Structural Mechanics, Help!!!

cgcsb

Can somebody please help me with this, I'm a first year in Civil Engineering and I've been tearing my hair out over it. I never done a subject like structural mechanics before in school, and they just kinda thwer us in the deep end in college, can someone just give me the basics. How do I go about answering this question?

Lu Tze

Its been a while since i have done these problems, 6-7 years, but if nobody else chimes in to give you a hand i'll try and work through it for you later. The first one though you should be able to resolve two forces at a time into one resultant with a magnitude and direction, and keep doing that until you have one left.

cgcsb

Lu Tze wrote: »

Its been a while since i have done these problems, 6-7 years, but if nobody else chimes in to give you a hand i'll try and work through it for you later. The first one though you should be able to resolve two forces at a time into one resultant with a magnitude and direction, and keep doing that until you have one left.

oh forgot to mention, I already done the first one it's just 2 and 3 that I cannot do. Your help is much appreciated

aurthurg

I feel your pain, structural mechanics aint to much fun in first year of engineering if you done nothing like it in school. I ll try give you a start anyway

Q2

Well first off you need the height of the frame to work out the internal angles i.e basic trig

Then you need to get the reactant forces at the support points by moments

You know (or you should) that Moment @ A is equal to 0

Right so sum of the forces in the x is equal to zero i.e

5+10+10+10+10 = Ra + Rb

Ra + Rb = 45

Now for the moments

Moment at A ( = 0) = 10(2.65)+10(5.25)+10(7.85)+10(10.5) - 10.5 (Rb)

Boil this down

You get

Rb = 25

and from Ra + Rb = 45

Ra = 20

Now you go to each node staring at the bottom left and work just as in q1 knowing that forces in both x directions or y directions when resolved are equal in a static frame

aurthurg

Q3

O.k so ye first have to recognise that there is two u.d.l s here and that you need to represent these as a point force acting through there centre of gravity. These two rectangular so the force acts through there mid point

So the 15kn/m u.d.l acts over 4.5m thus its force is = 4.5(15) = 67.5kn acting at a distance of (4.5/2) from the left edge

The 35kn/m acts over 6.5m so it is (35)(6.5)kn acting at (6.5/2) from the left edge

So now you will have two up arrows the reactions at the supports and two down arrows the loads due to the u.d.l using the moments and that the up forces are equal to the down forces just as in q2 you should be able to get the support reactions

After this you will use free body diagrams cutting at say every 1m along the beam to get the moment and shear etc
Hope these get you started anyway

godtabh

go back to your basics

Forces equal and opposite. Resolve into its components and sum up

You may need to use simultaneous equations as well

cgcsb

ArthurG. Thank you very much for your help. I kinda knew the reactions in question 2 from looking at the diagram. The problem with Q2 is that I have no idea how to get the force in each member. When you refer to question 1, I only solved that graphically to find the resultant force. I know there is a formula but I really dont remember it. I don't mean to sound ungrateful. Your explaination on Q3 got me started on it, and that's such a relief to me, you have no idea.

aurthurg

o.k so in q2 once you have the reactions at the supports you have to start resolving th forces about each node or connection. The frame is static so you know all forces must be equal and opposite

the first step would be the bottom left hand corner of the frame

o.k so you have the 5kn load acting down and the reaction of 20kn up

so you need to find the resolved force in the y direction that comes from the truss itself

we know these sUM to 0 thus our equation will be

20 = 5 + F1(sin (of the internal angle))

so now you know f1

now you do the same in the x direction at this node except you resolve by using F1 *cos(internal angle)

hope this helps but heres a good link to help you along

http://physics.uwstout.edu/Statstr/Strength/StatII/stat22.htm

cgcsb

I think I'm having trouble because I never really understood the basics so bear with me. If I do as you said, I get the force in that member as resolved into it's x and y componants by solving using cos and sin seperately, but what is the actual force in the individual members? also at the other nodes, all the forces are acting downwards, How do I calculate the forces in the members that are not conected to the raction points?

Turbulent Bill

cgcsb wrote: »

I think I'm having trouble because I never really understood the basics so bear with me. If I do as you said, I get the force in that member as resolved into it's x and y componants by solving using cos and sin seperately, but what is the actual force in the individual members? also at the other nodes, all the forces are acting downwards, How do I calculate the forces in the members that are not conected to the raction points?

To be honest, if you don't get the basics then you'll constantly struggle with this, but they're easy to pick up. The sums of forces at any node in the truss is zero, and the sum of the moments about any node is also zero. This will typically give you three simultaneous equations to solve for whatever unknown you need.

I'd recommend getting a copy of "Statics" by Meriam and Kraig or something similar which has plenty of worked examples. Do these by hand again to make sure you understand the solution (don't just look at them!).

cgcsb

Hey,

Just submitted the questions finished. Thanks alot everyone for your support