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Here is the complete solution for Problem 9.1 to Problem 9.25 in text format:
Problem 9.1
Problem Statement: A very long steel I-beam, 0.127 m deep, resting on
a foundation for which k = 1.4 MPa, is subjected to a concentrated load at
midlength. The flange is 0.0762 m wide, and the cross-sectional moment of
inertia is 5.04 × 10−6 m4 . What is the maximum load that can be applied
to the beam without causing the elastic limit to be exceeded? Assume that
E = 200 GPa and σyp = 210 MPa.
Solution:
1. Given Data:
• Depth of the beam, d = 0.127 m
2
• Foundation modulus, k = 1.4 MPa = 1.4 × 106 N/m
• Flange width, b = 0.0762 m
• Moment of inertia, I = 5.04 × 10−6 m4
2
• Modulus of elasticity, E = 200 GPa = 200 × 109 N/m
2
• Yield strength, σyp = 210 MPa = 210 × 106 N/m
2. Calculate the parameter β:
β=
k
4EI
1/4
• Substituting the values:
β=
β=
1.4 × 106
4 × 200 × 109 × 5.04 × 10−6
1.4 × 106
4.032 × 106
1/4
1/4
= (0.3472)1/4 = 0.769 m−1
3. Maximum bending moment Mmax : For an infinite beam with a concentrated load P at the center, the maximum bending moment is:
Mmax =
P
4β
• The maximum stress σmax is related to the bending moment by:
1
Mmax · c
I
• where c = d/2 = 0.127/2 = 0.0635 m is the distance from the neutral axis
to the outer fiber.
σmax =
Rearranging for P :
σmax =
P =
P ·c
4βI
4βIσmax
c
• Substituting the values:
P =
4 × 0.769 × 5.04 × 10−6 × 210 × 106
0.0635
3.24 × 103
= 51.02 kN
0.0635
4. Conclusion: The maximum load that can be applied without exceeding
the elastic limit is approximately 51.02 kN.
P =
Problem 9.2
Problem Statement: A long steel beam (E = 200 GPa) of depth 2.5b and
width b is to rest on an elastic foundation (k = 20 MPa) and support a 40-kN
load at its center (Fig. 9.2). Design the beam (compute b) if the bending stress
is not to exceed 250 MPa.
Solution:
1. Given Data:
• Depth of the beam, d = 2.5b
• Width of the beam, b
2
• Foundation modulus, k = 20 MPa = 20 × 106 N/m
2
9
• Modulus of elasticity, E = 200 GPa = 200 × 10 N/m
3
• Load, P = 40 kN = 40 × 10 N
2
• Maximum bending stress, σmax = 250 MPa = 250 × 106 N/m
2. Calculate the parameter β:
β=
k
4EI
2
1/4
• The moment of inertia I for a rectangular cross-section is:
b · d3
b · (2.5b)3
15.625b4
=
=
= 1.302b4
12
12
12
• Substituting I into the expression for β:
I=
β=
β=
20 × 106
4 × 200 × 109 × 1.302b4
20 × 106
1.0416 × 1012 b4
β=
1/4
20
1.0416 × 106
=
1/4
1/4
20
1.0416 × 106 b4
1/4
· b−1 = 0.112 · b−1
3. Maximum bending moment Mmax : For an infinite beam with a concentrated load P at the center, the maximum bending moment is:
Mmax =
P
4β
• Substituting β:
40 × 103
40 × 103
=
= 89.29 × 103 · b
−1
4 × 0.112 · b
0.448 · b−1
4. Maximum stress condition: The maximum stress is given by:
Mmax =
Mmax · c
I
• where c = d/2 = 1.25b. Substituting Mmax and I:
σmax =
250 × 106 =
250 × 106 =
b2 =
89.29 × 103 · b · 1.25b
1.302b4
111.61 × 103 · b2
111.61 × 103
=
4
1.302b
1.302b2
111.61 × 103
111.61 × 103
=
= 0.343 × 10−3
6
1.302 × 250 × 10
325.5 × 106
b=
p
0.343 × 10−3 = 0.0185 m = 18.5 mm
3
5. Conclusion: The required width b of the beam is approximately 18.5
mm.
Problem 9.3
Problem Statement: Calculate the maximum deflection, maximum bending
moment, and maximum
stress in the steel (E = 210 GPa) I-beam of moment of
inertia I = 40 106 mm4 subjected to a load P (Fig. 9.2). The distance from
the top of the beam to its centroid is 100 mm. Given: P = 150 kN, k = 15 MPa.
