• Welcome to engineeringclicks.com
• # Chassis/beam Calculations.

Discussion in 'Calculations' started by mechanicalorange, Feb 20, 2011.

1. ### mechanicalorangeNew Member

Joined:
Feb 2011
Posts:
1
0
Hi,

Iâ€™m in the second year of my B.ENG and have missed around 6 weeks of study due to illness. I have collected the notes I missed from a classmate. Iâ€™m currently struggling with Applied Mechanics. I canâ€™t understand one Example question. A question similar to this one WILL be in my end of year exam, so any help would be greatly appreciated!

Its asking to design a suitable chassis for an off road military vehicle. The initial specifications are as follows;
(1) Kerb weight of vehicle to be 4 tonnes.
(2) Must be capable of maintaining 80 KPH up a 1 in 5 incline.
(3) Must be capable of 160 KPH on level ground.
(4) It must be able to start from rest on a 1 in 2 incline when fully loaded.

â€¦A majority of that is to do with the Power, torque and gear calculations. Which Iâ€™m ok with. Just struggling with this questionâ€¦.

1. Chassis Calculation.
For initial calculation of the chassis section the following assumptions can be used;
â€¢ The weight of the engine/transmission is borne by the front sub frame and the front suspension and is therefore not carried by the chassis beams.
â€¢ The weight to be taken at normal loading is 3 tonnes. And there will be two beams taking half the load. These beams are supported at both ends.
â€¢ The load is assumed to be a point load acting in the middle of the beam.

(A)
Calculate a standard rectangular box section for each chassis section usng the following data;
The Beam length to be 3 meters.
The MAX allowable bending stress is 300 x 10? Nm-2
Under the worst case loading (going over an obstruction at speed) the vehicle experiences an upward acceleration equivalent to 10 times gravity.

(B)
Calculate the mass of each beam.

(C)
Assuming normal loading calculate the deflection of the beam at its midpoint. The material of the beam is Steel where E=210GPa.

(D)
A design change causes the loading of the beam to change so that the load is now a point load of 0.5 tonne situated 0.75m from the front of the beam and a point load of 1.0 tonne situated 2.0m from the front of the beam. Calculate the new position and magnitude of the maximum deflection under normal loading.

END

If anyone knows which calculations are to be used for each part of the question or has some worked examples I can use as reference that would be great.

Thanks in advance for your time and any input you have to offer.

2.
3. ### AndrewNewWell-Known Member

Joined:
Jul 2010
Posts:
73
0
The first step is to draw (or at least visualise) the bending moment diagram for your beam. You will find the methods for doing this in any basic textbook on mechanics of materials. Once you get familiar with thinking about beams, it becomes quite intuitive. I have Ryder's "Strength of Materials" on my shelf (which is an oldy but a goody), but there are many more. If you know the deflection formula for a simply supported beam, calculating deflections is just a question of plugging in your numbers. d = (Wa^2b^2)/(3 E I l) where d is deflection, W is the applied load, a and b are the distances of the point of application of the load from the left and right hand ends of the beam, E is Young's modulus, I is the second moment of area of the beam cross section and l is the length of the beam between the supports. The maximum stress, s, will occur where the bending moment is a maximum (at the point of load application in this case) and can be calculated from s = M y / I where M is the bending moment and y is the distance from the neutral axis (half the beam depth in your case). For a rectangular hollow section I = (BD^3/12) - (bd^3/12) where B and D are the breadth and depth of the outside of the beam section and b and d are the breadth and depth of the inside of the beam section. Mass can be calculated from volume (easy!)

Cheers

Andrew

4. ### rushinpatelActive Member

Joined:
Apr 2011
Posts:
25
0
hello,
you first divide total load in to 4 parts means w/4=total load on 1 beam

then after u find crippling load of the beam,(pc=n*3.14^2*E*I)/(Le^2)
le=Effective length of beam,(For this calculation you refer STRENGTH OF MATERIAL By.R.S.khurmi,)
then after u find deflection of beam,(For this calculation you refer chapter BIM load calculation STRENGTH OF MATERIAL By.R.S.khurmi)

then after u find cross section area of beam,

then after u find its weight with refrance to area of beam and density

5. ### maniacal_engineerWell-Known Member

Joined:
Jul 2009
Posts:
137
0
in this equation what is n?
is it a safety factor?

Joined:
Aug 2011
Posts:
13