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| help with forces question PLEASE!!!! | |||
| Post Reply | Engineering Forum | ||
| Posted by: martinedwards ® 04/15/2009, 15:52:28 Author Profile eMail author Edit |
Hello you clever people!! 25 years ago I did A-level Physics and I vaguely remember vector diagrams for loaded beams. I'm now teaching technology to GCSE level and I need to work out the following and I'm totally stumped!! I haven't used the scientific functions on a calculator since then, and according to the book I have it's all sin/tan etc. A symmetrical triangular span.
/I\
I need the magnitude & nature of forces in each bar. the two horizontals are in tension, the vertical is in tension & the two diagonals are in compression, right? the safety factor of 6 I understand and I can work out the deformation using young's modulous etc, but the vectors have me lost. anyone feel generous to help out an inumerate idiot? thanks!! |
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| : help with forces question PLEASE!!!! | |||
| : help with forces question PLEASE!!!! -- martinedwards | Post Reply | Top of thread | Engineering Forum |
| Posted by: martinedwards ® 04/15/2009, 15:54:31 Author Profile eMail author Edit |
aw, the diagram didn't work!! think a wide triangle with a vertical splitting it in two and the weight hanging from the bottom of the vertical....... |
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| : : help with forces question PLEASE!!!! | |||
| : : help with forces question PLEASE!!!! -- martinedwards | Post Reply | Top of thread | Engineering Forum |
| Posted by: bpd ® 04/15/2009, 16:28:07 Author Profile eMail author Edit |
Here is the deal. At the point where the load is connected, two -pinned- connections are horizontal, one is vertical=>, the force goes to the vertical member=> 180 kg(*g) N of force in the vertical member up top. 1766 Newts Up top, the forces sum to zero=> the vertical component of force in each diagonal is 180kg(*g)/2=883 Newts Calculate the total force based on the vertical force in diagonal member=> (Fvert=Ftot*sin(35))all diagonal member. Therefore
Sum at corner => Use horizontal component of diagonal, which will some up to zero with horizontal component of horizontal member. Fhor=Ftot cos(35)=1539*.819=1261Newts, which is your force of tension in horizontal member Modified by bpd at Wed, Apr 15, 2009, 16:30:05 |
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