Dynamic response of an RC beam to a drop weight impact (steel ball)

Ball mass - Ms = 2.1 t

Ball material - steel

Modulus of elasticity - Es = 206 GPa

Poisson′s ratio - νs = 0.3

Mass density - ρs = 7.85 tm3

Ball volume - Vs = Msρs = 2.1 t7.85 t/m3 = 0.268 m3 = 4/3πR³

Ball radius - Rs =  3  34·Vsπ = 400 mm

Height of bottom above the beam surface - H = 2 m

Structure type - simply supported beam

Beam length - L = 12 m

Material - reinforced concrete C20/25

Modulus of elasticity - E = 20 GPa

Poisson′s ratio - ν = 0.2

Shear modulus - G = E2· (1 + ν)  = 20 GPa2· (1 + 0.2)  = 8.33 GPa

Unit weight - γb = 25 kNm3

Cross section - rectangular with dimensions:

Width - b = 350 mm

Height - h = 650 mm

Area - A = b·h = 350 mm·650 mm = 2275 cm2

Second moment of area - I = b·h312 = 350 mm· (650 mm) 312 = 8009895833 mm4

Shear area - AQ = 56·A = 56·2275 cm2 = 1895.83 cm2

Self-weight - gb = A·γb = 2275 cm2·25 kN/m3 = 5.69 kN/m

Uniform load - q = 10 kNm

Gravity acceleration - g = 9.81 ms2

Uniform mass - m = gb + qg = 5.69 kN/m + 10 kN/m9.81 m/s2 = 1.6 t/m

L=12mH=2mM=2.1tg

Simple analytical solution

The structure is reduced to a SDOF system for simplicity

Dynamically equivalent mass - Md = 2·Lπ·m = 2·12 m3.14·1.6 t/m = 12.22 t

Potential energy of the ball before dropping

Ep = Ms·g·H = 2.1 t·9.81 m/s2·2 m = 41.28 kJ

Kinetic energy immediately before the impact - Ek = Ms·v022

The velocity at the moment before the impact is determined by the energy conservation law Ek = Ep :

v0 =   2·EpMs =   2·41.28 kJ2.1 t = 6.26 m/s

Perfectly inelastic collision model is assumed.

Total mass after contact - Mtot = Ms + Md = 2.1 t + 12.22 t = 14.33 t

The velocity immediately after the contact is determined by the law of conservation of momentum:

v1 = v0·MsMtot = 6.26 m/s·2.1 t14.33 t = 0.92 m/s

Structural stiffness for a vertical force applied at the middle point of the span

K = 48·E·IL3 = 48·20 GPa·8009895833 mm4 (12 m) 3 = 4449.94 kN/m

Deflection due to uniform load

z0 = 5· (gb + q) ·L4384·E·I = 5· (5.69 kN/m + 10 kN/m) · (12 m) 4384·20 GPa·8009895833 mm4 = 26.44 mm

Static displacement - zst = Mtot·gK = 14.33 t·9.81 m/s24449.94 kN/m = 31.57 mm

Natural circular frequency - ω1 =   KMtot =   4449.94 kN/m14.33 t = 17.62 s-1

Vibration period - T1 = 2·πω1 = 2·3.1417.62 s-1 = 0.356 s

Dynamic factor

μ = 1 +   1 + (v1·ω1g)2 = 1 +   1 + (0.92 m/s·17.62 s-19.81 m/s2)2 = 2.93

Dynamic displacement - zd = μ·zst = 2.93·31.57 mm = 92.58 mm

Dynamic force - Fd = μ·Ms·g = 2.93·2.1 t·9.81 m/s2 = 60.52 kN

(without self-weight and uniform load)

Simplified equation for the dynamic factor

μ1 = 1 + v1·ω1g = 1 + 0.92 m/s·17.62 s-19.81 m/s2 = 2.65

The difference will be smaller for greater heights.

