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I'm planning to install a wall-mounted TV with a very long arm on a plasterboard wall, although conventional wisdom advises against it. The TV mount is 45cm wide with holes for 6 screws. To address my concerns about wall strength, I intend to place a metal flat bar on top of the mount, increasing the total width to > 120cm and allowing for more screws, and thus also be able match the drilling points with the behind-wall metal studs framing, 4 in total, one every 40cm.

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The plug/anchors I've acquired are rated for a known strength of at least 430N.

Key Details:

  • Wall material: Plasterboard (2x15mm), but with attachment points on metal profiles. Not for study the strength of the profiles.
  • TV weight: 20kg, which is approximately 200 Newtons (N) for simplification.
  • The TV arm can be extended to either 10cm (rest) or 100cm (fully extended) from the wall.
  • The load capacity of my screws/plugs/anchors is 0.43kN (430N). However, I suspect this value is for direct attachment to the wall with no distance. enter image description here

I'm trying to understand the physics involved in this situation to ensure adequate support. When the TV is placed further from the wall, it exerts torque, and I'd like to calculate the number of plugs/anchors required for each scenario.

  • TV 1cm from the wall: Torque = 0.01m × 200N = 2 N⋅m
  • TV 10cm from the wall: Torque = 0.1m × 200N = 20 N⋅m
  • TV 100cm from the wall: Torque = 1m × 200N = 200 N⋅m

I've come up with some components of the formula, but I believe I'm missing a few:

Torque (τ) = Lever Arm (r) × Force (F)

number of screws required = Force applied / 430N

Essentially, I'm seeking guidance on how to account for the torque effect when determining the anchor requirements for the TV mount.

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  • $\begingroup$ They used studs even when walls were wood and plaster instead of drywall. Your load needs to reach those studs. Once that happens, you will be set. $\endgroup$
    – Abel
    Oct 24, 2023 at 0:26

1 Answer 1

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The important thing here is taking into account the dynamic load on the mount. Such as the impact of the vibration of banging a door or swinging the TV bracket into a new position. A dynamic load factor of 4 or 5 is reasonable. Adding the top flat bar helps but the effect is undeterminate because the flat bar is flexible in the plane you intend to use.

The Gypsum board is very brittle. Brittle materials create stress concentration through sharp crack development and break suddenly with no warning in an explosive fashion. So I would assume the entire load of tension on top of the bracket on the closest screw. If the closest screw fails the failure would propagate through the adjacent screws in a cascade of breaks.

Therefore, the crude, basic answer to your question is if we call the tension on the Top mount screw, T, and the height of the bracket, H:

$$T=\frac{5*200Nm}{H}$$ For example, assuming $ \ H=20cm= 0.2m$

$T=25*200N= 5000N>>430N $

Meaning the system will be woefully inadequate.

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  • $\begingroup$ Few things, although I have a gypsum wall, I will make sure the drilling points matches with the behind-wall metal studs framing. 430N is for a single anchor, based on your answer, 12 would be enough, right?. I didn't realised the height was so decisive, I could mount another vertical steel flat bar so to increase the system height. What is more effective, over the top, or the bottom screws line? $\endgroup$
    – Mikel
    Oct 24, 2023 at 7:27
  • $\begingroup$ on the bracket, you have to consider all member forces, momentum, shear, and torsion. Just a rough idea for a 1-meter-long swing mechanism you would consider the weight of the bracket and the overturning moment it imparts to the stud. you probably need to hire an engineer. $\endgroup$
    – kamran
    Oct 24, 2023 at 14:50

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