Roadmap to Membership of RICS - Pathways and Routes
Bef 2
1. BEAMS ON ELASTIC
FOUNDATION
Under the guidance
Of
Dr.M.V.RENUKA DEVI
Associate Professor
Department of Civil Engineering, RVCE
By
M.PRASANNA KUMAR
(1RV13CSE05)
2. INTRODUCTION
• Beneath the foundation soil mass is considered as
identical, independent, closely spaced, linearly elastic
springs which is known asWinkler foundation.
• Bending of beams on an elastic foundation is
developed on the assumption that the reaction forces of
the foundation are proportional at every point to the
deflection of the beam at that point .
• One of the most important deficiencies of the Winkler
model is that a displacement discontinuity appears
between the loaded and the unloaded part of the
foundation surface. In reality, the soil surface does not
show any discontinuity.
3. Classification of beams:
The beams on elastic foundation can be
classified into three types
• Short beams for which β퐿 ≤ 0.6
• Medium beams or semi-infinite beams for which
0.6< βL < 5
• Long beams or infinite beams β퐿 ≥ 5
L= length of the beam
4. Infinite beam subjected to point load
• As we know , EI
d2y
d푧2 = − M
d3y
d푧3 = − V
EI
EI
d4y
d푧4 = − q
5. • Where the distributed reaction force q is
positive when acting upward
• For linearly elastic foundation, the distributed
force q is linearly proportional to the deflection
y . Thus,
q = ky
k = bk0
Where k is the elastic coefficient, k0 is the elastic
foundation modulus, and b is the width of the
foundation.
6. EIx
d4y
dz4 = − ky
d4y
dz4 = −
k
EIx
y
• To solve this homogeneous, fourth order, linear
differential equation.
we will assume that
k
EIx
= 4β4,then
d4y
dz4 + 4β4y = 0
• By using method of differential equation the solution of
above equation is
y = eβz c1sinβz + c2cosβz + e−βz c3sinβz +
7. Since the deflection 푦 = 0 , z ∞ then the term eβz ∞
and e−βz 0
we obtain c1, = c2 = 0 , y = e−βz c3sinβz + c4cosβz
• By applying boundary conditions
1. dy
dz
= 0 at z = 0
dy
dz
= −βe−βz c3sinβz + c4cosβz +
βe−βz c3cosβz − c4sinβz = 0
we get
c3 = c4 = c ; y = ce−βz sinβz + cosβz
8. ∞
kydz = p
2. 2 0
∞
kce−βz sinβz + cosβz dz = p
2 0
∞
kce−βz sinβz) + 0
0
∞
e−βz cosβz dz =
P
2KC
1
2β
+
1
2β
=
P
2KC
C =
Pβ
2K
Then equation for deflection is
y =
Pβ
2K
e−βz sinβz + cosβz
9. θ =
dy
dz
=
Pβ
2K
[−2βe−βz sinβz ]
= −
Pβ2
K
[e−βz sinβz ]
Mx = −EIx
d2y
dz2 = EIx
Pβ2
K
d
dz
[e−βz sinβz ]
=
P
4β
e−βz cosβz − sinβz)
Vy = −
dM
dz
= −
d
dz
[
P
4β
e−βz cosβz − sinβz ]
=
P
2
e−βz cosβz
10. • Defining,
Aβz = e−βz cosβz + sinβz) Bβz = e−βz sinβz
Cβz = e−βz cosβz − sinβz) Dβz = e−βz cosβz)
Then, we have
y =
Pβ
2K
Aβz
θ = −
Pβ2
K
Bβz
Mx =
P
4β
Cβz
Vy =
P
2
Dβz
11. • A rail road uses steel rails (E = 200.0 GPa ) with a depth of
184 mm . The distance from the top of the rail to its centroid
is 99.1 mm , and the moment of inertia of the rail is 36 ×
106mm4. The rail is supported by ties, ballast, and a road bed
that together are assumed to act as an elastic foundation with
modulus of subgrade reaction k = 14.0N/mm2.length is 6m.
(a) Determine the maximum deflection, maximum bending
moment, and maximum flexural stress in the rail for a single
wheel load of 170 KN as shown in Fig
(b) If a locomotive has 3 wheels per truck equally spaced at 1.70
m, determine the maximum deflection, maximum bending
moment and maximum flexural stress in rail when the load
on each wheel is 170 kN.