Elastic time history response of the structure as an SDOF system

Damped vibration is assumed with factor - ξ = 0.05

Vibration amplitude - A = zd − zst = 92.58 mm − 31.57 mm = 61.01 mm or

A = v1ω1 = 0.92 m/s17.62 s-1 = 52.21 mm

Theoretical equation of motion

y (t)  = A·e-ξ·ω1·t·sin (ω1·t) 

Solution by direct integration of the impulse load

Duration of impulse transmission for a beam with infinite mass [1]

τL = 2.94·  5  (1516·Ms·(1 − ν2E + 1 − νs2Es))2Rs·v0 = 2.94·  5  (1516·2.1 t·(1 − 0.2220 GPa + 1 − 0.32206 GPa))2400 mm·6.26 m/s = 3.93 ms

Duration of impulse transmission for a beam with finite mass [2]

τL = 2.94·  5  (1516·Ms·MdMs + Md·(1 − ν2E + 1 − νs2Es))2Rs·v0 = 2.94·  5  (1516·2.1 t·12.22 t2.1 t + 12.22 t·(1 − 0.2220 GPa + 1 − 0.32206 GPa))2400 mm·6.26 m/s = 3.69 ms

The above values correspond well to the experimental data in [3], where the recorded durations are of a similar magnitude.

The impulse force function will be determined by using the recommended expressions (9.20) - (9.22) in [1]

The coefficient of restitution for perfectly inelastic collision is - e. = 0

F (t)  = Ms·v0· (1 + e.) ·π2·τL·sin(πτL·t)· (| t | ≤ τL) 

Impulse load diagram

0 2000 4000 0 5 10 15 20 25 30 35 40 45 50 x y [0; 0] [50; 5613.4]

Maximal impulse load value - Fmax = F(τL2) = F(3.69 ms2) = 5613.66 kN

The equation of motion is expressed by the Duhamel′s integral

yD (t)  = 1Mtot·ω1· min (t; τL) ∫0 ms F (τ) ·e-ξ·ω1· (t − τ) ·sin (ω1· (t − τ) )  dτ

Static displacement for the midpoint of the beam

y0 (t)  = z0 +  (zst − z0) ·{if t < T14: sin(2·π·tT1)
else: 1

Time history of the midpoint displacement, [mm]

-80 -70 -60 -50 -40 -30 -20 -10 0 10 0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 x y [0; -79.88] [5000; 9.72]

Elastic time history response of the structure as an MDOF system

Number of intermediate joints - nJ = 11 (odd)

Length of one segment - Δx = LnJ + 1 = 12 m11 + 1 = 1 m

Coordinate of joint j - xJ (j)  = Δx·j

Shear forces due to unit vertical load Fj = 1 at joint j

V1 (x; j)  = {if x < xJ (j) : 1 − xJ (j) L
else: -xJ (j) L

-0.25 0 0.25 0.5 0.75 0 1 2 3 4 5 6 7 8 9 10 11 12 x y [0; -0.25] [12; 0.917]

Bending moments due to unit vertical load Fj = 1 at joint j

M1,max (j)  = (xJ (j) L − 1)·xJ (j) 

M1 (x; j)  = M1,max (j) ·{if x < xJ (j) : xxJ (j) 
else: L − xL − xJ (j) 

0 0.75 1.5 2.25 0 1 2 3 4 5 6 7 8 9 10 11 12 x y [0; 0] [12; 2.25]

Flexibility matrix

D (i; j)  = ( L∫0 m M1 (x; i) ·M1 (x; j)  dx)·1E·I + ( L∫0 m V1 (x; i) ·V1 (x; j)  dx)·1G·AQ

D = 0.02160.03780.04890.05520.05740.0560.05140.04430.0350.0242⋯0.0124 0.03780.07040.0930.1060.1110.1090.10.08640.06850.0474⋯0.0242 0.04890.0930.1280.1490.1580.1550.1440.1240.09880.0685⋯0.035 0.05520.1060.1490.1790.1930.1930.180.1560.1240.0864⋯0.0443 0.05740.1110.1580.1930.2140.2170.2050.180.1440.1⋯0.0514 0.0560.1090.1550.1930.2170.2270.2170.1930.1550.109⋯0.056 0.05140.10.1440.180.2050.2170.2140.1930.1580.111⋯0.0574 0.04430.08640.1240.1560.180.1930.1930.1790.1490.106⋯0.0552 0.0350.06850.09880.1240.1440.1550.1580.1490.1280.093⋯0.0489 0.02420.04740.06850.08640.10.1090.1110.1060.0930.0704⋯0.0378 ⋮⋮⋮⋮⋮⋮⋮⋮⋮⋮⋱⋮0.01240.02420.0350.04430.05140.0560.05740.05520.04890.0378⋯0.0216 mm/kN

