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14.4 Design for Explosions/Shocks

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14.4.1 Column

A 3-m-tall viaduct column supporting a mass of 50 N sec2/cm is subjected to a shock acceleration that is

the same as the 1940 El Centro seismogram.

14.4.1.1 Load – Displacement Relationship

Using the same procedures presented in Section 14.3, the moment – curvature relationship of the cross

section and the force – displacement relationship of the column can be determined. It is assumed that the

force – displacement relationship of the column was found to be the same as that shown in Figure 14.25.

Reinforced Concrete Structures 14-21

14.4.1.2 Dynamic Response

To perform the dynamic response analysis, the computer program in Mo (1994) can be used. The

required input data may be determined as follows:

Stiffness 1 ¼

0:6175

2:6 ¼ 0:2375 T=mm

Stiffness 2 ¼

5:4994 2 0:6175

35:75 2 2:6 ¼ 0:1473 T=mm

Stiffness 3 ¼

5:7415 2 5:4994

45:6 2 35:75 ¼ 0:246 T=mm

dc ¼ 2:6 mm

dy ¼ 35:75 mm

du ¼ 45:6 mm

Result:

Maximum deflection ¼ 38.33 mm

14.4.2 Shear Walls

14.4.2.1 Design Approach

14.4.2.1.1 Design Considerations

To insure ductile failure, the two design limitations for overreinforcement and minimum reinforcement

can be derived in terms of the inclination angle, a; of the diagonal concrete struts.

1. Overreinforcement. Mo (1987) shows the inclination angle of the diagonal compression struts in

the “lower balanced reinforcement,” alb; to be

alb ¼ cos21 1

1ly

210

􀀏 􀀐 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1 þ

8120

2 1

vuut

; ð14:77Þ

alb is a function of only one variable — the yield strain of longitudinal steel 1ly : In underreinforced

design, the chosen angle a should be greater than alb:

2. Minimum reinforcement. Mo (1987) also shows the inclination angle for minimum reinforcement,

am; to be

am ¼ cos21 0:0025

fly

f 0c

􀀏 􀀐1=3

" #

ð14:78Þ

am is a function of two variables — the yield strain in longitudinal steel, 1ly, and the concrete

strength, f 0c: Equation 14.77 and Equation 14.78 give the lower and upper limits, respectively, for

the range of a; which insures ductile failure.

After the range of the inclination angle, a; is found, a preliminary angle, a, can be selected within this

range. The wall size, bd; can then be determined by substituting t ¼ V =bd and sd ¼ f 0c cos a into

t ¼ sd sin a cos a (Hsu and Mo, 1985; Mo, 1988).

bd ¼

f 0c sin a cos2a ð14:79Þ

The effective length, d; in Equation 14.79 is usually given in design and the web width, b; needs to be

chosen by the designer. After the wall size, bd; is determined, the actual inclination angle, a; can be

refined by solving a in Equation 14.79.

14-22 Vibration and Shock Handbook

The longitudinal steel can be determined from Alfl ¼ tbd cot a (Hsu and Mo, 1985; Mo, 1988), noting

that tbd becomes Vn at maximum shear

Alfly ¼ Vn cot a ð14:80Þ

The transverse (horizontal) steel can be determined by the specified minimum value (Hsu, 1993):

rt ¼ 0:0045 ð14:81Þ

14.4.2.1.2 Design Procedure

For a given dynamic loading history (or seismogram) and given material properties ðf 0c; fly ; jÞ; the

procedure to design the size of the wall, the vertical steel, and the horizontal steel is as follows:

1. Assume a shear force ðVnÞ:

2. Determine the range of a using Equation 14.77 and Equation 14.78.

3. Select an a value and find b and d using Equation 14.79. After b and d are selected, the actual

angle of inclination can be refined.

4. Determine the longitudinal steel using Equation 14.80 and the horizontal steel using Equation

14.81.

5. Determine the load – deflection relationship using the algorithm mentioned in Figure 14.18.

6. Calculate the stiffness at the ultimate state.

7. Determine the natural period at the ultimate state.

8. Select the time interval Dt # 0:1 times the natural period at the ultimate state.

9. Calculate the initial stiffness using the load – deflection relationship.

10. Calculate the damping coefficient using Equation 14.14.

11. Calculate the effective stiffness using Equation 14.72.

12. Calculate the incremental effective force using Equation 14.73.

13. Solve for the incremental displacement using Equation 14.71.

14. Calculate the incremental velocity using Equation 14.69.

15. Calculate displacement and velocity at the end of time interval using Equation 14.74 and

Equation 14.75.

