Chapter 5: CONCLUSION AND RECOMMENDATIONS
5.3 Recommendations for Future Study
63 d. Proposed Design Charts:
i. The proposed design charts are only applicable to the design of propped cantilever RC walls subjected to uniformly distributed boiler blast loads with a duration of blast of 40 seconds and more.
ii. The basis of the design is to ensure a satisfactory response of the wall by adjusting the safe standoff distance based on real life case scenarios.
iii. It is applicable to a wide range of water holding capacity of boilers making it a simple and suitable tool for the designers.
iv. For existing walls, these design charts will also help the designers to select a suitable standoff distance for which the wall will sustain the maximum blast pressure in a boiler blast event.
64
REFERENCES
J. Li and H. Hao, “A simplified numerical method for blast induced structural response analysis,” International Journal of Protective Structures, vol. 5, no. 3, pp. 323–
348, 2014.
Y. Xia, C. Wu, F. Zhang, Z. X. Li, and T. Bennett, “Numerical analysis of foam-protected RC members under blast loads,” International Journal of Protective Structures, vol. 5, no. 4, pp. 367–390, 2014.
X. Yin, X. Gu, F. Lin, and X. Kuang, “Numerical Analysis of Blast Loads inside Buildings,” Computational Structural Engineering, pp. 681–690, 2009.
W. Xiao, M. Andrae, L. Ruediger, and N. Gebbeken, “Numerical prediction of blast wall effectiveness for structural protection against air blast,” Procedia Engineering, vol.
199, pp. 2519–2524, 2017.
S. Schuldt and K. El-Rayes, “Quantifying Blast Effects on Constructed Facilities behind Blast Walls,” Journal of Performance of Constructed Facilities, vol. 31, no. 4, pp.
1–14, 2017.
X. Q. Zhou and H. Hao, “Numerical prediction of reinforced concrete exterior wall response to blast loading,” Advances in Structural Engineering, vol. 11, no. 4, pp.
355–367, 2008.
M. Abdel-Mooty, S. Alhayawei, and M. Issa, “Numerical evaluation of the performance of two-way RC panels under blast loads,” WIT Transactions on the Built Environment, vol. 141, no. June 2014, pp. 13–25, 2014.
M. Abdel-Mooty, S. Alhayawei, and M. Issa, “Performance of one-way reinforced concrete walls subjected to blast loads,” International Journal of Safety and Security Engineering, vol. 6, no. 2, pp. 406–417, 2016.
V. Karlos and G. Solomon, “Claculation of Blast Loads for Applications to Structural Components,” Publications Office of the European Union ( JRC Technical Reports), pp. 1–58, 2013.
M. F. Ibrahim, H. A. El-arabaty, and I. S. Moharram, “Effect of steam boiler explosion on boiler room and adjacent buildings structure,” International Journal of Engineering Science Invention (IJESI, vol. 8, no. 02, pp. 17–37, 2019.
65 T. Kim, D. Lee, H. Ko, and S. Cho, Design of Blast-Resistant Buildings in Petrochemical Facilities.2nd Edition, Task Committee on Blast Resistant Design of the Petrochemical Committee of the Energy Division of the American Society of Civil Engineers, 2006.
Razaqpur AG, Tolba A, Contestabile E. Blast loading response of reinforced concrete panels reinforced with externally bonded GFRP laminates. Elsevier, Composite, Part B38; 535–546, 2007.
Mander, J.B., M.J.N. Priestley, and R. Park 1984. Theoretical Stress-Strain Model for Confined Concrete. Journal of Structural Engineering. ASCE. 114(3). 1804-1826.
Islam, T., Experimental Investigation of Boiler Blast Load on Building Structures, 2022 (to be published)
CSI Analysis Reference Manual for SAP2000®, ETABS®, SAFE® and CSiBridge®, November 2017, Rev. 18.
66
Appendix A
EXPERIMENTAL BOILER BLAST DATA
0 5 10 15 20 25 30 35 40 45 50 55
-0.3 -0.2 -0.1 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7
-6.27 -4.18 -2.09 0.00 2.09 4.18 6.27 8.35 10.44 12.53 14.62
Explosion Pressure (psf)
Explosion Pressure (kPa)
Time (Seconds)
Pressure time-history of B40S8 Positive Pressure
Negative Pressure
Figure A.1: Experimental pressure time history of B40S8 (Islam, 2022).
0 5 10 15 20 25 30 35 40 45
-0.1 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8
-2.09 0.00 2.09 4.18 6.27 8.35 10.44 12.53 14.62 16.71
Explosion Pressure (psf)
Explosion Pressure (kPa)
Time (Seconds)
Pressure time-history of B40S10 Positive Pressure
Negative Pressure
Figure A.2: Experimental pressure time history of B40S10 (Islam, 2022).
