4"x4" openings in top&bottom flange of W14x68

[Heldbaum]

Hi folks
I have a problem and I would need your help to solve it - Contractor needs to do two openings in a hoist beam at elevator room level to run ropes from the motor - they encroaches onto the machine beam by 4" ( almost half of the flange width, beam is W14x68, 9' long).See attached sketch. They need to know if they can do that and what reinforcement would they need to use to make it possible (if any neeeded). I know reactions at the ends of the beam. I'd need to check if approx. 6" out of 10" overall flange width would work, if not, design the reinforcement, correct? Any thoughts? Do you know of any guides or examples that would be helpful ?
Thanks a lot for your help!

 

[oldestguy]

I'd suspect the supplier of the elevator (including hoist) would want to OK any modifications to the supports there.

 

[JLNJ]

An engineer would need the loads, not just the reactions. You compare the net section's properties to the required section derived from the loads and use some judgment.

If it doesn't check out, it can probably be reinforced for strength and stiffness. Since these appear to be head house/hoist room beams, fatigue would be a concern for any welded reinforcements.

 

[SlideRuleEra]

Do you know of any guides or examples that would be helpful?

No, each situation like this is unique. Show the locations of the cable interference and how all loads are applied to the beam. Any solution will start with those dimensions.

http://www.SlideRuleEra.net

 

[Heldbaum]

Thank you guys for your help.

SlideRuleEra - please take a look at attached screen shot. These are my loads that I've assumed. I've got an elevator specification for elevator company with all the loads and elevator capacity. Motor load and other equipment weights 16,500 lbs and elevator capacity is 3,500 lbs so I added those number together and multiplied by 2 as code recommends for impact load(dead load shall be multiplied by 2) so I got total suspended load of 40 kips supported at four points (2 points per one machine beams, 10 kips at each point) Then I assumed life load of 100 psf (there is grating between machine beams ) and tributary area is approx 4' so I have uniform load of 400 plf. Am I missing something ?
The openings 4"x4" in top&bottom flange that are under consideration will be located close to where point loads are(10" from the right side and then another one 4' away). See attached screenshot. MAchine beam at the right side is supported by RC shear wall and on the other side it sits on top of divider beam. Machine Beam length is approx. 8'-6", W14x68. It seems like an overkill to me, the beam size, under above loads the deflection is 0.02 whereas code says L/1666 which is 0.06 in.

I was thinking to run perpendicular beam or angle to tie these two machine beams together at the locations where the opening in flanges needs to be done, I think the one in the mid span is critical. And weld a flat bar to both flanges to compensate for what needs to be removed. What do you think ? I'd appreciate your help. Thanks

 

[SlideRuleEra]

Heldbaum - Need to find out what the modification (top & bottom flange width reduction) will do to the beam's properties before working on a "fix".

Per JLNJ's suggestion calculate the moment of inertia and section modulus of the remaining steel.
Assume the beam is a pair of rectangular flanges and the web is another rectangle:

Perform the composite section calcs on both an unmodified W14x68 and the modified section. Use the decimal dimensions of flange thickness, web thickness, beam height, etc. from the AISC Manual of Steel Construction. At this stage, do not round off any dimension, doing so will introduce unnecessary errors in the calcs. There are several ways to make these calcs. The simplified assumption of rectangular flanges and web will give results for "I" and "S" that are slightly low. By doing the calcs on an unmodified W14x68, we will know how low (there is a little more steel than represented by the simplified assumption). What do you get?

http://www.SlideRuleEra.net

 

[Heldbaum]

SlideRuleEra - Thanks for your help

Unmodified section properties Ix=722 in4 Iy=121 in4; Sx=103 in3 Sy=24.2 in3

For modified section I got Ix = 476.818 in4 Iy=38.426in4 so Sx=68.117 in3 and Sy=10.337 in3 . For loads that I figured (See attachment in my previous post) I got max. bending moment at the application of the force Mmax = 29 k-ft so the stress at this point Mmax/Sx seems to be around 5 ksi..

 

[SlideRuleEra]

Good. Solving by hand, I have I_x = 711 in⁴ and S_x = 101 in³ for the unmodified beam.
For the modified beam, Ix = 453 in⁴ and Sx = 65 in³.

If you are confident in your values (they look reasonable to me), round off to I_x = 477 in⁴ and S_x = 68 in³.

Assume the entire beam, all 8.5 feet, has the modified cross section. Calculate x-axis maximum shear stress, bending stress and deflection. Your results? We can discuss this tomorrow afternoon.

http://www.SlideRuleEra.net

 

[Heldbaum]

Ok will do that tomorrow morning. Thank you.

 

[mtu1972]

I agree with your section properties and the 5 ksi stress at the one set of openings. As the stresses are that low, I would not be overly concerned with reinforcing. I may add additional member(s) as you noted above to provide lateral bracing to the beam being modified. Maybe one at mid span would be enough.

gjc

 

[JLNJ]

Do you have a counterweight-type system? Did you add the cab self-weight to the lifted capacity? Are there traction force to consider?

The stress seems quite low. Make sure you have all your bases covered.