16. Calculate shear force using the load – deflection relationship.

17. Calculate acceleration at the end of time interval by Equation 14.76. Steps 9 to 17 will provide one

set of solutions.

18. Repeat Steps 9 to 17 for each time interval. This will provide a number of sets of solutions.

19. Determine the calculated maximum shear force in the entire loading history.

a. If the value of the calculated maximum shear force is not greater than the shear force

assumed, the design is finished.

b. If the value of the calculated maximum shear forces is greater than the shear force assumed,

reassume a new value for shear force Vn and repeat Steps 2 to 19 until Step 19a is satisfied.

The algorithm presented in this section has been extended to box tubes subjected to dynamic shear and

torsion and to hybrid reinforced concrete frame – steel wall systems.

14.4.2.2 Missile Impact

A shear wall 10 ft (3.05 m) high by 15 ft (4.57 m) wide (Figure 14.26(a)) is designed to withstand a

missile impact having the force – time relationship shown in Figure 14.26(b) and acting at the top of the

wall (Mo, 1988). For simplicity, the force – time relationship of the missile impact is used instead of

the earthquake seismogram. It is assumed that the mass at the top of the wall and one third of the mass of

the wall m are 0.5 kip sec2/in. (0.088 kN sec2/mm). The material properties are given as follows:

fly ¼ 60 ksi (414 MPa), f 0c ¼ 4000 psi (27.6 MPa), Es ¼ 29 £ 106 psi (2.0 £ 105 MPa), 10 ¼ 0:002;

and j ¼ 0:07: The wall thickness, b; and the reinforcement of the wall, rl and rt; are to be determined.

Reinforced Concrete Structures 14-23

The design of the boundary columns, which should be able to withstand the bending moment, will not be

considered in this example.

Solution

1. Assume Vn ¼ 1600 kips (7117 kN)

2. Determine the range of a

a . cos21 1

0:00207

2ð0:002Þ

􀀏 􀀐 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1 þ

8ð0:002Þ2

ð0:00207Þ2

2 1

; ð14:77Þ

a ¼ 60:48

a # cos21 ð0:0025Þð60;000Þ

4000

􀀏 􀀐1=3

" #

ð14:78Þ

a ¼ 70:48

[ 60:48 , a # 70:48

3. Select a ¼ 60:48 and d ¼ 180 in. (4572 mm)

b $

1;600;000

ð180Þð4000Þ sin 60:4 cos260:4 ð14:79Þ

b $ 10:48 in: ð266 mmÞ

m = 0.5 k.sec2/in

F(f)

0.02 0.01

t (sec)

1.2 × 103K

10'

(a) (b)

15'

Shearwall

1 k. sec/in = 0.175 kN.sec/mm

1 ft = 0.3048 m

1 k = 4.448 kN

1 in = 25.4 mm

1500

Shear force V (kips)

1000

500

0.1 0.2

Displacement y(in)

0.3 0.4

x = 0.07

FIGURE 14.26 Shear wall subjected to missile impact.

14-24 Vibration and Shock Handbook

Let us select b ¼ 12 in: (305 mm). Calculate a ¼ 62:98 from Equation 14.79.

a ¼ 62:98 , 70:48

4. Determine rl and rt

rl ¼

1;600;000 cot 62:9

ð12Þð180Þð60;000Þ ¼ 0:0063 ð14:80Þ

rt ¼ 0:0045 ð14:81Þ

5. Determine the load – deflection relationship using the algorithm mentioned in Figure 14.18. The

load – deflection relationship is determined as shown in Figure 14.26(c).