67
0 5 10 15 20 25 30 35 40 45 50 55 60 65
-0.2 -0.1 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7
-4.18 -2.09 0.00 2.09 4.18 6.27 8.35 10.44 12.53 14.62
Explosion Pressure (psf)
Explosion Pressure (kPa)
Time (Seconds)
Pressure time-history of B40S12 Positive Pressure
Negative Pressure
Figure A.3: Experimental pressure time history of B40S12 (Islam, 2022).
0 5 10 15 20 25 30 35 40 45 50 55
-0.2 -0.1 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7
-4.18 -2.09 0.00 2.09 4.18 6.27 8.35 10.44 12.53 14.62
Explosion Pressure (psf)
Explosion Pressure (kPa)
Time (Seconds)
Pressure time-history of B50S8 Positive Pressure
Negative Pressure
Figure A.4: Experimental pressure time history of B50S8 (Islam, 2022).
68
0 5 10 15 20 25 30 35 40 45 50 55 60 65
-0.2 -0.1 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7
-4.18 -2.09 0.00 2.09 4.18 6.27 8.35 10.44 12.53 14.62
Explosion Pressure (psf)
Explosion Pressure (kPa)
Time (Seconds)
Pressure time-history of B50S10 Positive Pressure
Negative Pressure
Figure A.5: Experimental pressure time history of B50S10 (Islam, 2022).
0 5 10 15 20 25 30 35 40 45 50 55 60 65
-0.2 -0.1 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7
-4.18 -2.09 0.00 2.09 4.18 6.27 8.35 10.44 12.53 14.62
Explosion Pressure (psf)
Explosion Pressure (kPa)
Time (Seconds)
Pressure time-history of B50S12 Positive Pressure
Negative Pressure
Figure A.6: Experimental pressure time history of B50S12 (Islam, 2022).
69
0 5 10 15 20 25 30 35 40 45 50 55 60 65
-0.1 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8
-2.09 0.00 2.09 4.18 6.27 8.35 10.44 12.53 14.62 16.71
Explosion Pressure (psf)
Explosion Pressure (kPa)
Time (Seconds)
Pressure time-history of B60S8 Positive Pressure
Negative Pressure
Figure A.7: Experimental pressure time history of B60S8 (Islam, 2022).
0 5 10 15 20 25 30 35 40 45 50 55 60 65 70
-0.3 -0.2 -0.1 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7
-6.27 -4.18 -2.09 0.00 2.09 4.18 6.27 8.35 10.44 12.53 14.62
Explosion Pressure (psf)
Explosion Pressure (kPa)
Time (Seconds)
Pressure time-history of B60S10 Positive Pressure
Negative Pressure
Figure A.8: Experimental pressure time history of B60S10 (Islam, 2022).
70
0 5 10 15 20 25 30 35 40 45 50
-0.2 -0.1 0.0 0.1 0.2 0.3 0.4 0.5 0.6
-4.18 -2.09 0.00 2.09 4.18 6.27 8.35 10.44 12.53
Explosion Pressure (psf)
Explosion Pressure (kPa)
Time (Seconds)
Pressure time-history of B60S12 Positive Pressure
Negative Pressure
Figure A.9: Experimental pressure time history of B60S12 (Islam, 2022).
71
Appendix B
SCALED UP BOILER BLAST LOAD DATA
Calculation of scale factor:
Experimental boiler blast pressures obtained for the following characteristics- Water holding capacity = 15 litres
Volume of boiler = 0.545 cft = 0.0154 m3
Scaling factor for pressures in this research has been calculated for an assumed full scale boiler with a water holding capacity of 2.5 metric ton or 2,500 litres.
Now,
Scaling factor = 2,500/15 = 166.67
This scaling factor has been multiplied to the pressure time-histories mentioned in Appendix A. The pressure- time histories were then converted to load-time history by multiplying the pressures with the area of the RC walls, i.e. (4.575 m x 4.575 m). The resulting load-time histories are shown in Figure B.1 to Figure B.9.
0 5 10 15 20 25 30 35 40 45 50 55
-10 0 10 20 30 40
-2.25 0.00 2.25 4.50 6.74 8.99
Explosion Load (kips)
Explosion Load (kN)
Time (Seconds)
Load time-history of B40S8
Figure B.10: Scaled up load time-history of B40S8.
72
0 5 10 15 20
-10 0 10 20 30 40
-2.25 0.00 2.25 4.50 6.74 8.99
Explosion Load (kips)
Explosion Load (kN)
Time (Seconds)
Load time-history of B40S10
Figure B.11: Scaled up load time-history of B40S10.
0 5 10 15 20 25 30 35 40 45 50 55 60 65
-20 -10 0 10 20 30
-4.50 -2.25 0.00 2.25 4.50 6.74
Explosion Load (kips)
Explosion Load (kN)
Time (Seconds)
Load time-history of B40S12
Figure B.12: Scaled up load time-history of B40S12.
73
0 5 10 15 20 25 30 35 40 45
-10 0 10 20 30 40 50
-2.25 0.00 2.25 4.50 6.74 8.99 11.24
Explosion Load (kips)
Explosion Load (kN)
Time (Seconds)
Load time-history of B50S8
Figure B.13: Scaled up load time-history of B50S8.
