Does Timber Beam Deflection Exceed Safe Limits?

[k_blow]

Will a timber beam deflect more than is safely permitted?
If a timber beam has dimensions of 225 mm by 75 mm and holds a UDL of 3.5 kN/m over a span of 4m. And If Young’s modulus for the timber is 10 kN/mm2, how much will the beam deflect?


A joist bending moment calculation?
A joist has section modulus 187.5 x 103 mm3 and the span is 3.8 m. It carries a Uniformly Distributed Load (UDL) of 1.7 kN/m.

What is the maximum bending moment for the joist

 

[PhanthomJay]

Will a timber beam deflect more than is safely permitted?

A beam is not safe if its maximum stress exceeds the allowable stress for that material. The deflection is not safe only if the stresses are not safe, or if a Code requires a deflection limit (to prevent plaster cracking, for example, if it is a floor beam supporting a ceiling, or for appearance,unsightliness, etc.)

If a timber beam has dimensions of 225 mm by 75 mm and holds a UDL of 3.5 kN/m over a span of 4m. And If Young’s modulus for the timber is 10 kN/mm2, how much will the beam deflect?

Please show your attempt at a solution. What are the end conditions? Which way is the beam oriented? Beam deflections for uniform loads are given in tables, as a function of the load, span, elasticity modulus, and geometric properties of the beam; or can be calculated using calculus.

A joist bending moment calculation?
A joist has section modulus 187.5 x 103 mm3 and the span is 3.8 m. It carries a Uniformly Distributed Load (UDL) of 1.7 kN/m.

What is the maximum bending moment for the joist

Please show your work attempt, please.

 

[piercas]

?

I would approach this question by first determining the maximum bending moment for the joist using the formula M = WL^2/8, where M is the maximum bending moment, W is the uniformly distributed load, and L is the span of the joist. Plugging in the given values, we get M = (1.7 kN/m)(3.8 m)^2/8 = 2.79 kNm.

Next, I would calculate the deflection of the beam using the formula δ = (5WL^4)/(384EI), where δ is the deflection, W is the uniformly distributed load, L is the span of the joist, E is the Young's modulus, and I is the moment of inertia. The moment of inertia for a rectangular beam is (bh^3)/12, where b is the width of the beam and h is the height. Plugging in the given values, we get I = (75 mm)(225 mm^3)/12 = 1.64 x 10^6 mm^4.

Substituting all values into the formula, we get δ = (5)(1.7 kN/m)(3.8 m)^4/(384)(10 kN/mm^2)(1.64 x 10^6 mm^4) = 0.012 mm.

Based on this calculation, it appears that the deflection of the beam is well within the safe limit. However, it is important to note that this is a theoretical calculation and other factors, such as the quality of the timber and any potential defects, should also be considered to ensure the beam can safely withstand the given load. It is always best to consult with a professional structural engineer for an accurate assessment and to ensure the safety and integrity of any structure.