about 5%. The strength is reduced by the same percentage.

In the Canadian code (other codes may be different), we use a resistance factor, Φ to account for uncertainty in predicting member resistance. In the case of bending, Φ is taken as 0.9 so that the strength of tube without hole is 0.9*Fy*1.191 and the strength of tube with hole is 0.9*Fy*1.129, or approximately 95% of the strength of the tube.

The actual Z taken from CISC Handbook of Steel Construction for a 76 x 51 x 3.2 HSS (3" x 2" x1/8" HSS) is 17000 mm³ or 1.037 in³. The reduced Z, accounting for the hole is (1.037 - 0.0625) = 0.975, a reduction of about 6.4% from the shape without a hole. Strength is reduced in the same proportion.

Thanks Bruce! I'm still trying to get over how relatively small the loss in strength is considering that 1/3 of the sidewall material on each side is gone. Supposing the hole was enlargened to the maximum such that there are essentially two horizontal strips of metal left (looking into the tube) above and below the hole, then the moment of inertia is:

Iresultant = Isolid - Iinside - Ihole

Iresultant = (2*3^3)/12 -(1.75*2.75^3)/12 - (0.125*2.75^3)/6

Iresultant = 4.5 - 3.033 -0.433 = 1.034 in^4.

Now if the moment of inertia is calculated for just two apparent horizontal strips above and below the hole, we get:

Iresultant = 2*0.125^3/6 = 0.000651 in^4 which is very small compared to the 1.034 in^4. Why does this not jive? (Sorry, my background is electrical engineering!).

Thanks for all the discussion so far, it's very enlightening.

Fred,

If the sidewalls are removed completely, leaving only a strip of 2" x 1/8" top and bottom, the resulting moment of inertia may be determined in a couple of ways.

1) I = (2*3³ - 2*2.75³)/12 = 1.033854 in⁴.

2) I = 2(2*0.125³)/12 + 2*0.125*{(3-0.125)/2}²

= 0.000651 + 1.033203 = 1.033854 in⁴.

I do not usually carry out calculations to such precision, but you can see that the two methods agree precisely.

The first method simply considers a rectangle of 2 x 3 and subtracts a rectangle of 2 x 2.75 . No correction has to be made because both rectangles are centered on the neutral axis.

The first term in the second method results in a quantity so small, it is usually ignored. But the expression I = t*h³/12 is valid for the moment of inertia of a rectangle about its centroid. If you want its moment of inertia about some other point, you must add the term A*d². In this case, the centroids of the two plates are separated by a distance 3-0.125 = 2.875" so each plate is 1.4375" from the neutral axis.

Hope this clears it up for you.

Thanks so much Bruce! Nothing like exploring a real problem. Yes, that's the missing part which explains the difference between two separate strips and a pair connected as a structure. Thanks again!

The deflection depends on the art the beam is fixed at the left end. You can neglect the effect of the holes for the deflection and use the inertia of the full section. Assuming that at left end the beam can rotate in its "bearing" the deflection at the right end will be about 0.06" for the load you gave. If the rotating freedom at the left end is limited the deflection at right end will be less than above given figure. The effect represents a loss of only about 1.5%. As long as you stay in the elastic domain the effect of the holes is totally neglectable for the loading capability of the beam.

I hope I have not a computing error since I am less accustomed to deal with inches and lbs. If there is any I shall be glad to have the correct value.

nick name,

You are quite right. The moment at the fixed end is exactly half of the moment at the pin support. M_pin = 121*17 = 2057"#. M_fixed = 1028.5"# (both moments clockwise). I agree that the effect of the hole on deflection is negligible, but I thought Fred was interested in the method of calculating it so that he might apply it to other situations.