0 5 10 15 20 25 30 35 40 45
-10 0 10 20 30 40
-2.25 0.00 2.25 4.50 6.74 8.99
Explosion Load (kips)
Explosion Load (kN)
Time (Seconds)
Load time-history of B50S10
Figure B.14: Scaled up load time-history of B50S10.
74
0 5 10 15 20 25 30 35 40
-20 -10 0 10 20 30 40
-4.50 -2.25 0.00 2.25 4.50 6.74 8.99
Explosion Load (kips)
Explosion Load (kN)
Time (Seconds)
Load time-history of B50S12
Figure B.15: Scaled up load time-history of B50S12.
0 5 10 15 20 25 30 35
-10 0 10 20 30 40
-2.25 0.00 2.25 4.50 6.74 8.99
Explosion Load (kips)
Explosion Load (kN)
Time (Seconds)
Load time-history of B60S8
Figure B.16: Scaled up load time-history of B60S8.
75
0 5 10 15 20 25 30 35 40 45 50
-20 -10 0 10 20 30 40
-4.50 -2.25 0.00 2.25 4.50 6.74 8.99
Explosion Load (kips)
Explosion Load (kN)
Time (Seconds)
Load time-history of B60S10
Figure B.17: Scaled up load time-history of B60S10.
0 5 10 15 20 25 30 35 40 45
-10 0 10 20 30
-2.25 0.00 2.25 4.50 6.74
Explosion Load (kips)
Explosion Load (kN)
Time (Seconds)
Load time-history of B60S12
Figure B.18: Scaled up load time-history of B60S12.
76
Appendix C
DYNAMIC DESIGN OF RC WALL
[Note: All the tables, figures, equations along with design approach have been taken from
“Design of Blast-Resistant Buildings in Petrochemical Facilities.2nd Edition, Task Committee on Blast Resistant Design of the Petrochemical Committee of the Energy Division of the American Society of Civil Engineers, 2006” unless otherwwise mentioned]
1. Design of B40-6 Design data:
Length of Wall, L = 15 ft
Width of Wall, B = 15 ft
Unit width of Wall, b = 12 inch
Thickness of wall, t = 6 inch
Compressive strength of concrete, f’c = 4 ksi
Yield Strength of rebar, fy = 60 ksi
Vertical reinforcement, As,ver = 0.40 in2
Horizontal reinforcement, As,hor = 0.40 in2
Dynamic Increase Factor for
Concrete in bending, DIFcb = 1.19
Dynamic Increase Factor for
Concrete in shear, DIFcs = 1
Dynamic Increase Factor for
Reinforcing Steel in bending, DIFs = 1.17 Stress Increase Factor for Concrete, SIFc = 1 Stress Increase Factor for Reinforcing Steel, SIFs = 1.1 Effective duration of blast, td = 50 sec
Peak load = 38038.8 N
Peak Pressure, Fo = 0.2639 psi
77 Computation of Bending and Shear Resistance
Variable Formula Result Unit Remarks
Fdy = SIFs×DIFsb×fy 77.22 ksi
f'dc = SIFc×DIFcb×f'c 4.76 ksi
Calculation of Bending Resistance
As,ver = 0.40 sq. inch
dsb = As,ver×Fdy/0.85×f'dc×
b 0.64 inch
d = t/2 3 inch
As,min = 0.10 sq. inch Okay
As,ver = 0.40 sq. inch
Mp = As,ver×Fdy/(d-dsb/2) 82.84 kip-inch
Bending
Resistance, Rb = 8Mp/L 3681.72 lb Table 6.1
Unit Resistance,
Rb,unit = Rb/(L×b) 1.70 psi
Calculation of Shear Resistance
f'dc = SIFc×DIFcs×f'c 4 ksi
Vn = 2√f'dc×bd 4553.679831 lb
Shear at distance
"d", Rs = Vn×L/(0.5L-d) 9421.406546 lb Unit Resistance,
Rs,unit = Rs/L×b 4.36 psi
Bending Controls
Design unit Resistance, Ru = 1.70 psi
Allowable ductility
ratio μa N/A Table 5.B.3
Allowable support
rotation θa 1 degrees Table 5.B.3
(Low)
Computation of SDOF Equivalent System
Variable Formula Result Unit Remarks
Gross moment of
inertia, Ig = bh3/12 216 in4
Ec = 57000√f'c 3604996.53 psi
Es = 29000000 psi
n = Es/Ec 8.04
Transformed rebar
area n×As 3.22 sq. inch
Location of
transformed neutral axis, dna =
[((n×As(n×As+2bd))^0.