12. Solution: β =
4 14
4×200×103×36.0×106 = 0.000835mm−1
β × 푙 = 0.000835 × 6 = 5
(a) The maximum deflection and the maximum
bending moment occur under the load Where z =0
therefore Aβz = Cβz = 1.0
13. ymax =
Pβ
2K
Aβz =
170 × 103 × 0.000835
2 × 14
= 5.069 푚푚
Mmax =
P
4β
Cβz =
170 × 103
4 × 0.000835
= 50.89 × 106 N − m
σmax =
Mmax × C
Ix
=
50.89 × 106 × 99.1
36.9 × 106 = 136.6 Mpa
(b) Case (i):
14. • For case 1, let the origin of the coordinate be located
under one of the end wheel. The distance to the first
wheel 푧1 = 0 , we have
Aβz1 = Cβz1 = 1.0
The distance from the origin to the next wheel is
z2 = 1700mm
Aβz2 = 0.2797 Cβz2 = −0.2018
The distance from the origin to the next wheel is
z3 = 3400mm
Aβz3 = −0.0377 Cβz3 = −0.0752
• we get the maximum deflection and the maximum
bending moment equal to
18. Each spring has same spring constant K . The
reaction force R exerted on the beam is directly
proportional to the deflection y
R = ky
the load R can be idealized as uniformly distributed
over a total span L
Ky = kyl
k = K
l
k is the elastic coefficient
However it has to satisfy the condition
Check for spacing of the spring l ≤
π
4β
19. To obtain a reasonable approximate solution
Check for length of the beam is
L′′≥
3π
2β
20. • Problem2: An aluminium alloy I-beam (depth =
100푚푚, Ix = 2.45 × 106mm4 , E = 72.0GPa) as shown
in figure has a length of 7m and is supported by 7
springs k = 110N mm. Spaced at a distance of 1.1m
centre to centre along the beam. A load P = 12.0KN is
applied at the centre of the beam. Determine the
maximum deflection of the beam, the maximum bending
moment, and the maximum bending stress in the beam.
21. Solution: The elastic coefficient
k =
110
1100
= 0.1N mm2
the value of β
β =
4 0.1
4 × 72 × 103 × 2.45 × 106 = 0.000614 mm−1
Check the spacing of the spring.
l <
π
4β
=
π
4×0.000614
= 1279 mm
Check the length of the beam
• L′′ = 7000 + 1100 = 8100 >
3π
2β
=
3π
2 0.000614)
= 7675
22. The maximum deflection and the maximum
bending moment of the beam occur under the load
where, Aβz= Cβz = 1.0
ymax =
Pβ
2K
Aβz =
12 × 103 × 0.000614
2 × 0.1
= 36.84 mm
Mmax =
P
4β
Cβz =
12 × 103
4 × 0.0006140
= 4.886 × 106 N − m
σmax =
Mmax × C
Ix
= 99.7 MPa
23. Infinite Beam Subjected to a Distributed
Load Segment
From the displacement solution of the beam
subjected to concentrated load
y =
Pβ
2K
e−βz sinβz + cosβz
24. dyh =
wβ
2K
e−βz sinβz + cosβz dz
By using the principle of superposition, the total
deflection due to the distributed load is
a wβ
yh = 0
2K
e−βz sinβz + cosβz dz +
b wβ
0
2K
e−βz sinβz + cosβz dz
yh =
wβ
2K
[
1
β
1 − e−aβcosβa +
1
β
1 − e−bβcosβb ]
=
w
2K
[2 − e−aβcosβa − e−bβcosβb]
θH =
a dyh
0
dz
dz +
b dyh
0
dz
dz =
wβ
2K
[Aβa − Aβb]
MH =
w
4β2 [Bβa + Bβb] VH =
w
4β
[Cβa − Cβb]
26. • Problem 3: A long wood beam (E = 10.0 GPa) has a
rectangular cross section with a depth of 200 mm and a
width of 100 mm . It rests on an earth foundation having
spring constant of k0 = 0.040N mm3and is subjected
to a uniformly distributed load 35 N/mm extending over
a lengthL′ = 3.61m. Taking the origin of the coordinate
at the centre of the segment L′ = 3.61m. determine the
maximum deflection, the maximum bending stress in the
beam, and the maximum pressure between the beam and
the foundation. The moment of inertia of the beam about
x -axis isIx = 66.67 × 106mm4.
