imtek lehrstuhl konstruktion von mikrosystemen micromechanics – ws 2011/2012/ exercise 1 / sheet 1...
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![Page 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](https://reader030.vdocument.in/reader030/viewer/2022020306/551436655503466d1a8b52cc/html5/thumbnails/1.jpg)
IMTEKLehrstuhl Konstruktion von Mikrosystemen
Micromechanics – WS 2011/2012/ Exercise 1 / sheet 1
Supplementary problem 1
A B
0.1 m 0.1 m 0.1 m 0.1 m
100 N 160 N200 Nm
A B
0.1 m 0.1 m 0.1 m 0.1 m
100 N 160 N200 Nm
AxAz Bz
2. Replace bearings with reactions2. Replace bearings with reactions
1. Assign the co-ordinate system1. Assign the co-ordinate system
A
B
0.1 m 0.1 m 0.1 m 0.1 m
100 N 160 N200 Nm
Z
XY
3. Use equilibrium conditions3. Use equilibrium conditions
∑Fx = 0 → Ax = 0∑My = 0 → 0.4Bz-0.3×160-0.2×100- 200 = 0 → Bz = 670 N∑Fz = 0 → Az+Bz = 100+160 = 260 N → Az = -410 N
∑Fx = 0 → Ax = 0∑My = 0 → 0.4Bz-0.3×160-0.2×100- 200 = 0 → Bz = 670 N∑Fz = 0 → Az+Bz = 100+160 = 260 N → Az = -410 N
![Page 2: 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](https://reader030.vdocument.in/reader030/viewer/2022020306/551436655503466d1a8b52cc/html5/thumbnails/2.jpg)
IMTEKLehrstuhl Konstruktion von Mikrosystemen
Micromechanics – WS 2011/2012/ lecture 1 / sheet 2
A B
3 m 9 m
50 N
34
11
AB
3 m 9 m
50 N
34
11
Z
XY
A B
3 m 9 m
50 N
34
Ax AzB xz
4. Resolve forces along X and Y directions
4. Resolve forces along X and Y directions
11
22
33
A B
3 m 9 m
40 N
Ax Az Bz
Bx
30 N
Supplementary problem 2
![Page 3: 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](https://reader030.vdocument.in/reader030/viewer/2022020306/551436655503466d1a8b52cc/html5/thumbnails/3.jpg)
IMTEKLehrstuhl Konstruktion von Mikrosystemen
Micromechanics – WS 2011/2012/ lecture 1 / sheet 3
5. Use equilibrium conditions5. Use equilibrium conditions
•∑Fx = 0 → Ax-Bx-30 = 0•∑My = 0 → 3×40-12Bz = 0 → Bz = 10 N•∑Fz = 0 → Az+Bz = 40 N → Az = 30 N•Bx = Bz→ Bx = 10 N → Ax = 40 N
•∑Fx = 0 → Ax-Bx-30 = 0•∑My = 0 → 3×40-12Bz = 0 → Bz = 10 N•∑Fz = 0 → Az+Bz = 40 N → Az = 30 N•Bx = Bz→ Bx = 10 N → Ax = 40 N
Supplementary problem 2
A B
3 m 9 m
40 N
Ax Az Bz
Bx
30 N
![Page 4: 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](https://reader030.vdocument.in/reader030/viewer/2022020306/551436655503466d1a8b52cc/html5/thumbnails/4.jpg)
IMTEKLehrstuhl Konstruktion von Mikrosystemen
Supplementary problem 3
Micromechanics – WS 2011/2012/ lecture 1 / sheet 4
p
al/2 l/2
A CB
p
al/2 l/2
A CB
Z
XY
p
al/2 l/2
A CB
Ax Az
Cx
Cz
11
22
33
![Page 5: 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](https://reader030.vdocument.in/reader030/viewer/2022020306/551436655503466d1a8b52cc/html5/thumbnails/5.jpg)
IMTEKLehrstuhl Konstruktion von Mikrosystemen
Supplementary problem 3
Micromechanics – WS 2011/2012/ lecture 1 / sheet 5
4. Make free body diagrams4. Make free body diagrams
p
al/2 l/2
A CB
Ax Az
Cx
Cz
Bx
Bz Bx
Bz
CMy
B
55•∑Fx = 0 → Ax = Bx = Cx
•∑My = 0 → Bz = -p/2 • → Az = p/2 •→ CMy = -pa/2
•∑Fz = 0 → Bz+Cz = 0 → Cz = p/2
•∑Fx = 0 → Ax = Bx = Cx
•∑My = 0 → Bz = -p/2 • → Az = p/2 •→ CMy = -pa/2
•∑Fz = 0 → Bz+Cz = 0 → Cz = p/2
![Page 6: 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](https://reader030.vdocument.in/reader030/viewer/2022020306/551436655503466d1a8b52cc/html5/thumbnails/6.jpg)
IMTEKLehrstuhl 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 10 kN/m
A
B
2 m3 m
3×2/3 m
Ax Az
Bz
• 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
• Bz = 6 kN, Az = 9 kN, Ax = 0
• Bz = 6 kN, Az = 9 kN, Ax = 0
15 kN
![Page 7: 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](https://reader030.vdocument.in/reader030/viewer/2022020306/551436655503466d1a8b52cc/html5/thumbnails/7.jpg)
IMTEKLehrstuhl Konstruktion von Mikrosystemen
Problem 3
Micromechanics – WS 2011/2012/ lecture 1 / sheet 7
Z
XY
Z
XY
Z
XY
Z
XY
1. Movement of axes along body axis1. Movement of axes along body axis
A
B C
D
![Page 8: 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](https://reader030.vdocument.in/reader030/viewer/2022020306/551436655503466d1a8b52cc/html5/thumbnails/8.jpg)
IMTEKLehrstuhl Konstruktion von Mikrosystemen
Problem 3
Micromechanics – WS 2011/2012/ lecture 1 / sheet 8
Z
XY
2. Dissection of body in segment CD2. Dissection of body in segment CD
Z
XY
A
B
C
Q
LM
D
Q
LM
3. Use equilibrium conditions for the segment and calculate IFVs
3. Use equilibrium conditions for the segment and calculate IFVs
F3
F2
x
c-x
negative area
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IMTEKLehrstuhl Konstruktion von Mikrosystemen
Micromechanics – WS 2011/2012/ lecture 1 / sheet 9
4. Repeat for segments BC and AB4. Repeat for segments BC and AB
5. Draw the diagrams for Q,L and M5. Draw the diagrams for Q,L and M
a. Draw the body axisb. Draw the IFV distributionc. Show the sign of IFV in the diagramd. Show the important IFV magnitudes
a. Draw the body axisb. Draw the IFV distributionc. Show the sign of IFV in the diagramd. Show the important IFV magnitudes
F1 only
+ Q
F1