Introduction
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Fig 1: Schematic representation of beam on elastic foundation |
For instance, the beam shown in Fig. 1 will deflect due to the externally applied load, and produce continuously distributed reaction forces in the supporting medium. The intensity of these reaction forces at any point is proportional to the deflection of the beam y(x) at that point via the constant ks:
Where ks is the soil’s modulus of subgrade reaction which is the pressure per unit settlement of the soil (unit in kN/m2/m).
If we assume that the beam under consideration has a constant cross section with constant width b which is supported by the foundation. A unit deflection of this beam will cause reaction equal to ks.b in the foundation, therefore the intensity of distributed reaction (per unit length of the beam) will be:
R(x) = b.ks.y(x) = k.y(x) (2)
where k = k0.b is the constant of the foundation, known as Winkler’s constant, which includes the effect of the width of the beam, and has unit of kN/m/m.
The general 4th order differential equation for beam on elastic foundation is given by equation (3);
The homogenous equation is given by;
EI(d4y)/dx4 + k.y = 0
(d4y)/dx4 + 4β4y = 0
Where β = ∜(k/4EI) = (k/EI)(1/4)
The general solution for the equation is available, which is given by;
y = eβx (C1sinβx + C2cosβx) + e-βx(C3sinβx + C4cosβx)
Warren and Richard (2002) published tables containing equations for analysing beams on elastic foundation subjected to different loads. The method described in the book has been employed to analyse a beam on elastic foundation and compare the results with Staad Pro software.
Solved Example
A 600mm x 400mm rectangular beam is resting on a homogenous soil of modulus of subgrade reaction ks = 10000 kN/m2/m. The beam is 10m long, and carrying a concentrated load of W = 300 kN from the left hand side. Neglecting the self weight of the beam, and assuming freely supported ends, obtain the bending moment at the point of the concentrated load. Take modulus of elasticity of concrete Ec = 21.7 x 106 kN/m2.
Solution
Second moment of area of concrete beam IB = (bh3)/12 = (0.4 × 0.63)/12 = 7.2 x 10-3 m4
Flexural rigidity of the beam EcIB = 21.7 × 106 × 7.2 × 10-3 = 156240 kN.m2
β = (bks/4EI)(1/4) = [(0.4 × 10000)/(4 × 156240)](1/4) = 0.2828
βl = 0.2828 × 10 = 2.828 m; β(l – a) = 0.282(10 – 3) = 1.979
Where a is the distance of the concentrated load from the left end of the beam.
Since βl < 6.0, we can use Table 8.5 of Roark’s Table for Stress and Strain (Warren and Richard, 2002).
For a beam with both ends free;
RA = 0; MA = 0
The equations for bending moment and shear force along the beam is as given below;
Mx = MAF1 + RA/2βF2 – yA2EIβ2F3 – θAEIβF4 – W/2βFa2
Vx = RAF1 – yA2EIβ3F2 – θAEIβ2F3 – MAβF4 – WFa1
Where;
θA (rotation at point A) = [W/(2EIβ2)] × [(C2Ca2 – 2C3Ca1)/C11]
yA (vertical deflection at point A) = [W/(2EIβ3)] × [(C4Ca1 – C3Ca2)/C11]
We can therefore compute the constants as follows;
C2 = coshβl.sinβl + sinhβl.cosβl = [cosh(2.828) × sin(2.828)] + [sinh(2.828) × cos(2.828)] = (8.485 × 0.3084) + (8.426 × –0.9512 ) = –5.398
C3 = sinhβl.sinβl = (8.426 × 0.3084) = 2.5985
C4 = coshβl.sinβl – sinhβl.cosβl = [cosh(2.828) × sin(2.828)] + [sinh(2.828) × cos(2.828)] = (8.485 × 0.3084) – (8.426 × –0.9512) = 10.631
Ca1 = coshβ(l – a).cosβ(l – a) = cosh(1.974) × cos(1.974) = 3.669 × -0.392 = -1.438
Ca2 = coshβ(l – a).sinβ(l – a) + sinhβ(l – a).cosβ(l – a) = [cosh(1.974) × sin(1.974)] + [sinh(1.974) × cos(1.974)] = (3.669 × 0.9198) + (3.530 × -0.392) = 1.9909
Ca3 = sinhβ(l – a).sinβ(l – a) = (3.530 × 0.9198) = 3.247
C11 = sinh2βl – sin2βl = 8.4262 – 0.30842 = 70.902
θA = W/(2EIβ2) × [(C2Ca2 – 2C3Ca1)/C11]
θA = [300/(2 × 156240 × 0.28282 ) × [(–5.398 × 1.9909) – (2 × 2.5985 × –1.438)/70.902] = (0.012 × –0.0462) = -0.0005544 radians
yA (vertical deformation at point A) = [W/(2EIβ3)] × [(C4Ca1 – C3Ca2)/C11]
y = [300/(2 × 156240 × 0.28283)] × [(10.631 × –1.438) –2.5985 × 1.9909)/70.902] = (0.0424 × –0.2885) = –0.01223 m = -12.23 mm
Bending moment at point C
Substituting the values of deflection and slope into the equation for bending moment (note that the first and second terms of the equation goes to zero since RA = MA = 0);
Mx = – yAEIβ2F3 – θAEIβF4 – W/2βFa2
Mx = – (–0.01223 × 2 × 156240 × 0.28282)F3 – (– 0.0005544 × 156240 × 0.2828)F4 – 300/(2 × 0.2828)Fa2
Mx = 305.638F3 + 24.495F4 – 530.410Fa2
Substituting F3, F4, and Fa2 (see Table 8.5, Warren and Richard, 2002) into the equation;
Mx = 305.638(sinhβx.sinβx) + 24.495(coshβx.sinβx – sinhβx.cosβx) – 530.410[coshβ(x –a).sin β(x –a) + sinh β(x –a).cos β(x –a)]
The bending moment under the concentrated load (point C);
x = 3m; (x – a) = 3 – 3 = 0
βx = 0.8484
Mx = 305.638[sinh(0.8484) × sin(0.8484)] + 24.495[cosh(0.8484) × sin(0.8484) – sinh(0.8484) × cos(0.8484)]
Mc = 305.638 (0.953 × 0.750) + 24.495(1.382 × 0.750 – 0.9539 × 0.6611) = 218.45 + 9.942 = 228.392 kN.m
Verification
This manual calculation has been verified using Staad Pro software. The steps adopted were as follows:
(1) Modelling
The 10m beam was modelled as a one dimensional line element connected by nodes at 1m length interval. This was to represent/attach soil springs at 1m interval.
Hi
May you kindly elaborate on the results received for soil deformation, especial from 8.5m to 10m?
Do the positive values not imply tension forces within the supporting soil medium?
Thanks and Regards,
ML
When a beam is resting on an elastic medium, it is assumed that it is not anchored to the soil. The physical interpretation is that there is a slight uplift of the beam at those points, thereby preventing contact with the soil. As this is undesirable, the designer can adopt some measures to ensure ‘no tension’ in the soil.
In such case, are there any list of remedial measures in order to cut-off the tesnion
Hi
I think a in Roark’s formula is the distance from left to the intermediate of the beam. So, a=10/2=5 and x-a=3-5