5)-(n×As)]/b 1.03 inch
Cracked moment of
inertia, Icr = b×dna3/3 + n×As(d-dna)2 16.86 in4
78 Computation of SDOF Equivalent System
Variable Formula Result Unit Remarks
Avg. moment of
inertia, Ia = (Ig + Icr)/2 116.43 in4
Effective stiffness,
Ke = 185Ec×Ia/5L3 2.66 kip/inch
Beam mass, M = Wall weight/g 0.0029 k-s2/inch Unit beam mass,
Munit = M/L×b 0.0013 psi-s2/in
Elastic Mass Factor KM,elastic 0.45 Table 6.3
Elastic Load Factor KL,elastic 0.58 Table 6.3
Elastic Load-Mass
Factor, KLM,elastic = KM,elastic/KL,elastic 0.78
Plastic Mass Factor KM,plastic 0.33 Table 6.3
Plastic Load Factor KL,plastic 0.5 Table 6.3
Plastic Load-Mass
Factor, KLM,plastic = KM,plastic/KL,plastic 0.66 Avg. Load-Mass
Factor, KLM,avg. = (KLM,elastic+KLM,plastic)/2 0.72 Equivalent Mass,
Me = KLM,avg.×Munit 0.0010 psi-s2/in
Unit Effective
Stiffness, Ke,unit = Ke/L×b 1.23 psi/in
Period of
vibration,tn = 2π√(Me/Ke) 0.18 sec.
T = td/tn = 283.88
Ru/Fo = 6.46
Ductility demand,
μd 0.59880
By Iteration using eq.
6.10 and 6.11 Elastic deflection,
ye = Ru/Ke,unit 1.38 inch
Max. deflection, ym
= μd×ye 0.8279 inch
Max. support
rotation, θd = arctan (ym/0.5L) 0.527 degree Okay
Iteration for Ductility Demand
Fo/Rm 0.1650 Inverse of Ru/Fo
as initial value for iteration
μd 1/(2(1- Fo/Rm)) 0.59880
Fo/Rm ((2μd -1)^0.5/(πT))+(((2μd -1)×T)/(2μd (T+0.7))) 0.1651
79 2. Design of B40-8
Design data:
Length of Wall, L = 15 ft
Width of Wall, B = 15 ft
Unit width of Wall, b = 12 inch
Thickness of wall, t = 8 inch
Compressive strength of concrete, f’c = 4 ksi
Yield Strength of rebar, fy = 60 ksi
Vertical reinforcement, As,ver = 0.53 in2
Horizontal reinforcement, As,hor = 0.53 in2
Dynamic Increase Factor for
Concrete in bending, DIFcb = 1.19
Dynamic Increase Factor for
Concrete in shear, DIFcs = 1
Dynamic Increase Factor for
Reinforcing Steel in bending, DIFs = 1.17 Stress Increase Factor for Concrete, SIFc = 1 Stress Increase Factor for Reinforcing Steel, SIFs = 1.1 Effective duration of blast, td = 50 sec
Peak load = 38038.8 N
Peak Pressure, Fo = 0.2639 psi
Computation of Bending and Shear Resistance
Variable Formula Result Unit Remarks
Fdy = SIFs×DIFsb×fy 77.22 ksi
f'dc = SIFc×DIFcb×f'c 4.76 ksi
Calculation of Bending Resistance
As,ver = 0.53 sq. inch
dsb = As,ver×Fdy/0.85×f'dc×
b 0.84 inch
d = t/2 4 inch
As,min = 0.13 sq. inch Okay
As,ver = 0.53 sq. inch
Mp = As,ver×Fdy/(d-dsb/2) 146.46 kip-inch
Bending
Resistance, Rb = 8Mp/L 6509.20 lb Table 6.1
80 Computation of Bending and Shear Resistance
Variable Formula Result Unit Remarks
Unit Resistance,
Rb,unit = Rb/(L×b) 3.01 psi
Calculation of Shear Resistance
f'dc = SIFc×DIFcs×f'c 4 ksi
Vn = 2√f'dc×bd 6071.573108 lb
Shear at distance
"d", Rs = Vn×L/(0.5L-d) 12707.94371 lb Unit Resistance,
Rs,unit = Rs/L×b 5.88 psi
Bending Controls
Design unit Resistance, Ru = 3.01 psi
Allowable ductility
ratio μa N/A Table 5.B.3
Allowable support
rotation θa 1 degrees Table 5.B.3
(Low)
Computation of SDOF Equivalent System
Variable Formula Result Unit Remarks
Gross moment of
inertia, Ig = bh3/12 512 in4
Ec = 57000√f'c 3604996.53 psi
Es = 29000000 psi
n = Es/Ec 8.04
Transformed rebar
area n×As 4.26 sq. inch
Location of
transformed neutral axis, dna =
[((n×As(n×As+2bd))^0.
