How to find maximum bending moment and maximum bending stress – GATE 2019 preparation


In this video we will try to understand how to calculate maximum bending moment & bending stress. This question we have taken from GATE 2010 question paper. A mass less beam has loading pattern as shown in the figure. The beam is of rectangular cross section with width of 30 mm & height of 100 mm. Now we have to find out the maximum bending moment occurs at so we have to find out what is a location of maximum bending moment & and the maximum magnitude of bending stress. So we can say that over here at A & C , there will be reactions RA and RC. Then this is the u.d.l. of 3000 N/m So uniformly distributed load of 3000 N per m length is given over a length of BC & the length is given as 2000 mm . Now here the u.d.l. is of N per m. So we will convert this 2000 mm into meter so it is 2 m that we have shown over here. Similarly this 2000 mm that also we have converted into m that is 2m. Now as it is a rectangular distribution we will convert this uniformly distributed load into a point load. So it is 3000 N/m so for 1m length the load acting is 3000 then for for 2 m length what will be the load that is acting. So it is 3000 X 2 that is 6000 N or 6 X 10 ^ 3 N and it will be acting at the center of 2 m length because it is a rectangular distribution. so it will act at the center of this particular 2 m length that is 1 m . Now our first job is to calculate these reactions RA & RC so we will apply the conditions of equilibrium. Summation of forces in y direction is equal to 0 . We will consider upward forces +ve & downward forces -Ve. RA & RC they are acting in the upward direction so they +ve whereas 6 X 10 ^ 3 or 6000 that is acting in the downward direction so that is -ve. Therefore we can say that RA + RC is equal to 6 X 10 ^ 3 N Now we will apply the another condition of equilibrium that is summation of moments is equal to zero. Clockwise +ve & anti-clockwise -ve. Now we will take moments about point A We will start with RC the arrow of this RC will be producing anti-clockwise moment about A so it is -ve and perpendicular distance between C & A is this 2 + 2 that is 4 So it is -RC X 4 whereas this uniformly distributed load which we have converted into point load will produce the clockwise moment about point A so it is 6 X 10 ^ 3 and the total distance from this point up to point A will be 1 + 2 that is 3 Now transfer this -ve term on the other side So it is RC X 4 is equal to 6 X 10 ^ 3 X 3. Or we can say that RC will be equal to 4500 N. Now we can substitute this value of RC in equation (i) so that we will get the value of RA .. So RA will be equal to 6 X 10 ^ 3 – RC So RA will equal to 6 X 10 ^ 3 value of RC is 4500 .. so therefore RA will be equal to 1500 N. Now we will consider any section XX at a distance of x from point A because we have to find out what will be the location of maximum bending moment. Now if this distance is x and this distance is 2m then the remaining distance will be (x-2) meter. and the u.d.l. for this particular length that is from B up to this section XX will be 3000 X (x-2) because 3000 N/m & length is now (x-2) and this will be acting at the center of (x-2) so it will be (x-2)/2. Now we will take the moment about this particular section XX Now this RA will produce the clockwise moment about this section XX so it is RA into x whereas this u.d.l. of 3000 X (x-2) this arrow will produce anti-clockwise moment about this section XX So load is 3000 X (x-2) into (x-2)/2 3000 divided by 2 that is 1500 and (x-2) X (x-2) that is (x-2)^2. Now we have to find out the location of maximum bending moment & the maximum bending moment So we will differentiate this Mxx w.r.t. x & equate it to 0. So derivative of Mxx & value of Mxx is 1500 x minus 1500 (x-2)^2 & we have equated it to 0. Now derivative of x w.r.t. x is 1 and derivative of (x-2)^2 is 2 times (x-2) So we can transfer this -ve term on the other side so it will become +ve. Now cancel out this 1500 & 1500 so it is 1 is equal to 2 (x-2) transfer this 2 on this side so it is 1/2 is equal to x-2 or 0.5 .. 1 upon 2 is 0.5 transfer this -2 on this side so it is plus 2 is equal to x. So therefore x=2.5 m from A or 2500 mm from A . So that is our first answer. Now we have to find out the maximum bending stress so for that we will require to find out maximum bending moment. Now substitute this value of x=2.5 in this equation of Mxx … So in place of x it is 2.5 that we have substituted here also in place of x we have substituted 2.5 So the value that we have obtained is 3375 Nm or 3375 X 1000 Nmm Now we know that maximum bending stress ‘ Sigma b ‘ is given by M upon Z or maximum bending moment upon Z where Z is the section modulus & by default this particular section modulus is about xx axis. So it is 1/6 b d^2 So width is given as 30 mm & depth or height is given as 100 mm so those values we have substituted & we have calculated our second answer. So bending stress is given by 67.5 MPa. So in this way we can find out the location of maximum bending moment then value of maximum bending moment that is Mmax & maximum bending stress induced in a simply supported beam carrying uniformly distributed load. Thank you very much for watching.

17 thoughts on “How to find maximum bending moment and maximum bending stress – GATE 2019 preparation

  1. superb explanation. Pls, cover lope and deflection models for gate. Also shear stress in beams concept and problems. it would be of great help

  2. I tried this method on a different problem which had other forces between A and xx and ended up getting a value that doesn't quite make sense. I have checked my math and do not see an error. Is it perhaps this method doesn't apply for all max bending moment problems?

  3. To find maximum B.M location
    Why not we just find the location where Shear force is Zero .
    That will be more easier then this.

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