Solution:
1. Given Data:
2
• Modulus of elasticity, E = 210 GPa = 210 × 109 N/m
• Moment of inertia, I = 40 × 106 mm4 = 40 × 10−6 m4
• Load, P = 150 kN = 150 × 103 N
2
• Foundation modulus, k = 15 MPa = 15 × 106 N/m
• Distance from top to centroid, c = 100 mm = 0.1 m
2. Calculate the parameter β:
β=
k
4EI
1/4
• Substituting the values:
β=
β=
15 × 106
4 × 210 × 109 × 40 × 10−6
15 × 106
33.6 × 106
1/4
1/4
= (0.4464)1/4 = 0.818 m−1
3. Maximum deflection vmax : For an infinite beam with a concentrated
load P at the center, the maximum deflection is:
vmax =
Pβ
2k
• Substituting the values:
150 × 103 × 0.818
122.7 × 103
=
= 0.00409 m = 4.09 mm
2 × 15 × 106
30 × 106
4. Maximum bending moment Mmax : The maximum bending moment
is:
vmax =
4
Mmax =
P
4β
• Substituting the values:
150 × 103
150 × 103
=
= 45.84 kN · m
4 × 0.818
3.272
5. Maximum stress σmax : The maximum stress is given by:
Mmax =
σmax =
Mmax · c
I
• Substituting the values:
σmax =
45.84 × 103 × 0.1
4.584 × 103
=
= 114.6 MPa
40 × 10−6
40 × 10−6
6. Conclusion:
• Maximum deflection: 4.09 mm
• Maximum bending moment: 45.84 kN·m
• Maximum stress: 114.6 MPa
Problem 9.4
Problem Statement: A long wooden beam with a cross-section b × h rests
on an elastic foundation and carries a uniform load p over a region L (Fig.
9.3). Given: b = 90 mm, h = 180 mm, p = 40 kN/m, k = 5 MPa, L = 4 m,
E = 12 GPa. Taking the origin of the coordinates at the center of the beam
(a = b = L/2), calculate the maximum deflection.
Solution:
1. Given Data:
• Width of the beam, b = 90 mm = 0.09 m
• Height of the beam, h = 180 mm = 0.18 m
• Uniform load, p = 40 kN/m = 40 × 103 N/m
2
• Foundation modulus, k = 5 MPa = 5 × 106 N/m
• Length of the loaded region, L = 4 m
2
• Modulus of elasticity, E = 12 GPa = 12 × 109 N/m
2. Calculate the moment of inertia I: For a rectangular cross-section:
bh3
0.09 × (0.18)3
0.09 × 0.005832
=
=
= 4.374 × 10−5 m4
12
12
12
3. Calculate the parameter β:
I=
5
β=
k
4EI
1/4
• Substituting the values:
β=
β=
5 × 106
4 × 12 × 109 × 4.374 × 10−5
5 × 106
2.1 × 106
1/4
1/4
= (2.381)1/4 = 1.24 m−1
4. Maximum deflection vmax : For a uniformly distributed load over a finite
length L on an infinite beam, the maximum deflection occurs at the center
of the loaded region. The deflection at any point x is given by:
p 2 − e−βa cos(βa) − e−βb cos(βb)
2k
• Since a = b = L/2 = 2 m, we have:
v(x) =
vmax =
p 2 − e−β·2 cos(β · 2) − e−β·2 cos(β · 2)
2k
p 2 − 2e−2.48 cos(2.48)
2k
• Using e−2.48 ≈ 0.083 and cos(2.48) ≈ −0.79:
vmax =
vmax =
vmax =
40 × 103
[2 − 2 × 0.083 × (−0.79)]
2 × 5 × 106
40 × 103
[2 + 0.131] = 4 × 10−3 × 2.131 = 0.00852 m = 8.52 mm
10 × 106
5. Conclusion: The maximum deflection of the beam is approximately 8.52
mm.
Problem 9.5
Problem Statement: A long beam on an elastic foundation is subjected to
a sinusoidal loading p = p1 sin(2πx/L), where p1 and L are the peak intensity
and wavelength of loading, respectively. Determine the equation of the elastic
deflection curve in terms of k and β.
