# Mechanical Engineering - Strength of Materials - Discussion

Discussion Forum : Strength of Materials - Section 1 (Q.No. 33)

33.

The maximum bending moment for the beam shown in the below figure, lies at a distance of __________ from the end

*B*.Discussion:

18 comments Page 1 of 2.
Salahuddin said:
5 months ago

No need to solve the bending moment equation.

BM is max where SF is zero. Therefore take a section x-x at a distance x from end B. Write SF equation and equate it to zero.

Thank you.

BM is max where SF is zero. Therefore take a section x-x at a distance x from end B. Write SF equation and equate it to zero.

Thank you.

(2)

Sweetu said:
6 years ago

Bending moment will max. Where s.f is zero.

Take distance X m from B.

Nd s.f. = (wl/6)-(wX*X/2)=0.

So X=l/√3.

Take distance X m from B.

Nd s.f. = (wl/6)-(wX*X/2)=0.

So X=l/√3.

(1)

Vikki said:
6 years ago

Bending moment will maximum where shear force is zero.

We have shear for at dist X from point A is,

F= WL/6-[Wx^2/(2L)]. Since F=0.

WL/6=Wx^2/(2L),

On simplification we get x=L/√3.

We have shear for at dist X from point A is,

F= WL/6-[Wx^2/(2L)]. Since F=0.

WL/6=Wx^2/(2L),

On simplification we get x=L/√3.

(1)

Vikesh said:
6 years ago

How you got -WX?

In equation (WL/3)-Wx+[ Wx^2/(2L)].

Here WL/3 is Ra.

And Wx^2/(2L) for UVL.

Then Remaining one is for what?

In equation (WL/3)-Wx+[ Wx^2/(2L)].

Here WL/3 is Ra.

And Wx^2/(2L) for UVL.

Then Remaining one is for what?

AMOL said:
6 years ago

First calculates reactions.

Ra = WL/3.

Rb = WL/6.

If we take section at distance x from A.

And shear force is calculated for this section.

We get S.F. at this section= (WL/3) - Wx + [ W(x2) / (2L)].

For maximum B.M. shear force at this section should be zero.

therefore,

(WL/3) - Wx + [ W(x2) / (2L) ] =0.

Cancel W from each term.

We get second order equation.

(L/3) - x + [ (x2) / (2L) ] =0.

[ (x2) / (2L) ] -x + (L/3) = 0.

If we solve this equation.

We get,

x = [ 1(+or-)(1/sqroot 3)] L.

Where x is the distance of point where S.F. is zero from end A.

We have to find the distance from pt B.

Therefore distance from point B = L-x.

= L - [ 1(+or-) (1/sqroot 3)] L.

= [1 - 1(-or+) (1/sqroot 3)] L.

= [(-or+) (1/sqroot 3)]L.

This distance will never -ve.

therefore = [(1/sqroot 3)]L= (L/sqroot 3) from B end.

Ra = WL/3.

Rb = WL/6.

If we take section at distance x from A.

And shear force is calculated for this section.

We get S.F. at this section= (WL/3) - Wx + [ W(x2) / (2L)].

For maximum B.M. shear force at this section should be zero.

therefore,

(WL/3) - Wx + [ W(x2) / (2L) ] =0.

Cancel W from each term.

We get second order equation.

(L/3) - x + [ (x2) / (2L) ] =0.

[ (x2) / (2L) ] -x + (L/3) = 0.

If we solve this equation.

We get,

x = [ 1(+or-)(1/sqroot 3)] L.

Where x is the distance of point where S.F. is zero from end A.

We have to find the distance from pt B.

Therefore distance from point B = L-x.

= L - [ 1(+or-) (1/sqroot 3)] L.

= [1 - 1(-or+) (1/sqroot 3)] L.

= [(-or+) (1/sqroot 3)]L.

This distance will never -ve.

therefore = [(1/sqroot 3)]L= (L/sqroot 3) from B end.

(1)

Chandra prakash said:
6 years ago

Please explain me clearly.

Anamol said:
6 years ago

The point where shear stress will change its sign will indicate maximum bending moment so.

So Mx = Rb * x - WX * x * x/(l * 2 * 3).

or, (WLX/6)-(WX^3)/(6l) = 0.

or, WX^2/(2 * l) = Wl/6.

or x = L/√3.

So Mx = Rb * x - WX * x * x/(l * 2 * 3).

or, (WLX/6)-(WX^3)/(6l) = 0.

or, WX^2/(2 * l) = Wl/6.

or x = L/√3.

Gopi Krishna Bese said:
6 years ago

M(a)=0 => L* Ra =(WL/2) * (2L/3) = WL/3.

=> Rb= WL/2 - WL/3 = WL/6.

Fx= Rb - load on lenght Ax.

= WL/6 - (wx/L) * x/2.

= WL/6 - wx^2/2L -->(1).

Apply x = 0 and x = L.

At some point S.F must be zero then (1) equal to zero.

X = L/√3. Ans.

=> Rb= WL/2 - WL/3 = WL/6.

Fx= Rb - load on lenght Ax.

= WL/6 - (wx/L) * x/2.

= WL/6 - wx^2/2L -->(1).

Apply x = 0 and x = L.

At some point S.F must be zero then (1) equal to zero.

X = L/√3. Ans.

Sharad said:
6 years ago

The Given answer is right.

RAJESH said:
6 years ago

L/3 FROM BASE, 2L/3 FROM APEX.

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