5)-(n×As)]/b 1.37 inch
Cracked moment of
inertia, Icr = b×dna3/3 + n×As(d-dna)2 39.78 in4 Avg. moment of
inertia, Ia = (Ig + Icr)/2 275.89 in4
Effective stiffness,
Ke = 185Ec×Ia/5L3 6.31 kip/inch
Beam mass, M = Wall weight/g 0.0039 k-s2/inch Unit beam mass,
Munit = M/L×b 0.0018 psi-s2/in
Elastic Mass Factor KM,elastic 0.45 Table 6.3
Elastic Load Factor KL,elastic 0.58 Table 6.3
Elastic Load-Mass
Factor, KLM,elastic = KM,elastic/KL,elastic 0.78
Plastic Mass Factor KM,plastic 0.33 Table 6.3
Plastic Load Factor KL,plastic 0.5 Table 6.3
81 Computation of SDOF Equivalent System
Variable Formula Result Unit Remarks
Plastic Load-Mass
Factor, KLM,plastic = KM,plastic/KL,plastic 0.66 Avg. Load-Mass
Factor, KLM,avg. = (KLM,elastic+KLM,plastic)/2 0.72 Equivalent Mass,
Me = KLM,avg.×Munit 0.0013 psi-s2/in
Unit Effective
Stiffness, Ke,unit = Ke/L×b 2.92 psi/in
Period of
vibration,tn = 2π√(Me/Ke) 0.13 sec.
T = td/tn = 378.45
Ru/Fo = 11.42
Ductility demand,
μd 0.54801
By Iteration using eq.
6.10 and 6.11 Elastic deflection,
ye = Ru/Ke,unit 1.03 inch
Max. deflection, ym
= μd×ye 0.5653 inch
Max. support
rotation, θd = arctan (ym/0.5L) 0.360 degree Okay
Iteration for Ductility Demand
Fo/Rm 0.0876 Inverse of Ru/Fo
as initial value for iteration
μd 1/(2(1- Fo/Rm)) 0.54801
Fo/Rm ((2μd -1)^0.5/(πT))+(((2μd -1)×T)/(2μd (T+0.7))) 0.0877
82 3. Design of B40-10
Design data:
Length of Wall, L = 15 ft
Width of Wall, B = 15 ft
Unit width of Wall, b = 12 inch
Thickness of wall, t = 10 inch
Compressive strength of concrete, f’c = 4 ksi
Yield Strength of rebar, fy = 60 ksi
Vertical reinforcement, As,ver = 0.67 in2
Horizontal reinforcement, As,hor = 0.67 in2
Dynamic Increase Factor for
Concrete in bending, DIFcb = 1.19
Dynamic Increase Factor for
Concrete in shear, DIFcs = 1
Dynamic Increase Factor for
Reinforcing Steel in bending, DIFs = 1.17 Stress Increase Factor for Concrete, SIFc = 1 Stress Increase Factor for Reinforcing Steel, SIFs = 1.1 Effective duration of blast, td = 50 sec
Peak load = 38038.8 N
Peak Pressure, Fo = 0.2639 psi
Computation of Bending and Shear Resistance
Variable Formula Result Unit Remarks
Fdy = SIFs×DIFsb×fy 77.22 ksi
f'dc = SIFc×DIFcb×f'c 4.76 ksi
Calculation of Bending Resistance
As,ver = 0.67 sq. inch
dsb = As,ver×Fdy/0.85×f'dc×
b 1.07 inch
d = t/2 5 inch
As,min = 0.16 sq. inch Okay
As,ver = 0.67 sq. inch
Mp = As,ver×Fdy/(d-dsb/2) 231.12 kip-inch
Bending
Resistance, Rb = 8Mp/L 10272.05 lb Table 6.1
83 Computation of Bending and Shear Resistance
Variable Formula Result Unit Remarks
Unit Resistance,
Rb,unit = Rb/(L×b) 4.76 psi
Calculation of Shear Resistance
f'dc = SIFc×DIFcs×f'c 7589.466384 ksi
Vn = 2√f'dc×bd 16071.81117 lb
Shear at distance
"d", Rs = Vn×L/(0.5L-d) 7.44 lb Unit Resistance,
Rs,unit = Rs/L×b 7589.466384 psi
Bending Controls
Design unit Resistance, Ru = 4.76 psi
Allowable ductility
ratio μa N/A Table 5.B.3
Allowable support
rotation θa 1 degrees Table 5.B.3
(Low)
Computation of SDOF Equivalent System
Variable Formula Result Unit Remarks
Gross moment of
inertia, Ig = bh3/12 1000 in4
Ec = 57000√f'c 3604996.53 psi
Es = 29000000 psi
n = Es/Ec 8.04
Transformed rebar
area n×As 5.39 sq. inch
Location of
transformed neutral axis, dna =
[((n×As(n×As+2bd))^0.
