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

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

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

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

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

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

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

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

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