14.3 Beams under Harmonic Excitations

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14.3.1 Mechanical Properties

A 6-m-long beam is simply supported. The cross section and material properties are shown in

Figure 14.2(b).

f 0c ¼ 39 N=mm2

A0s

¼ 142 mm2

d ¼ 305 2 12:7 2 6:35 2

1

2 ð19:05Þ ¼ 276 mm

As ¼ 852 mm2

Find the moment ðMÞ– curvature ðf Þ relationship of the cross section at the midspan and the force

ðPÞ– displacement ðdÞ relationship of the beam.

14-18 Vibration and Shock Handbook

© 2005 by Taylor & Francis Group, LLC

Solution

1. Moment – curvature relationship

E ¼ 1500

ffiffiffi

f 0c

q

ðN=mm2Þ ¼ 29614:9 N=mm2

Using the derived equations in Section 14.2.2

gives

Mu ¼ 86;122;000 N mm;

cu ¼ 5:63 £ 1025 rad=mm

My ¼ 82;491;700 N mm;

cy ¼ 1:22 £ 1025 rad=mm

Mc ¼ 9;263;000 N mm cc ¼ 8:66 £ 1027 rad=mm

They are plotted in Figure 14.21, and the corresponding loads are determined below.

2. Force – displacement relationship

PU ¼

4 £ 86;122;000

6000 ¼ 57;415 N

PY ¼

4 £ 82;491;700

6000 ¼ 54;994 N

PC ¼

4 £ 9;263;000

6000 ¼ 6175 N

The corresponding deflections can be

determined as follows:

1. Cracking state (Figure 14.22)

dc ¼

McL2

12EI

¼

9;263;000 £ð6000Þ2

12 £ð29;614:9Þ£

1

12 £ð152:4Þ£ð305Þ3

¼2:6 mm

2. Yielding state (Figure 14.23)

L1 ¼3000 £

9;263;000

82;491;700 ¼336:9 mm

(122, 82.5)

(563, 86.1)

M (kN m)

(8.66, 9.26)

y(×10−7 rad/mm)

FIGURE 14.21 Moment – curvature curve for the

example beam.

L1 L2 C

A B

L3

Yc

Yy

RA

FIGURE 14.23 Curvature diagram at yielding state.

A C

L

ψc

B

FIGURE 14.22 Curvature diagram at cracking state.

Reinforced Concrete Structures 14-19

© 2005 by Taylor & Francis Group, LLC

L2 ¼3000 2 336:9 ¼2663:1 mm

L3 ¼336:9 þ

8:66 £1027 £2663:1 £

2663:1

2 þð1:22 £1025 2 8:66 £1027Þ£

2663:1

2 £

2

3 £2663:1

8:66 £1027 £2663:1 þ

1

2 £2663:1 £ð1:22 £1025 2 8:66 £1025Þ

¼2053 mm

dy ¼ 13

£8:66 £1027 £ð342:2Þ2 þ 12

£ð8:66 £1027 þ1:22 £1025Þ£ð2663:1Þ£2053 ¼35:75 mm

3. Ultimate state (Figure 14.24)

L1 ¼3000 £

9;263;000

86;122;000 ¼322:7 mm

L2 ¼3000 £

82;491;600

86;122;000

2 322:7

¼2550:8 mm

L4 ¼3000 2 322:7 2 2550:8 ¼126:5 mm

L3 ¼322:7 þ

8:66 £1027 £2550:8 £

2550:8

2 þð1:22 £1025 2 8:66 £1027Þ£

2550:8

2 £

2

3 £2550:8

8:66 £1027 £2550:8 þ

1

2 £2550:8 £ð1:22 £1025 2 8:66 £1027Þ

¼1967:3 mm

L5 ¼322:7 þ2550:8 þ

1:22 £1025 £126:5 £

126:5

2 þð5:63 £1025 2 1:22 £1025Þ£

126:5

2 £

2

3 £126:5

1:22 £1025 £126:5 þ

1

2 £126:5 £ð5:63 £1025 2 1:22 £1025Þ

¼2950:4 mm

du ¼

1

3 £8:66 £1027 £ð322:7Þ2

þ

1

2 £ð8:66 £1027 þ1:22 £1025Þ

£ 2550:8 £1967:3

þ

1

2 £ð5:63 £1025 þ1:22 £1025Þ

£ 126:5 £2950:4 ¼45:6 mm

The load – deflection curve is shown in

Figure 14.25.

L1 L2 L4 C

A

L3

L5

Yc

Yy

Yu

B

FIGURE 14.24 Curvature diagram at yielding state.

d (mm)

(2.6, 6175)

P (N)

(35.75, 54994)

(45.6, 57415)

FIGURE 14.25 Load – deflection curve.

14-20 Vibration and Shock Handbook

© 2005 by Taylor & Francis Group, LLC

14.3.2 Design for Machine Vibration

The same beam as described in Section 14.3.1 is supporting a machine with a rotating frequency of 9 Hz

at the midspan of the beam. The mass of the rotating machine and the beam is 1 N sec2/mm. Check if

resonance will occur.

Solution

Find the stiffness, k; the natural frequency, f ; and the frequency ratio, r; for each of the three states,

namely the elastic, cracking, and yield states.

1. Elastic state

k ¼

617:5

2:6 ¼ 2375 N=mm

f ¼

1

2p

ffiffiffiffiffiffiffiffi

2375

1

r

¼ 7:76 Hz

r ¼

7:76

9 ¼ 0:862

Since 0:8 , r , 1:2; it is in the resonance zone. The beam needs to be redesigned to avoid

resonance.

2. Cracking state

k ¼

5499:4 2 617:5

35:75 2 2:6 ¼ 1472:7 N=mm

f ¼

1

2p

ffiffiffiffiffiffiffiffiffiffi

1472:7

1

r

¼ 6:11 Hz

r ¼

6:11

9 ¼ 0:68 , 0:8

OK

3. Yield state

k ¼

5741:5 2 5499:4

45:6 2 35:75 ¼ 245:8 N=mm

f ¼

1

2p

ffiffiffiffiffiffiffiffi

245:8

1

r

¼ 2:50 Hz

r ¼

2:50

9 ¼ 0:28 , 0:8

OK

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