5)-(n×As)]/b 1.72 inch
Cracked moment of
inertia, Icr = b×dna3/3 + n×As(d-dna)2 78.34 in4 Avg. moment of
inertia, Ia = (Ig + Icr)/2 539.17 in4
Effective stiffness,
Ke = 185Ec×Ia/5L3 12.33 kip/inch
Beam mass, M = Wall weight/g 0.0049 k-s2/inch Unit beam mass,
Munit = M/L×b 0.0022 psi-s2/in
Elastic Mass Factor KM,elastic 0.45 Table 6.3
Elastic Load Factor KL,elastic 0.58 Table 6.3
Elastic Load-Mass
Factor, KLM,elastic = KM,elastic/KL,elastic 0.78
Plastic Mass Factor KM,plastic 0.33 Table 6.3
Plastic Load Factor KL,plastic 0.5 Table 6.3
84 Computation of SDOF Equivalent System
Variable Formula Result Unit Remarks
Plastic Load-Mass
Factor, KLM,plastic = KM,plastic/KL,plastic 0.66 Avg. Load-Mass
Factor, KLM,avg. = (KLM,elastic+KLM,plastic)/2 0.72 Equivalent Mass,
Me = KLM,avg.×Munit 0.0016 psi-s2/in
Unit Effective
Stiffness, Ke,unit = Ke/L×b 5.71 psi/in
Period of
vibration,tn = 2π√(Me/Ke) 0.11 sec.
T = td/tn = 473.20
Ru/Fo = 18.02
Ductility demand,
μd 0.52938
By Iteration using eq.
6.10 and 6.11 Elastic deflection,
ye = Ru/Ke,unit 0.83 inch
Max. deflection, ym
= μd×ye 0.4410 inch
Max. support
rotation, θd = arctan (ym/0.5L) 0.281 degree Okay
Iteration for Ductility Demand
Fo/Rm 0.0555 Inverse of Ru/Fo
as initial value for iteration
μd 1/(2(1- Fo/Rm)) 0.52938
Fo/Rm ((2μd -1)^0.5/(πT))+(((2μd -1)×T)/(2μd (T+0.7))) 0.0556
85 4. Design of B50-6
Design data:
Length of Wall, L = 15 ft
Width of Wall, B = 15 ft
Unit width of Wall, b = 12 inch
Thickness of wall, t = 6 inch
Compressive strength of concrete, f’c = 4 ksi
Yield Strength of rebar, fy = 60 ksi
Vertical reinforcement, As,ver = 0.40 in2
Horizontal reinforcement, As,hor = 0.40 in2
Dynamic Increase Factor for
Concrete in bending, DIFcb = 1.19
Dynamic Increase Factor for
Concrete in shear, DIFcs = 1
Dynamic Increase Factor for
Reinforcing Steel in bending, DIFs = 1.17 Stress Increase Factor for Concrete, SIFc = 1 Stress Increase Factor for Reinforcing Steel, SIFs = 1.1 Effective duration of blast, td = 40 sec
Peak load = 41380.3 N
Peak Pressure, Fo = 0.2871 psi
Computation of Bending and Shear Resistance
Variable Formula Result Unit Remarks
Fdy = SIFs×DIFsb×fy 77.22 ksi
f'dc = SIFc×DIFcb×f'c 4.76 ksi
Calculation of Bending Resistance
As,ver = 0.40 sq. inch
dsb = As,ver×Fdy/0.85×f'dc×
b 0.64 inch
d = t/2 3 inch
As,min = 0.10 sq. inch Okay
As,ver = 0.40 sq. inch
Mp = As,ver×Fdy/(d-dsb/2) 82.84 kip-inch
Bending
Resistance, Rb = 8Mp/L 3681.72 lb Table 6.1
86 Computation of Bending and Shear Resistance
Variable Formula Result Unit Remarks
Unit Resistance,
Rb,unit = Rb/(L×b) 1.70 psi
Calculation of Shear Resistance
f'dc = SIFc×DIFcs×f'c 4 ksi
Vn = 2√f'dc×bd 4553.679831 lb
Shear at distance
"d", Rs = Vn×L/(0.5L-d) 9421.406546 lb Unit Resistance,
Rs,unit = Rs/L×b 4.36 psi
Bending Controls
Design unit Resistance, Ru = 1.70 psi
Allowable ductility
ratio μa N/A Table 5.B.3
Allowable support
rotation θa 1 degrees Table 5.B.3
(Low)
Computation of SDOF Equivalent System
Variable Formula Result Unit Remarks
Gross moment of
inertia, Ig = bh3/12 216 in4
Ec = 57000√f'c 3604996.53 psi
Es = 29000000 psi
n = Es/Ec 8.04
Transformed rebar
area n×As 3.22 sq. inch
Location of
transformed neutral axis, dna =
[((n×As(n×As+2bd))^0.