6. Calculate the stiffness at the ultimate state

ku ¼

1609:9

0:415 ¼ 3879:3 kip=in: ð679 kN=mmÞ

7. Determine the natural period at the ultimate state

Tu ¼ 2p

ffiffiffiffiffiffiffiffiffiffi

0:5

3879:3

¼ 0:071 sec ð14:82Þ

8. Select Dt

Dt ¼ ð0:1Þð0:071Þ ¼ 0:0071 sec

Let us select Dt ¼ 0:005 sec

Dt ¼ 0:005 , 0:0071 sec

9. Calculate the initial stiffness k

k ¼

572:2

0:0778 ¼ 7354:8 kip=in: ð1288 kN=mmÞ

10. Calculate the damping coefficient C

C ¼ ð0:07Þð2Þ½ð0:5Þð7354:8Þ􀀉1=2 ¼ 8:49 kip sec=in: ð1:49 kN sec=mmÞ ð14:83Þ

11. Calculate the effective stiffness k􀀊

k􀀊 ¼ 7354:8 þ

6ð0:5Þ

ð0:005Þ2 þ

3ð8:49Þ

0:005 ¼ 132;448:8 kip=in: ð23194 kN=mmÞ

12. Calculate the incremental effective force

DF ¼ 300 kips ð1334 kNÞ

DF􀀊 ¼ 300 þ ð0:5Þ

6ð0Þ

0:005 þ 3ð0Þ

􀀒 􀀓

þ ð8:49Þ 3ð0Þ þ

0:005ð0Þ

􀀒 􀀓

¼ 300 kips ð1334 kNÞ

13. Determine the incremental displacement Dy

Dy ¼

300

132;448:8 ¼ 0:002265 in: ð0:058 mmÞ ð14:84Þ

Reinforced Concrete Structures 14-25

14. Determine the incremental velocity Dy_

Dy_ ¼

0:005 ð0:002265Þ 2 3ð0Þ 2

0:005

2 ð0Þ ¼ 1:359 in:=sec ð34:52 mm=secÞ ð14:85Þ

15. Calculate the displacement and velocity at the end of the time interval

yiþ1 ¼ 0 þ 0:002265 ¼ 0:002265 in: ð0:058 mmÞ ð14:86Þ

y_iþ1 ¼ 0 þ 1:359 ¼ 1:359 in:=sec ð34:52 mm=secÞ ð14:87Þ

16. Determine the shear force V : From Figure 14.26(c), V ¼ 16:7 kips (74.3 kN) when

y ¼ 0:002265 in. (0.058 mm).

17. Calculate the acceleration at the end of the time interval

y€iþ1 ¼

0:5 ½300 2 ð8:49Þð1:359Þ 2 16:7􀀉 ¼ 543:5 in:=sec2 ð13;805 mm=sec2Þ

Steps 9 to 17 provide one set of solutions.

18. Repeating Steps 9 to 17 for each time interval gives a number of sets of solutions, as illustrated in

Table 14.2.

TABLE 14.2 Nonlinear Response — Linear Acceleration Step-by-Step Method for Design Example

(sec)

(kip)

(in.)

(in./sec)

(kip)

y€

(in./sec)

(kip/in.)

􀀊 k

(kip/in.)

(kip)

DF􀀊

(kip)

(in.)

Dy_

(in./sec)