Solution:
6
1. Given Data:
• Sinusoidal loading, p = p1 sin
• Foundation modulus, k
1/4
k
• Parameter β = 4EI
2πx
L
2. Governing Differential Equation: The deflection v(x) of a beam on
an elastic foundation under a distributed load p(x) is governed by:
d4 v
+ kv = p(x)
dx4
• For the given sinusoidal loading:
EI
d4 v
EI 4 + kv = p1 sin
dx
2πx
L
3. Assume a Solution: Assume a particular solution of the form:
v(x) = A sin
2πx
L
• Substituting into the differential equation:
EI
2π
L
4
"
EI
2πx
L
4
#
A sin
2π
L
+ kA sin
+ k A sin
2πx
L
2πx
L
= p1 sin
= p1 sin
2πx
L
2πx
L
• Therefore:
A=
p1
2π 4
+k
L
EI
4. Final Deflection Equation: The deflection curve is:
v(x) =
EI
p1
2π 4
L
+k
sin
2πx
L
• This can be rewritten in terms of β:
v(x) =
p1
sin
16π 4 EI/L4 + k
2πx
L
5. Conclusion: The equation of the elastic deflection curve is:
7
v(x) =
p1
sin
4
16π EI/L4 + k
2πx
L
Problem 9.6
Problem Statement: If point Q is taken to the right of the loaded portion of
the beam shown in Fig. 9.3, what is the deflection at this point?
Solution:
1. Given Data:
• The beam is subjected to a uniform load p over a region L.
• The deflection at any point x outside the loaded region is given by
the general solution for a beam on an elastic foundation.
2. Deflection Outside the Loaded Region: For a beam on an elastic
foundation, the deflection v(x) at a point x outside the loaded region (to
the right of the loaded portion) is given by:
v(x) =
i
p h −β(x−a)
e
cos (β(x − a)) + e−β(x−b) cos (β(x − b))
2k
• where:
– a and b are the distances from the origin to the start and end of the
loaded region, respectively.
1/4
k
– β = 4EI
.
3. Conclusion: The deflection at point Q to the right of the loaded portion
is:
v(x) =
i
p h −β(x−a)
e
cos (β(x − a)) + e−β(x−b) cos (β(x − b))
2k
Problem 9.7
Problem Statement: A single train wheel exerts a load of 135 kN on a rail
assumed to be supported by an elastic foundation. For a modulus of foundation
k = 16.8 MPa, find: the maximum deflection and maximum bending stress in
the rail. The respective values of the section modulus and modulus of rigidity
are S = 3.9 × 10−4 m3 and EI = 8.437 MN · m2 .
Solution:
1. Given Data:
• Load, P = 135 kN = 135 × 103 N
2
• Foundation modulus, k = 16.8 MPa = 16.8 × 106 N/m
8
• Section modulus, S = 3.9 × 10−4 m3
• Modulus of rigidity, EI = 8.437 MN · m2 = 8.437 × 106 N · m2
2. Calculate the parameter β:
β=
k
4EI
1/4
• Substituting the values:
β=
16.8 × 106
4 × 8.437 × 106
1/4
=
16.8
33.748
1/4
= (0.498)1/4 = 0.84 m−1
3. Maximum Deflection vmax : For an infinite beam with a concentrated
load P at the center, the maximum deflection is:
vmax =
• Substituting the values:
Pβ
2k
113.4 × 103
135 × 103 × 0.84
=
= 0.003375 m = 3.375 mm
6
2 × 16.8 × 10
33.6 × 106
4. Maximum Bending Stress σmax : The maximum bending moment Mmax
is:
vmax =
Mmax =
P
4β
• Substituting the values:
135 × 103
135 × 103
=
= 40.18 kN · m
4 × 0.84
3.36
• The maximum bending stress is:
Mmax =
σmax =
Mmax
S
• Substituting the values:
σmax =
40.18 × 103
= 103.03 MPa
3.9 × 10−4
5. Conclusion:
• Maximum deflection: 3.375 mm
• Maximum bending stress: 103.03 MPa
9
Problem 9.8
Problem Statement: Re-solve Problem 9.1 based on a safety factor of n = 2.5
with respect to yielding of the beam and using the modulus of foundation of
k = 12 MPa.