5)-(n×As)]/b 1.03 inch
Cracked moment of
inertia, Icr = b×dna3/3 + n×As(d-dna)2 16.86 in4 Avg. moment of
inertia, Ia = (Ig + Icr)/2 116.43 in4
Effective stiffness,
Ke = 185Ec×Ia/5L3 2.66 kip/inch
Beam mass, M = Wall weight/g 0.0029 k-s2/inch Unit beam mass,
Munit = M/L×b 0.0013 psi-s2/in
Elastic Mass Factor KM,elastic 0.45 Table 6.3
Elastic Load Factor KL,elastic 0.58 Table 6.3
Elastic Load-Mass
Factor, KLM,elastic = KM,elastic/KL,elastic 0.78
Plastic Mass Factor KM,plastic 0.33 Table 6.3
Plastic Load Factor KL,plastic 0.5 Table 6.3
87 Computation of SDOF Equivalent System
Variable Formula Result Unit Remarks
Plastic Load-Mass
Factor, KLM,plastic = KM,plastic/KL,plastic 0.66 Avg. Load-Mass
Factor, KLM,avg. = (KLM,elastic+KLM,plastic)/2 0.72 Equivalent Mass,
Me = KLM,avg.×Munit 0.0010 psi-s2/in
Unit Effective
Stiffness, Ke,unit = Ke/L×b 1.23 psi/in
Period of
vibration,tn = 2π√(Me/Ke) 0.18 sec.
T = td/tn = 227.11
Ru/Fo = 5.94
Ductility demand,
μd 0.60096
By Iteration using eq.
6.10 and 6.11 Elastic deflection,
ye = Ru/Ke,unit 1.38 inch
Max. deflection, ym
= μd×ye 0.8309 inch
Max. support
rotation, θd = arctan (ym/0.5L) 0.529 degree Okay
Iteration for Ductility Demand
Fo/Rm 0.1680 Inverse of Ru/Fo
as initial value for iteration
μd 1/(2(1- Fo/Rm)) 0.60096
Fo/Rm ((2μd -1)^0.5/(πT))+(((2μd -1)×T)/(2μd (T+0.7))) 0.1680
88 5. Design of B50-8
Design data:
Length of Wall, L = 15 ft
Width of Wall, B = 15 ft
Unit width of Wall, b = 12 inch
Thickness of wall, t = 8 inch
Compressive strength of concrete, f’c = 4 ksi
Yield Strength of rebar, fy = 60 ksi
Vertical reinforcement, As,ver = 0.53 in2
Horizontal reinforcement, As,hor = 0.53 in2
Dynamic Increase Factor for
Concrete in bending, DIFcb = 1.19
Dynamic Increase Factor for
Concrete in shear, DIFcs = 1
Dynamic Increase Factor for
Reinforcing Steel in bending, DIFs = 1.17 Stress Increase Factor for Concrete, SIFc = 1 Stress Increase Factor for Reinforcing Steel, SIFs = 1.1 Effective duration of blast, td = 40 sec
Peak load = 41380.3 N
Peak Pressure, Fo = 0.2871 psi
Computation of Bending and Shear Resistance
Variable Formula Result Unit Remarks
Fdy = SIFs×DIFsb×fy 77.22 ksi
f'dc = SIFc×DIFcb×f'c 4.76 ksi
Calculation of Bending Resistance
As,ver = 0.53 sq. inch
dsb = As,ver×Fdy/0.85×f'dc×
b 0.84 inch
d = t/2 4 inch
As,min = 0.13 sq. inch Okay
As,ver = 0.53 sq. inch
Mp = As,ver×Fdy/(d-dsb/2) 146.46 kip-inch
Bending
Resistance, Rb = 8Mp/L 6509.20 lb Table 6.1
89 Computation of Bending and Shear Resistance
Variable Formula Result Unit Remarks
Unit Resistance,
Rb,unit = Rb/(L×b) 3.01 psi
Calculation of Shear Resistance
f'dc = SIFc×DIFcs×f'c 4 ksi
Vn = 2√f'dc×bd 6071.573108 lb
Shear at distance
"d", Rs = Vn×L/(0.5L-d) 12707.94371 lb Unit Resistance,
Rs,unit = Rs/L×b 5.88 psi
Bending Controls
Design unit Resistance, Ru = 3.01 psi
Allowable ductility
ratio μa N/A Table 5.B.3
Allowable support
rotation θa 1 degrees Table 5.B.3
(Low)
Computation of SDOF Equivalent System
Variable Formula Result Unit Remarks
Gross moment of
inertia, Ig = bh3/12 512 in4
Ec = 57000√f'c 3604996.53 psi
Es = 29000000 psi
n = Es/Ec 8.04
Transformed rebar
area n×As 4.26 sq. inch
Location of
transformed neutral axis, dna =
[((n×As(n×As+2bd))^0.
5)-(n×As)]/b 1.37 inch
Cracked moment of
inertia, Icr = b×dna3/3 + n×As(d-dna)2 39.78 in4 Avg. moment of
inertia, Ia = (Ig + Icr)/2 275.89 in4
Effective stiffness,
Ke = 185Ec×Ia/5L3 6.31 kip/inch
Beam mass, M = Wall weight/g 0.0039 k-s2/inch Unit beam mass,
Munit = M/L×b 0.0018 psi-s2/in
Elastic Mass Factor KM,elastic 0.45 Table 6.3
Elastic Load Factor KL,elastic 0.58 Table 6.3
Elastic Load-Mass
Factor, KLM,elastic = KM,elastic/KL,elastic 0.78
Plastic Mass Factor KM,plastic 0.33 Table 6.3
Plastic Load Factor KL,plastic 0.5 Table 6.3
90 Computation of SDOF Equivalent System
Variable Formula Result Unit Remarks
Plastic Load-Mass
Factor, KLM,plastic = KM,plastic/KL,plastic 0.66 Avg. Load-Mass
Factor, KLM,avg. = (KLM,elastic+KLM,plastic)/2 0.72 Equivalent Mass,
Me = KLM,avg.×Munit 0.0013 psi-s2/in
Unit Effective
Stiffness, Ke,unit = Ke/L×b 2.92 psi/in
Period of
vibration,tn = 2π√(Me/Ke) 0.13 sec.