Mass matrix

dM.j = m·Δxt = 1.6 t/m·1 mt = 1.6

M = 1.6000000000⋯0 01.600000000⋯0 001.60000000⋯0 0001.6000000⋯0 00001.600000⋯0 000003.70000⋯0 0000001.6000⋯0 00000001.600⋯0 000000001.60⋯0 0000000001.6⋯0 ⋮⋮⋮⋮⋮⋮⋮⋮⋮⋮⋱⋮0000000000⋯1.6

Total mass of the structure - sum (⃗dM)  = 19.7 t

Eigenvalues

Msq =   √ M = 1.26000000000⋯0 01.2600000000⋯0 001.260000000⋯0 0001.26000000⋯0 00001.2600000⋯0 000001.920000⋯0 0000001.26000⋯0 00000001.2600⋯0 000000001.260⋯0 0000000001.26⋯0 ⋮⋮⋮⋮⋮⋮⋮⋮⋮⋮⋱⋮0000000000⋯1.26

C = copy (Msq·D·Msq; symmetric (nJ) ; 1; 1)  = copy (Msq·D·Msq; symmetric (11) ; 1; 1)  = 0.03450.06050.07810.08830.09180.1360.08220.07080.0560.0387⋯0.0198 0.06050.1130.1490.170.1780.2650.160.1380.110.0758⋯0.0387 0.07810.1490.2040.2380.2520.3780.230.1990.1580.11⋯0.056 0.08830.170.2380.2870.3090.4690.2870.250.1990.138⋯0.0708 0.09180.1780.2520.3090.3430.5290.3280.2870.230.16⋯0.0822 0.1360.2650.3780.4690.5290.8390.5290.4690.3780.265⋯0.136 0.08220.160.230.2870.3280.5290.3430.3090.2520.178⋯0.0918 0.07080.1380.1990.250.2870.4690.3090.2870.2380.17⋯0.0883 0.0560.110.1580.1990.230.3780.2520.2380.2040.149⋯0.0781 0.03870.07580.110.1380.160.2650.1780.170.1490.113⋯0.0605 ⋮⋮⋮⋮⋮⋮⋮⋮⋮⋮⋱⋮0.01980.03870.0560.07080.08220.1360.09180.08830.07810.0605⋯0.0345

⃗λ = reverse (last (eigenvals (C·10-3) ; 7) )  = [0.00261 0.000137 3.32×10-5 9.33×10-6 4.78×10-6 2.17×10-6 1.51×10-6]

Natural circular frequences - ⃗ω =   1⃗λ = [19.57 85.55 173.53 327.32 457.58 678.76 814.79] s⁻¹

Natural vibration frequences - ⃗f = ⃗ω2·π·Hz = ⃗ω2·3.14·Hz = [3.11 Hz 13.61 Hz 27.62 Hz 52.09 Hz 72.83 Hz 108.03 Hz 129.68 Hz]

Natural vibration periods - ⃗T = 1⃗f = [0.321 s 0.0734 s 0.0362 s 0.0192 s 0.0137 s 0.00926 s 0.00771 s]

Eigenvectors

Φ = inverse (Msq) ·extractcols (eigenvecs (C·10-3) ; range (nJ; nJ − 7 + 1; -1) )  = inverse (Msq) ·extractcols (eigenvecs (C·10-3) ; range (11; 11 − 7 + 1; -1) )  = 0.07520.1610.2170.280.3010.3230.309 0.1450.280.3190.280.186-9.74×10-13-0.113 0.2060.3230.2522.84×10-13-0.187-0.323-0.266 0.2530.280.0565-0.28-0.3091.15×10-130.216 0.2830.161-0.155-0.28-0.0270.3230.209 0.2930-0.251-7.46×10-150.225.01×10-14-0.193 0.283-0.161-0.1550.28-0.027-0.3230.209 0.253-0.280.05650.28-0.309-2.16×10-130.216 0.206-0.3230.252-2.77×10-13-0.1870.323-0.266 0.145-0.280.319-0.280.1861.09×10-12-0.113 ⋮⋮⋮⋮⋮⋮⋮0.0752-0.1610.217-0.280.301-0.3230.309

