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IMTEK Lehrstuhl Konstruktion von Mikrosystemen Micromechanics – WS 2011/2012/ Exercise 1 / sheet 1 Problem 1 F1F1 F2F2 F3F3 F1F1 F R1 F R2 F1F1 F2F2 F3F3.

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Presentation on theme: "IMTEK Lehrstuhl Konstruktion von Mikrosystemen Micromechanics – WS 2011/2012/ Exercise 1 / sheet 1 Problem 1 F1F1 F2F2 F3F3 F1F1 F R1 F R2 F1F1 F2F2 F3F3."— Presentation transcript:

1 IMTEK Lehrstuhl Konstruktion von Mikrosystemen Micromechanics – WS 2011/2012/ Exercise 1 / sheet 1 Problem 1 F1F1 F2F2 F3F3 F1F1 F R1 F R2 F1F1 F2F2 F3F3 F R1 F1F1 Forces can be moved along their lines of action

2 IMTEK Lehrstuhl Konstruktion von Mikrosystemen Problem 1 Micromechanics – WS 2011/2012/ Exercise 1 / sheet 2 F1F1 F R1 F R2 F1F1 F2F2 F3F3 F1F1 F2F2 F3F3 F R1 Correct ? No F1F1 F R1 F R2

3 IMTEK Lehrstuhl Konstruktion von Mikrosystemen Micromechanics – WS 2011/2012/ Exercise 1 / sheet 3 Problem 1 F1F1 F2F2 F3F3 F1F1 F2F2 F3F3 F R1 F2F2 F R2 F1F1 F R1 F R2

4 IMTEK Lehrstuhl Konstruktion von Mikrosystemen Problem 2 Micromechanics – WS 2011/2012/ Exercise 1 / sheet 4 α β F2F2 F1F1 Calculate α and β graphically

5 IMTEK Lehrstuhl Konstruktion von Mikrosystemen Problem 2 Micromechanics – WS 2011/2012/ Exercise 1 / sheet 5 α β F2F2 F1F1 Draw the free body diagrams FaFa FbFb FaFa FbFb FcFc FcFc FdFd FdFd

6 IMTEK Lehrstuhl Konstruktion von Mikrosystemen Problem 2 Micromechanics – WS 2011/2012/ Exercise 1 / sheet 6 Draw the force systems α F2F2 FaFa FbFb β F1F1 FaFa FcFc FbFb F2F2 FaFa α β F1F1 FaFa β FcFc FdFd

7 IMTEK Lehrstuhl Konstruktion von Mikrosystemen Problem 2 Micromechanics – WS 2011/2012/ Exercise 1 / sheet 7 α F2F2 FaFa FbFb FbFb F2F2 FaFa α β Body is in equilibrium : Sum of all force vectors is zero Ropes can transfer only uni-axial load F b = F 2 F a = F 1 F b + F 2 + F a = 0 FbFb F2F2 FaFa α β Construct the triangle

8 IMTEK Lehrstuhl Konstruktion von Mikrosystemen Problem 2 Micromechanics – WS 2011/2012/ Exercise 1 / sheet 8 CalculateCalculate α and β numerically to verify results FbFb F2F2 FaFa α β F1F1 FaFa β FcFc Resolve forces into orthogonal components Sum of all forces in vertical (Y) direction is zero Sum of all forces in horizontal (X) direction is zero Compare the answers with those from the graphical solution FdFd


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