T = td/tn = 302.76
Ru/Fo = 10.50
Ductility demand,
μd 0.55261
By Iteration using eq.
6.10 and 6.11 Elastic deflection,
ye = Ru/Ke,unit 1.03 inch
Max. deflection, ym
= μd×ye 0.5701 inch
Max. support
rotation, θd = arctan (ym/0.5L) 0.363 degree Okay
Iteration for Ductility Demand
Fo/Rm 0.0952 Inverse of Ru/Fo
as initial value for iteration
μd 1/(2(1- Fo/Rm)) 0.55261
Fo/Rm ((2μd -1)^0.5/(πT))+(((2μd -1)×T)/(2μd (T+0.7))) 0.0953
91 6. Design of B50-10
Design data:
Length of Wall, L = 15 ft
Width of Wall, B = 15 ft
Unit width of Wall, b = 12 inch
Thickness of wall, t = 10 inch
Compressive strength of concrete, f’c = 4 ksi
Yield Strength of rebar, fy = 60 ksi
Vertical reinforcement, As,ver = 0.67 in2
Horizontal reinforcement, As,hor = 0.67 in2
Dynamic Increase Factor for
Concrete in bending, DIFcb = 1.19
Dynamic Increase Factor for
Concrete in shear, DIFcs = 1
Dynamic Increase Factor for
Reinforcing Steel in bending, DIFs = 1.17 Stress Increase Factor for Concrete, SIFc = 1 Stress Increase Factor for Reinforcing Steel, SIFs = 1.1 Effective duration of blast, td = 40 sec
Peak load = 41380.3 N
Peak Pressure, Fo = 0.2871 psi
Computation of Bending and Shear Resistance
Variable Formula Result Unit Remarks
Fdy = SIFs×DIFsb×fy 77.22 ksi
f'dc = SIFc×DIFcb×f'c 4.76 ksi
Calculation of Bending Resistance
As,ver = 0.67 sq. inch
dsb = As,ver×Fdy/0.85×f'dc×
b 1.07 inch
d = t/2 5 inch
As,min = 0.16 sq. inch Okay
As,ver = 0.67 sq. inch
Mp = As,ver×Fdy/(d-dsb/2) 231.12 kip-inch
Bending
Resistance, Rb = 8Mp/L 10272.05 lb Table 6.1
92 Computation of Bending and Shear Resistance
Variable Formula Result Unit Remarks
Unit Resistance,
Rb,unit = Rb/(L×b) 4.76 psi
Calculation of Shear Resistance
f'dc = SIFc×DIFcs×f'c 4 ksi
Vn = 2√f'dc×bd 7589.466384 lb
Shear at distance
"d", Rs = Vn×L/(0.5L-d) 16071.81117 lb Unit Resistance,
Rs,unit = Rs/L×b 7.44 psi
Bending Controls
Design unit Resistance, Ru = 4.76 psi
Allowable ductility
ratio μa N/A Table 5.B.3
Allowable support
rotation θa 1 degrees Table 5.B.3
(Low)
Computation of SDOF Equivalent System
Variable Formula Result Unit Remarks
Gross moment of
inertia, Ig = bh3/12 1000 in4
Ec = 57000√f'c 3604996.53 psi
Es = 29000000 psi
n = Es/Ec 8.04
Transformed rebar
area n×As 5.39 sq. inch
Location of
transformed neutral axis, dna =
[((n×As(n×As+2bd))^0.
5)-(n×As)]/b 1.72 inch
Cracked moment of
inertia, Icr = b×dna3/3 + n×As(d-dna)2 78.34 in4 Avg. moment of
inertia, Ia = (Ig + Icr)/2 539.17 in4
Effective stiffness,
Ke = 185Ec×Ia/5L3 12.33 kip/inch
Beam mass, M = Wall weight/g 0.0049 k-s2/inch Unit beam mass,
Munit = M/L×b 0.0022 psi-s2/in
Elastic Mass Factor KM,elastic 0.45 Table 6.3
Elastic Load Factor KL,elastic 0.58 Table 6.3
Elastic Load-Mass
Factor, KLM,elastic = KM,elastic/KL,elastic 0.78
Plastic Mass Factor KM,plastic 0.33 Table 6.3
Plastic Load Factor KL,plastic 0.5 Table 6.3