X = stack (matrix (1; 3) ; Φ; matrix (1; 3) )  = 0000000 0.07520.1610.2170.280.3010.3230.309 0.1450.280.3190.280.186-9.74×10-13-0.113 0.2060.3230.2522.84×10-13-0.187-0.323-0.266 0.2530.280.0565-0.28-0.3091.15×10-130.216 0.2830.161-0.155-0.28-0.0270.3230.209 0.2930-0.251-7.46×10-150.225.01×10-14-0.193 0.283-0.161-0.1550.28-0.027-0.3230.209 0.253-0.280.05650.28-0.309-2.16×10-130.216 0.206-0.3230.252-2.77×10-13-0.1870.323-0.266 ⋮⋮⋮⋮⋮⋮⋮0000000

-0.2 0 0.2 0 1 2 3 4 5 6 7 8 9 10 11 12 x y [0; -0.323] [12; 0.323]

Modal masses - ⃗mΦ = diag2vec (transp (Φ) ·M·Φ) ·t = [1 t 1 t 1 t 1 t 1 t 1 t 1 t]

Rayleigh damping model is assumed

β = 2·ξ⃗ω1 + ⃗ω2 = 2·0.0519.57 + 85.55 = 0.000951 , α = β·⃗ω1·⃗ω2 = 0.000951·19.57·85.55 = 1.59

ξ (ω)  = α2·ω + β·ω2

Modal damping factors - ⃗ξΦ = ξ (⃗ω)  = [0.05 0.05 0.0871 0.158 0.219 0.324 0.389]

0 0.1 0.2 0.3 0 100 200 300 400 500 600 700 800 x y [0; 0] [814.79; 0.389]

Damped natural frequences

⃗ωD = ⃗ω·  √ 1 − ⃗ξΦ^2·s-1 = [19.54 s-1 85.44 s-1 172.87 s-1 323.2 s-1 446.43 s-1 642.14 s-1 750.77 s-1]

Dynamic load vector

FΦ (i; t)  = Φjm, i·F (t) 

The equations of modal dynamic displacements are expressed by the Duhamel′s integral

yΦ (i; t)  = 1⃗mΦ.i·ωD.i· min (t; τL) ∫0 ms FΦ (i; τ) ·e-⃗ξΦ.i·⃗ωi·s-1· (t − τ) ·sin (ωD.i· (t − τ) )  dτ

Joint displacements

yJ (j; t)  = 7∑i= 1Φj, i·yΦ (i; t) 

Comparison of time history records of the midpoint displacements for SDOF and MDOF systems, [mm]

-80 -70 -60 -50 -40 -30 -20 -10 0 10 0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 x y [0; -85.91] [5000; 14.2]

Time history records for the amplitudes of separate joints, [mm]

-50 -40 -30 -20 -10 0 10 20 30 40 0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 x y [0; -52.21] [5000; 44.12]

Beam deflections for the first five time steps at Δt = 17.7 ms

-30 -20 -10 0 0 1 2 3 4 5 6 7 8 9 10 11 12 x y [0; -37.32] [12; 0]

Animation of beam response (slowed down)

-100 -75 -50 -25 0 0 1 2 3 4 5 6 7 8 9 10 11 12 x y [0; -100] [12; 0]

[1] Harris C. M., Piersol A.G., HARRIS’ SHOCK AND VIBRATION HANDBOOK, Fifth Edition, McGraw-Hill 2002, ISBN 0-07-137081-1

[2] Qing Peng, Xiaoming Liu, Yueguang Wei, Elastic impact of sphere on large plate, Journal of the Mechanics and Physics of Solids, Volume 156, 2021, 104604, ISSN 0022 - 5096, https://doi.org/10.1016/j.jmps.2021.104604

[3] Hong Hao and Thong M. Pham, Performance of RC Beams with or without FRP Strengthening Subjected to Impact Loading, Proceedings of the 2nd World Congress on Civil, Structural, and Environmental Engineering (CSEE’17) Barcelona, Spain – April 3– 4, 2017 ISSN:2371 - 5294 DOI:10.11159/icsenm17.1

[4] Gugan, D. “Inelastic collision and the Hertz theory of impact.” American Journal of Physics 68 (2000): 920-924., http://www.oxfordcroquet.com/tech/gugan/index.asp