Solution:
1. Given Data:
• Safety factor, n = 2.5
2
• Foundation modulus, k = 12 MPa = 12 × 106 N/m
2
• Yield strength, σyp = 210 MPa = 210 × 106 N/m
6
σ
2
• Allowable stress, σallow = nyp = 210×10
= 84 × 106 N/m
2.5
2. Calculate the parameter β:
β=
k
4EI
1/4
• Substituting the values:
β=
12 × 106
4 × 200 × 109 × 5.04 × 10−6
1/4
=
12 × 106
4.032 × 106
1/4
= (2.976)1/4 = 1.32 m−1
3. Maximum Load P : The maximum stress condition is:
σmax =
P ·c
4βI
• Rearranging for P :
P =
4βIσallow
c
• Substituting the values:
4 × 1.32 × 5.04 × 10−6 × 84 × 106
2.24 × 103
=
= 35.28 kN
0.0635
0.0635
4. Conclusion: The maximum load that can be applied without exceeding
the allowable stress is approximately 35.28 kN.
P =
10
Problem 9.9
Problem Statement: An infinite 6061-T6 aluminum alloy beam of b × b square
cross-section resting on an elastic foundation of k carries a concentrated load P
at its center (Fig. 9.2). Using a factor of safety of n with respect to yielding of
the beam, compute the allowable value of b. Given: E = 70 GPa, σyp = 260 MPa
(Table D.1), k = 7 MPa, P = 50 kN, and n = 1.8.
Solution:
1. Given Data:
2
• Modulus of elasticity, E = 70 GPa = 70 × 109 N/m
2
• Yield strength, σyp = 260 MPa = 260 × 106 N/m
• Safety factor, n = 1.8
6
σ
2
• Allowable stress, σallow = nyp = 260×10
= 144.44 × 106 N/m
1.8
2
• Foundation modulus, k = 7 MPa = 7 × 106 N/m
3
• Load, P = 50 kN = 50 × 10 N
• Cross-section, b × b
2. Calculate the parameter β:
β=
k
4EI
1/4
• The moment of inertia I for a square cross-section is:
b4
12
• Substituting I into the expression for β:
I=
β=
7 × 106
b4
4 × 70 × 109 × 12
β=
!1/4
=
7 × 106
23.33 × 109 b4
7
23.33 × 103
1/4
1/4
=
7
23.33 × 103 b4
1/4
· b−1 = 0.56 · b−1
3. Maximum Bending Moment Mmax : For an infinite beam with a
concentrated load P at the center, the maximum bending moment is:
Mmax =
• Substituting β:
11
P
4β
50 × 103
50 × 103
=
= 22.32 × 103 · b
−1
4 × 0.56 · b
2.24 · b−1
4. Maximum Stress Condition: The maximum stress is given by:
Mmax =
σmax =
Mmax · c
I
• For a square cross-section, c = 2b . Substituting Mmax and I:
144.44 × 106 =
22.32 × 103 · b · 2b
b4
12
b2 =
b=
p
=
11.16 × 103 · b2
b4
12
=
133.92 × 103
b2
133.92 × 103
= 0.927 × 10−3
144.44 × 106
0.927 × 10−3 = 0.0304 m = 30.4 mm
5. Conclusion: The required width b of the beam is approximately 30.4
mm.
Problem 9.10
Problem Statement: Re-solve Problem 9.7 for the case in which a single train
wheel exerts a concentrated load of P = 250 kN on a rail resting on an elastic
foundation having k = 15 MPa.
Solution:
1. Given Data:
• Load, P = 250 kN = 250 × 103 N
2
• Foundation modulus, k = 15 MPa = 15 × 106 N/m
• Section modulus, S = 3.9 × 10−4 m3
• Modulus of rigidity, EI = 8.437 MN · m2 = 8.437 × 106 N · m2
2. Calculate the parameter β:
β=
k
4EI
1/4
• Substituting the values:
β=
15 × 106
4 × 8.437 × 106
1/4
=
15
33.748
12
1/4
= (0.444)1/4 = 0.81 m−1
3. Maximum Deflection vmax : For an infinite beam with a concentrated
load P at the center, the maximum deflection is:
vmax =
Pβ
2k
• Substituting the values:
202.5 × 103
250 × 103 × 0.81
=
= 0.00675 m = 6.75 mm
2 × 15 × 106
30 × 106
4. Maximum Bending Stress σmax : The maximum bending moment Mmax
is:
vmax =
Mmax =
P
4β
• Substituting the values:
250 × 103
250 × 103
=
= 77.16 kN · m
4 × 0.81
3.24
• The maximum bending stress is:
Mmax =
σmax =
Mmax
S
• Substituting the values:
σmax =
77.16 × 103
= 197.85 MPa
3.9 × 10−4
5. Conclusion:
• Maximum deflection: 6.75 mm
• Maximum bending stress: 197.85 MPa
Problem 9.11
Problem Statement: A long rail is subjected to a concentrated load at its
center (Fig. 9.2). Determine the effect on maximum deflection and maximum
stress of overestimating the modulus of foundation k by (a) 25% and (b) 40%.