27. Solution: The elastic coefficient k = bk0 = 100 × 0.040 =
4N mm2
value of β =
4 4
4×104×66.67×106 = 0.001107mm−1
The maximum deflection occurs at the centre of segment
L′ since a = b = L′/2
βa = βb = β
L′
2
= 2.0, Dβa = Dβb = −0.0563
yh =
w
2K
2 − Dβa − Dβb =
35
4
2 − 2 −0.0563
= 9.243mm
28. • The maximum pressure between the beam and the
foundation occurs at the point of the maximum
deflection
qmax = k0ymax = 0.040 × 9.243 = 0.370MPa
Maximum bending occurs at where shear force is zero
VH =
w
4β
Cβa − Cβb = 0
Cβa = Cβb
e−aβ cosβa − sinβa = e−bβ cosβb − sinβb)
β
L′
2
= 2.0 (L′= a+b) βb = 4 − βa
βa = 0.858 , −0.777
βb = 3.142 , 4.777
29. Mmax =
w
4β2 [Bβa + Bβb]
=
35
4 0.001072)
[0.3233 + 0.0086]
= 2.363 KN-m
Bending stress is σ =
M
I
× Y = 3.544 MPa
30. Semi-infinite beam Subjected to Loads at
Its End
Consider the semi-infinite beam subjected to a point load
P and a positive bending moment M0 at its end
y = e−βz c3sinβz + c4cosβz
32. c3 + c4 =
p
2β3EIx
=
2βP
k
c4 =
2βP
k
−
2β2M0
k
the deflection of the beam is
y =
2βe−βz
k
[Pcosβz − βM0 cosβz − sinβz)]
By rearranging, y =
2βP
k
Dβz −
2β2M0
k
Cβz
θ = −
2Pβ2
K
Aβz +
4β3M0
k
Dβz
Mx = −
P
β
Bβz + M0 Aβz
Vy = −PCβz − 2M0β Bβz
33. A steel I-beam (E = 200GPa ) has a depth of 102mm , a
width of 68mm , a moment of inertia of Ix = 2.53 ×
106mm4 and a length of 4m. It is attached to a rubber
foundation for which k0 = 0.350 N mm3. A concentrated
load p = 30.0 KN is applied at one end of the beam.
Determine the maximum deflection, the maximum bending
stress in the beam, and their locations.
Solution: The spring coefficient, k = bk0
= 68 × 0.350 = 23.58N mm2
value of β =
4 23.8
4×200×103×2.53×106 = 0.001852mm−1
The maximum deflection occurs at the end where load P is
applied (z =0 ), sinceDβz is maximum we have βz = 0 and
Dβz = 1
34. Mx = −
P
β
Bβz = −
30 × 103
0.001852
× 0.3224 = −5.22 KN − m
σ =
M
I
× Y =
5.22 × 106 × 4.67
2.53 × 106 = 9.63MPa
FINITE BEAMS:
• Finite beams are defined as beams for which 훽푙 ≤ 0.6
• Finite beams can be analysed using the analysis results
of infinite beams
• Finite beam can be split into loading case namely
symmetric and anti-symmetric load acting on infinite
beam.
35. PRACTICAL SIGNIFICANCE
• Historically the first application of this theory was to
rail road track
• Another application of this soon after the rail road
track is grid works of beams
• Mat foundation under certain structures, such as silos
water storage tanks, coal storage tanks, etc., and
footing foundation supporting group of columns, are
frequently designed and constructed in the form of
beams resting on soil.
• No doubt it satisfies the actual conditions of real
elastic theory of soil.
36. REFERENCES
• Arthur P. Boresi, Richard J.Schindt, “Advanced Mechanics of materials”,
Sixth edition John Wiley &Sons. Inc., New Delhi, 2005
• Thimoshenko & J N Goodier, “Mechanics Of Solids”, Tata McGraw-Hill
publishing Co.Ltd, New Delhi, 1997
• Seely Fred B. and Smith James O., “ Advanced Mechanics of Materials”, 2nd
edition, John Wiley & Sons Inc, New York, 1952, pp.112-136
• Srinath L.S., “Advanced Mechanics of Solids”, Tata McGraw-Hill
publishing Co.Ltd, New Delhi, 1980, pp180-191
• Thimoshenko S., “Strength of Materials”, Part-1, Elementary Theory and
Problems, 3rd Edition, D. Van Nostrand company Inc., New York, 1955,
pp.227-244
• Boresi A.P and Chong K.P(2000), “Elasticity In Engineering Mechanics”
2nd edition New York ; Wiley – Interscience.
• N Krishna Raju & D R Gururaja, “Advance Mechanics Of Solids &
Structures”, 1997
• B C Punmia & A K Jain. “Strength of Materials and Theory of Structures”,
Vol.2 Lakshmi publications (P) Ltd.
