Any structural or civil engineers here?

[danielbriggs]

Try this book, it may explain it better: Engineering Maths (6th Ed. K. Stroud) pg. 1163
If I had my copy with my, I'd scan it in...

I could see how it could be done from a histogram, but would be an approximate solution. Stating your distribution as a normal distribution, you can achieve much more accurate answers... (As the histogram would have discrete steps, whereas a normal distribution has a continuos 'bell shaped' curve)

If you need to know anything else, just shout.

Dan

[andy_con]

will do many thanks

[andy_con]

Daniel could we just run through this part one more time –

(Line 1) ∅ (-0.9984) = 1 - ∅ (0.99684)

(line 2) = 1 – 0.8413

(line 3) = 0.1587

now i dont understand how you get from line 1 to line 2. well i dont actually get what line 1 is?

i understand how you get from line 2 to line 3 .

[andy_con]

when i do 1 - 0.8413 x (-0.99684) i get 1.838

[danielbriggs]

Daniel could we just run through this part one more time –

(Line 1) ∅ (-0.9984) = 1 - ∅ (0.99684)

(line 2) = 1 – 0.8413

(line 3) = 0.1587

now i dont understand how you get from line 1 to line 2. well i dont actually get what line 1 is?

i understand how you get from line 2 to line 3 .

Phi (the circle with the line through) is the cumulative probability function of a given value of Z.
It's like saying "sin(x)", you need to put a number in to get the answer.

On some previous lines we calculated our Z value to be -0.9968
(close enough to -1)

Now, we need the probability less than Z= -1
See fig. 1

However the stats tables only have values for positive Z values.
Now the Phi function is cumulative, so looking up the value of Z=1 gives us a probability of 0.8413
(Fig 2.)

But the normal distribution is symmetrical. So if we find that small part above the 1 value on fig 2.
(Fig 3.) We find the same part of the other side of the graph effectivly. i.e. the higlighted section in Fig 1.

So, the reason for the 1 minus part is to find the other area under the graph. (As total probabilities always add up to 1)

How's that?

Dan

[andy_con]

yes hlepful but i still get this LOL

when i do 1 - 0.8413 x (-0.99684) i get 1.838 ??????????

[danielbriggs]

The phi function of 0.99684 equals 0.8413. So you need to substitute that value, not multiply the two.
Phi of Z = probability
Phi(0.99684)=0.8413


So: 1 - 0.8413 = 15.87%
Everytime you get an answer, just check it is less than 1.00, as probabilities are ALWAYS less than or equal to 1

Ta,
Dan

[T0mmm]

Hi all, yet another beam related question that has got me stumped!

i've attached the question as a picture.. so hope it turns up



i understand how to do point load moments, but not UDL's so much


i know the 2kN * 4m bit

any pointers would be greatly apprecated, i.e. to which books, etc to get from the library.

our lecturer is about as helpful as a broom, and my tutor is a chemisty lecturer... although im studying mat.eng


Cheers,
Tom

[danielbriggs]

UDL's are just what they say on the tin.
In your case, for every 1m, there is 3kN of force applied. So over the span, there is a 24kN force applied.

Try solving it like this:
- Resolve vertically with the reactions for supports, say P and Q.
- Take moments about the first support, P; and the same about Q (2nd support post)
- Then consider a section of the beam cut at point 'x' from the left side. And generate a function from that for the bending moment. Then to find the bending moment at point A, just stick in the distance of 6m and out pops the answer.

If you still get stuck, I'll write it out tonight; but it's the whole 'give a man a fish...' story

Dan

Edit: Hibbler's Mechanics of Materials may be of great use (7th Edition), should be in most libraries.

[T0mmm]

yeahhhhh, i've sorted it now, thanks for that!

and believe it or not, they dont have that book in the vast library that we have!

cheers for the pointers dan

Tom