(a) Explain why the typical shape of an \"T\" section compared to a rectangular

Question

(a) Explain why the typical shape of an "T" section compared to a rectangular section makes (15 marks) (b) A simply supported 10.5 m long beam is required to carry a uniformly distributed dead load of 2.5 kN/m (including self-weight of the beam) and a live load of 4.2 kN/m. It is proposed that the 460UB82.1 may be a suitable section for the beam. If the deflection of this beam to be limited to span/250, check whether the deflection of the beam satisfies (35 marks) more suitable for bending applications. this requirement. Data: Density of the steel can be taken as 7850 kg/m Hint: Maximum deflection for a simply supported beam under uniformly distributed load 5wL is given by 384EI Load combimation for serviceability lim it state is G+0.7Q. (c) You are required to assess the tensile strength of a connection shown in Figure B3. As shown in the figure a 100x100x12 EA welded to a 10 mm thick gusset plate (Grade 300 to AS/NZS 3678). Determine the minimum value for 1 as shown in Figure B3 (30 marks) There is no need to consider or calculate the capacity of the gusset plate or the angle. Data: Design load (N*) 600 kN Use the leg size of the fillet weld -6 mm. Use Nominal tensile strength of weld metal ( w) as 430 MPa Weld N* Figure B3

Explanation / Answer

Answer to part a-

As we know, from bending equation theory –

?/ y = M/ I

Where ? = bending stress, y = distance of the section from neutral axis on which bending stress is being calculated.

M = Bending moment, I = Moment of inertia

Hence, ? = My/ I

So, ? is directly proportional to y, distance from neutral axis.

Hence, we can state that maximum bending moment is located at 2 ends of cross section which are at maximum distance from the neutral axis.

Hence, maximum area should be allocated at the far end and least at the middle.

I section suffice the criteria maximum out of the all section.

In circular section maximum mass is concentrate at centre opposite of the ideal condition.

The detailed derivation of bending moment equation is shown below-

Derivation of bending moment equation-

Suppose bending moment M is applied on a beam, the resultant beam would be something like this-

As you can see, bottom section of the beam is extended from its original self hence, length of the section is more as compared to original one. And length of upper section of the beam is reduced compared to the previous length. Hence, somewhere between these sections, there will be zero deflection line. Which is called neutral axis. Where length of the section will remain same as previous length. Please note that we are considering this section in pure bending.

Now we consider some irregular figure shown below,

Now consider section shown here at y distance from neutral axis.

Say this section is AB.

After bending this will stretch to A’B’.

Now, as we know

Strain = change in length/ original length

Strain = A’B’-AB/AB

Now, consider Neutral axis as CD before bending, and length of neutral axis C’D’ after bending.

From definition of neutral axis we know,

CD = C’D’

Hence, strain = A’B’-C’D’/AB

Now in above figure, beam will be bent till the radius R and project angle ?.

Hence, as distance of AB is y, after bending A’B’ = (R+y) ?

CD = C’D’ = R ?

Strain = (R+y) ?-R ?/ R ? = y/ R

As we know, strain = Stress/ E

Hence

Stress/ E = y/ R

Or

?/ y = E/ R

Now the termis the property of the material and is called as a second moment of area of the cross-section and is denoted by a symbol I.

Therefore

This equation is known as the Bending Theory Equation.The above proof has involved the assumption of pure bending without any shear force being present. Therefore this termed as the pure bending equation. This equation gives distribution of stresses which are normal to cross-section i.e. in x-direction.

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