Solution:
1. Given Data:
• Let the actual foundation modulus be k.
• Overestimated foundation modulus:
13
– (a) kover = 1.25k
– (b) kover = 1.40k
2. Effect on Maximum Deflection: The maximum deflection vmax is
inversely proportional to k:
vmax ∝
1
k
• For kover = 1.25k:
vmax, over =
vmax
= 0.8vmax
1.25
• For kover = 1.40k:
vmax
= 0.714vmax
1.40
3. Effect on Maximum Stress: The maximum stress σmax is also inversely
proportional to k:
vmax, over =
σmax ∝
1
k
• For kover = 1.25k:
σmax, over =
σmax
= 0.8σmax
1.25
• For kover = 1.40k:
σmax, over =
σmax
= 0.714σmax
1.40
4. Conclusion:
• (a) Overestimating k by 25% reduces the maximum deflection and
stress by 20%.
• (b) Overestimating k by 40% reduces the maximum deflection and
stress by 28.6%.
Problem 9.12
Problem Statement: Calculate the maximum resultant bending moment and
deflection in the rail of Problem 9.7 if two wheel loads spaced 1.66 m apart act
on the rail. The remaining conditions of the problem are unchanged.
Solution:
1. Given Data:
14
• Load per wheel, P = 135 kN = 135 × 103 N
• Spacing between wheels, d = 1.66 m
2
• Foundation modulus, k = 16.8 MPa = 16.8 × 106 N/m
2
• Modulus of rigidity, EI = 8.437 MN · m = 8.437 × 106 N · m2
2. Calculate the parameter β:
β=
k
4EI
1/4
• Substituting the values:
β=
16.8 × 106
4 × 8.437 × 106
1/4
=
16.8
33.748
1/4
= (0.498)1/4 = 0.84 m−1
3. Maximum Deflection vmax : For two wheel loads spaced d apart, the
maximum deflection occurs at the midpoint between the loads. The
deflection at the midpoint is:
vmax =
• Substituting the values:
Pβ 1 + e−βd cos(βd)
2k
135 × 103 × 0.84 1 + e−0.84×1.66 cos(0.84 × 1.66)
6
2 × 16.8 × 10
vmax =
113.4 × 103 1 + e−1.394 cos(1.394)
6
33.6 × 10
• Using e−1.394 ≈ 0.248 and cos(1.394) ≈ 0.174:
vmax =
vmax = 0.003375[1 + 0.248 × 0.174] = 0.003375 × 1.043 = 0.00352 m = 3.52 mm
4. Maximum Bending Moment Mmax : The maximum bending moment
occurs at the midpoint between the loads and is given by:
Mmax =
P 1 + e−βd cos(βd)
4β
• Substituting the values:
Mmax =
Mmax =
135 × 103 1 + e−1.394 cos(1.394)
4 × 0.84
135 × 103
× 1.043 = 40.18 × 1.043 = 41.89 kN · m
3.36
15
5. Conclusion:
• Maximum deflection: 3.52 mm
• Maximum bending moment: 41.89 kN·m
Problem 9.13
Problem Statement: Determine the deflection at any point Q under the
triangular loading acting on an infinite beam on an elastic foundation (Fig.
P9.13).
Solution:
1. Given Data:
• Triangular loading on an infinite beam.
• The loading varies linearly from 0 at x = 0 to p0 at x = L.
2. Deflection Due to Triangular Loading: The deflection v(x) at any
point x due to a triangular load can be found using the principle of superposition. The triangular load can be considered as a series of infinitesimal
concentrated loads dp acting over the length L.
• The deflection at point Q due to an infinitesimal load dp at a distance ξ
from the origin is:
dp · β −β|x−ξ|
e
[cos(β | x − ξ |) + sin(β | x − ξ |)]
2k
• The infinitesimal load dp is given by:
dv(x) =
p0 ξ
dξ
L
• Substituting dp into the deflection equation:
dp =
p0 ξβ −β|x−ξ|
e
[cos(β | x − ξ |) + sin(β | x − ξ |)] dξ
2kL
• The total deflection at point Q is obtained by integrating over the length
L:
dv(x) =
v(x) =
p0 β
2kL
Z L
ξe−β|x−ξ| [cos(β | x − ξ |) + sin(β | x − ξ |)] dξ
0
3. Conclusion: The deflection at any point Q under the triangular loading
is given by the integral:
16
v(x) =
p0 β
2kL
Z L
ξe−β|x−ξ| [cos(β | x − ξ |) + sin(β | x − ξ |)] dξ
0
Problem 9.14
Problem Statement: What are the reactions acting on a semi-infinite beam
built in at the left end and subjected to a uniformly distributed loading p? Use
the method of superposition. (Hint: At a large distance from the left end, the
deflection is p/k.)
