In your initial sketch, allegedly without sway, you have included some sway of the left hand column, but not of the right hand column. The line AF produced meets the line DC produced at a point 4 m above C. This is the instantaneous centre of rotation of the beam segment FC, say call it point...
There are at least two ways of learning about influence lines. The first way is algebraic, as given in the solution quoted. If you have learned about shear force diagrams before coming to influence lines, it is easy to get the two things confused. It is necessary to keep returning to the...
I think you seem intent on using elastic analysis, which is fine at working loads, if can do it. However, plastic analysis is easier, and arguably safer, whilst admittedly not giving you deflections.
You know the forces at the ends of the cable. You know those forces are tangential to the cable ends. You know the drag. That makes 3 forces. If they are in equilibrium they must all meet at one point. You don't have enough information about the cable self-weight to take account of that. Can you...
Still looks funny. I am struggling to visualise 3.174 mm2. You haven't declared your units. Can you explain the 10^6 and the 10^3 in your corrected equation?
Your answer of 0.52 mm2 seems improbable, if not impractical, and so, without looking at your working, I am inclined to try a different approach. The usual problem here is often the units; so stick to, say N and mm. Try a concrete area of say 50 mm x 50 mm. At 14 MPa (ie N/mm2) , its resistance...
The assumption that a point of zero moment occurs half-way up the columns is unconvincing, because the conditions at the top and the bottom of the columns are not the same; at the bottom it is encastre and no rotation is assumed to occur, but at the top of the columns, the fixity is provided by...
The diagram is quite a good representation of what I mean, although I am struggling to understand your language: "800Nm is the diffrence in length of the triangle"
You are right to recognise that the 2400 has a different orientation to the -2000. But, just as you can add a positive number (say, +6) to a negative number (say, -2) to give a result of +4, so you can add areas on a graph. At a point just to the right of C, the result is +2400 -2000= +400 (the...
Because the expression for M is the summation of moments to the LEFT of section X, but the 12 kN load is to the right. If you were to write down an expression for the moment to the RIGHT of x, you would get an expression that may at first sight look different, but, if correctly drawn up, will be...
Unfortunately the M diagram you gave us is not drawn to scale. I suggest you redraw it to scale on graph paper. Let's call C the point of application of the 400 Nm moment, and point D the point of application of the 1000 N load. If you redraw the 2000 Nm triangle above the horizontal axis, as if...
The so-called 'missing' 400 Nm is the difference between 2400 and -2000. This is a graph preparing for gaphical summation. If you like, you can imagine folding the lower triangle about the horizontal axis, to cancel a good bit of the right hand upper triangle. The zero line would then, in that...