Recent content by mariechap89

Hi I have calculated the point loads on the girder MN (Ithink). I have used a 310UB40.4 beam. So 40.4 kg x 6 m = 242.4 kg 242.4 kg/2 = 121.1 kg (Asthe girder only supports half the weight of the beams) 121.1 x 9.81 = 1188.0 N = 1.188 kN And there would be 3 point loads at 3m apart as girder...

Hi. Thanks for your help, this stuff is finally starting to make sense. I understand how you've done the calcs for part a of the question and am pretty sure I know where I was going wrong. I'm not too sure how to handle part b of the question though ??

Here is what I have now: Volume of concrete= 6x6x0.175=6.3 m^3 Then the unit weight for concrete, dense aggregate, un-reinforced from a table is 24.0 kN/m^3 and then adding 0.6 for each 1% by volume of steel reinforcement the unit weight is 24.6015 kN/m^3. Therefore G= 6.3x24.6015= 155 kN/m^2...

Thanks for your help, I realize I have got the calculation very wrong not even including the weight of the steel beam. I am pretty sure we are supposed to ignore the self weight of the beam for these calculations. I'm not sure how to use the weight of the concrete slab in the calculation for the...

Hi. I have started to do this question (the question it's self is attached with it's relevant diagrams) but I am coming up with a too large a maximum design load, the value I am looking for isn't on the tables I have so I must have made a calculation error somewhere. Any help would be much...

a) Masses, m1,m2,...mk are at points P1,P2,...Pk with position vestors P1,P2,...Pk (these should be in bold) respectively. The position vector g of the centre mass is given by: (m1+m2+...mk)g=m1,p1+m2p2+...+mkpk Find the position vector of the centre of mass of the masses 3,4, and 5kg placed...

An architectural feature of a building is a flat wall that is angled out over the footpath. Measurements (in meters) of three vertices of the wall were recorded against the x,y and z axes: x- axis is horizontal, the straight boundary of the footpath and the property on which the building sits...

Express the following in cartesian curves in polar form i) 4x-5y=2 Not sure how to do this ii) (x-3)^2+(y-4)^2=25 r=9cos16(theta) Is this correct ? Any help would be great

For the following curves i) y=x^2+4x-1 ii) y=-2+or-Square root(x+5) a) Sketch both the curves on the same sheet of graph paper- against the same axis I have done this, although I have not shown it here b) Determine with proof, whether the above curves are related. Not sure how to do this...

Hi. Does this look any better?? BMe would be 8(5)-20+15(4)-2(Vd)=0 therefore Vd=40 Sum of vertical forces: 5+40-15-15(4)+VF=0 therefore Vf=30 Sum of moment about F + clockwise: Mf-15(4)(2)+40(6)-15(8)-5(12)=0, therefore Mf=60kNm I am also having trouble drawing the shear force and...

Hi. I have redone my first equation and got: BMc=4(Va)+20=0, therefore Va=5 kn acting upwards Then for my second equation I have: BMe=-8(5)+20+15(4)-2(Vd)=0, therefore Vd=20kN acting upwards Then the sum of the vertical forces: 5+20-15-15(4)+Vf=0, therefore Vf=50kN acting upwards Sum of...

I have attached the original question and my attempt at the solution. I think I am on the right path but I am not sure. Can some one please direct me on if I am right so far and what to do if i am/ am not. Thanks

Beam analysis Help! I have attached the question and my attempt at a solution, although I am pretty sure I have calculated the reactions wrong and therefore my sfd and bmd are wrong as well. Any help to solve this would be great. Thanks.

Hi. I have attached the question and my attempt at the answer. I am not sure if I have the original equation correct and am not sure where to go from the last equation I have shown. Any help would be great. Thanks.