1. IMTEK Lehrstuhl Konstruktion von Mikrosystemen Micromechanics – WS 2011/2012/ Exercise 1 / sheet 1 Supplementary problem 1 AB 0.1 m 100 N160 N200 Nm AB 0.1 m 100 N160 N200 Nm AxAx AzAz BzBz 2. Replace bearings with reactions 1. Assign the co-ordinate system A B 0.1 m 100 N160 N200 Nm Z X Y 3. Use equilibrium conditions F x = 0 A x = 0 M y = 0 0.4B z -0.3×160-0.2×100- 200 = 0 B z = 670 N F z = 0 A z +B z = 100+160 = 260 N A z = -410 N F x = 0 A x = 0 M y = 0 0.4B z -0.3×160-0.2×100- 200 = 0 B z = 670 N F z = 0 A z +B z = 100+160 = 260 N A z = -410 N

2. IMTEK Lehrstuhl Konstruktion von Mikrosystemen Micromechanics – WS 2011/2012/ lecture 1 / sheet 2 AB 3 m 9 m 50 N 3 4 1 1 A B 3 m 9 m 50 N 3 4 1 1 Z X Y A B 3 m 9 m 50 N 3 4 AxAx AzAz B xz 4. Resolve forces along X and Y directions 1 1 2 2 3 3 A B 3 m 9 m 40 N AxAx AzAz BzBz BxBx 30 N Supplementary problem 2

3. IMTEK Lehrstuhl Konstruktion von Mikrosystemen Micromechanics – WS 2011/2012/ lecture 1 / sheet 3 5. Use equilibrium conditions F x = 0 A x -B x -30 = 0 M y = 0 3×40-12B z = 0 B z = 10 N F z = 0 A z +B z = 40 N A z = 30 N B x = B z B x = 10 N A x = 40 N F x = 0 A x -B x -30 = 0 M y = 0 3×40-12B z = 0 B z = 10 N F z = 0 A z +B z = 40 N A z = 30 N B x = B z B x = 10 N A x = 40 N Supplementary problem 2 A B 3 m 9 m 40 N AxAx AzAz BzBz BxBx 30 N

4. IMTEK Lehrstuhl Konstruktion von Mikrosystemen Supplementary problem 3 Micromechanics – WS 2011/2012/ lecture 1 / sheet 4 p a l/2 A C B p a A CB Z X Y p a A CB AxAx AzAz CxCx CzCz 1 1 2 2 3 3

5. IMTEK Lehrstuhl Konstruktion von Mikrosystemen Supplementary problem 3 Micromechanics – WS 2011/2012/ lecture 1 / sheet 5 4. Make free body diagrams p a l/2 A C B AxAx AzAz CxCx CzCz BxBx BzBz BxBx BzBz C My B 5 5 F x = 0 A x = B x = C x M y = 0 B z = -p/2 A z = p/2 C My = -pa/2 F z = 0 B z +C z = 0 C z = p/2 F x = 0 A x = B x = C x M y = 0 B z = -p/2 A z = p/2 C My = -pa/2 F z = 0 B z +C z = 0 C z = p/2

6. IMTEK Lehrstuhl Konstruktion von Mikrosystemen Supplementary problem 4 Micromechanics – WS 2011/2012/ lecture 1 / sheet 6 10 kN/m A B 2 m3 m 10 kN/m A B 2 m3 m 3×2/3 m AxAx AzAz BzBz Equivalent force generated by line load = area under the force distribution curve Line of action – through centroid of the force distribution curve Equivalent force generated by line load = area under the force distribution curve Line of action – through centroid of the force distribution curve B z = 6 kN, A z = 9 kN, A x = 0 15 kN

7. IMTEK Lehrstuhl Konstruktion von Mikrosystemen Problem 3 Micromechanics – WS 2011/2012/ lecture 1 / sheet 7 Z X Y Z X Y Z X Y Z X Y 1. Movement of axes along body axis A B C D

8. IMTEK Lehrstuhl Konstruktion von Mikrosystemen Problem 3 Micromechanics – WS 2011/2012/ lecture 1 / sheet 8 Z X Y 2. Dissection of body in segment CD Z X Y A B C Q L M D Q L M 3. Use equilibrium conditions for the segment and calculate IFVs F3F3 F2F2 x c-x negative area

9. IMTEK Lehrstuhl Konstruktion von Mikrosystemen Micromechanics – WS 2011/2012/ lecture 1 / sheet 9 4. Repeat for segments BC and AB 5. Draw the diagrams for Q,L and M a.Draw the body axis b.Draw the IFV distribution c.Show the sign of IFV in the diagram d.Show the important IFV magnitudes a.Draw the body axis b.Draw the IFV distribution c.Show the sign of IFV in the diagram d.Show the important IFV magnitudes F 1 only + Q F1F1