0 0 0 0 0 0 7355 132,449 300 300 0.0023 1.36

0.005 300 0.0023 1.36 17 544 7355 132,449 300 1977 0.0149 3.50

0.010 600 0.0172 4.86 126 865 7355 132,449 300 4658 0.0352 4.37

0.015 900 0.0524 9.23 385 947 7355 132,449 300 7514 0.0567 3.96

0.020 1200 0.1091 13.19 700 776 5549 129,971 2 75 9309 0.0716 1.45

0.025 1125 0.1807 14.64 1015 4 4399 128,341 2 75 9004 0.0702 2 1.81

0.030 1050 0.2509 12.83 1300 2 669 4060 127,846 2 75 6852 0.0536 2 4.66

0.035 975 0.3045 8.17 1455 2 1063 2892 126,084 2 75 3349 0.0266 2 5.89

0.040 900 0.3311 2.28 1530 2 1347 2820 125,976 2 75 2 709 2 0.0056 2 6.83

0.045 825 0.3255 2 4.55 989 2 280 4621 128,659 2 75 2 3322 2 0.0258 2 1.13

0.050 750 0.2997 2 5.68 870 2 164 4621 128,659 2 75 2 3846 2 0.0299 2 0.49

0.055 675 0.2698 2 6.17 732 2 31 4621 128,659 2 75 2 3948 2 0.0307 0.17

0.060 600 0.2391 2 6.00 590 101 4621 128,659 2 75 2 3645 2 0.0283 0.78

0.065 525 0.2108 2 5.22 459 202 4621 128,659 2 75 2 3009 2 0.0234 1.13

0.070 450 0.1874 2 4.10 351 253 4621 128,659 2 75 2 2232 2 0.0173 1.28

0.075 375 0.1701 2 2.82 271 246 4621 128,659 2 75 2 1449 2 0.0113 1.05

0.080 300 0.1588 2 1.77 219 186 4621 128,659 2 75 2 888 2 0.0069 0.69

0.085 225 0.1519 2 1.07 187 90 4621 128,659 2 75 2 603 2 0.0047 0.17

0.090 150 0.1472 2 0.90 165 2 19 4621 128,659 2 75 2 661 2 0.0051 2 0.34

0.095 75 0.1421 2 1.24 142 2 116 4621 128,659 2 75 2 1019 2 0.0079 2 0.74

0.100 0 0.1342 2 1.97 105 2 183 4621 128,659 0 2 1502 2 0.0117 2 0.64

0.105 0 0.1225 2 2.62 51 2 67 4621 128,659 0 2 1723 2 0.0134 2 0.03

0.110 0 0.1091 2 2.64 0 36 4621 120,000 0 2 1532 2 0.0128 0.16

0.115 0 0.0963 2 2.49 0 0 0 120,000 0 2 1491 2 0.0124 0

0.120 0 0.839 2 2.49 0 0 0 120,000 0 2 1491 2 0.0124 0

0.125 0 0.0715 2 2.49 0 0 0 120,000 0 2 1491 2 0.0124 0

0.130 0 0.0591 2 2.49 0 0 0 120,000 0 2 1491 2 0.0124 0

1 kip ¼ 4.448 kN; 1 in. ¼ 25.4 mm; 1 in./sec ¼ 25.4 mm/sec; 1 in./sec2 ¼ 25.4 mm/sec2; 1 kip/in. ¼ 0.175 kN/mm.

14-26 Vibration and Shock Handbook

19. Determine the calculated maximum shear force. From Table 14.2, Vmax ¼ 1530 kips

(6805 kN) , assumed Vn ¼ 1600 kips (7117 kN).

The dynamic inelastic response calculated in Table 14.2 is plotted in Figure 14.27. Also plotted for

comparison is the linear elastic response obtained by a similar step-by-step analysis. The effect of the

inelastic response on the displacement shows up clearly in this comparison.

14.4.2.3 Explosion Interaction

The basic algorithm of Section 14.2.4 is modified to allow the analysis of a framed shear wall which

interacts with a steel structure as shown in Figure 14.28.

Time T (10−3 sec)

Inelastic response

Elastic

response

1 in. = 2.54 cm

10 20 30 40 50 60 70 80 90 100 110 120 130

−4

−8

Displacement Y (10−2 in.)

FIGURE 14.27 Comparison of inelastic with elastic response.

Shearwall Steel Structure

4.57 m

3.05 m

thickness = 0.225m I= 5.0 × 10−3 m4

fy = 414 Mpa

f′c = 27.6 Mpa

p1 = 0.0042

pt = 0.0025

mass = 0.105 kN· sec2/mm mass = 0.1232 kN· sec2/mm

FIGURE 14.28 A framed shear wall that interacts with a steel structure.

Reinforced Concrete Structures 14-27

When a collision between the two structures is assumed to be elastic, the new velocities y_i1p and y_i2p can

be computed by assuming the conservation of momentum and energy:

m1y_i1 þ m2y_i2 ¼ m1y_i1p þ m2y_i2p ð14:88Þ

and

m1y_2i

1 þ

m2y_2i

2 ¼

m1y_2i

1p þ

m2y_2i

2p ð14:89Þ

These equations give the post-impact velocity as a function of the preimpact velocity:

y_2i

1p ¼ y_i1 2

1 þ R ð_yi1 2 y_i2Þ ð14:90Þ

and

y_2i

2p ¼ y_i2 2

1 þ R ð_yi2 2 y_i1Þ ð14:91Þ

where R is the mass ratio m2=m1 and y_i1p and y_i2p are the relative postimpact velocities for systems 1 and 2

at time ti:

The algorithm in Section 14.2.4 may be modified to accommodate explosion interaction by allowing

for two structures instead of one and inserting the following step. If the two structures collide, (a)

determine the postimpact velocities using Equation 14.90 and Equation 14.91, and (b) recalculate the

acceleration using the postimpact velocities.

When the explosion interaction occurs, the response is shown in Figure 14.29(c) and Figure 14.29(d)

with the assumption that the initial gap between the two structures is zero. For Figure 14.29(a) and Figure

14.29(b), the gap is so large that the two structures cannot collide. A comparison of Figure 14.29(b) and

Figure 14.29(d) shows that the maximum force is significantly greater when the explosion interaction

occurs than it is when no explosion interaction occurs.

0 1 2 3 4 5 6

(a) Time (sec) (b)

Deflection (mm)

0 1 2 3

Time (sec)

Force (kN) Force (kN)

4 5 6

3000

1000

−1000

−3000

−5

0 1 2 3

Deflection (mm)

4 5 6 0 1 2 3

Time (sec)

4 5 6

3000

1000

−1000

−3000

−5

FIGURE 14.29 In (a) and (b) no explosion interaction occurs, but in; (c) and (d) explosion interaction does occur.

14-28 Vibration and Shock Handbook

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