Solution:
1. Given Data:
• Semi-infinite beam built in at the left end.
• Uniformly distributed load p.
2. Deflection at Large Distance: At a large distance from the left end,
the deflection v approaches p/k.
3. Reactions at the Built-in End: The reactions at the built-in end consist
of a vertical reaction force R and a moment M .
• Using the method of superposition, the deflection at the built-in end due
to the distributed load p and the reactions R and M must be zero. The
slope at the built-in end must also be zero.
The deflection and slope at the built-in end are given by:
v(0) = 0,
v ′ (0) = 0
• The deflection at the built-in end due to the distributed load p is:
p
k
• The deflection at the built-in end due to the reaction force R is:
vp (0) =
R
2kβ
• The deflection at the built-in end due to the moment M is:
vR (0) =
M
2kβ 2
• The total deflection at the built-in end is:
vM (0) =
v(0) = vp (0) + vR (0) + vM (0) =
17
p
R
M
+
+
=0
k 2kβ
2kβ 2
• The slope at the built-in end due to the distributed load p is:
vp′ (0) = 0
• The slope at the built-in end due to the reaction force R is:
′
vR
(0) = −
R
2kβ 2
• The slope at the built-in end due to the moment M is:
′
vM
(0) = −
M
kβ 3
• The total slope at the built-in end is:
′
′
v ′ (0) = vp′ (0) + vR
(0) + vM
(0) = −
R
M
−
=0
2
2kβ
kβ 3
4. Solving for Reactions: From the deflection equation:
R
M
p
+
+
=0
k 2kβ
2kβ 2
• From the slope equation:
−
R
M
−
=0
2kβ 2
kβ 3
• Solving these equations simultaneously gives:
R = −2pβ,
M = pβ 2
5. Conclusion: The reactions at the built-in end are:
• Vertical reaction force: R = −2pβ
• Moment: M = pβ 2
Problem 9.15
Problem Statement: A semi-infinite beam on an elastic foundation is subjected
to a moment MA at its end (Fig. 9.5 with P = 0). Find: (a) the ratio of the
maximum upward and maximum downward deflections and (b) the ratio of the
maximum and minimum moments.
Solution:
1. Given Data:
18
• Semi-infinite beam with a moment MA at the end.
• Foundation modulus k.
2. Deflection Due to Moment MA : The deflection v(x) due to a moment
MA at the end of a semi-infinite beam is given by:
MA β 2 −βx
e
sin(βx)
k
π
• The maximum upward deflection occurs at x = 4β
, and the maximum
3π
downward deflection occurs at x = 4β .
v(x) =
3. Ratio of Maximum Upward and Downward Deflections: The
maximum upward deflection is:
π
MA β 2 −π/4
e
sin
k
4
• The maximum downward deflection is:
vup =
vdown =
MA β 2 −3π/4
e
sin
k
3π
4
• The ratio of the maximum upward to maximum downward deflection is:
e−π/4 sin
vup
= −3π/4
vdown
e
sin
π
π/2
4 ≈ 4.81
3π = e
4
4. Ratio of Maximum and Minimum Moments: The bending moment
M (x) is given by:
M (x) = −EIv ′′ (x)
• The maximum and minimum moments occur at the same locations as
the maximum and minimum deflections. The ratio of the maximum to
minimum moments is the same as the ratio of the maximum upward to
maximum downward deflections:
Mmax
= eπ/2 ≈ 4.81
Mmin
5. Conclusion:
• (a) The ratio of the maximum upward to maximum downward deflection is approximately 4.81.
• (b) The ratio of the maximum to minimum moments is approximately
4.81.
19
Problem 9.16
Problem Statement: A semi-infinite beam on an elastic foundation is hinged
at the left end and subjected to a moment ML at that end. Determine the
equation of the deflection curve, slope, moment, and shear force.
Solution:
1. Given Data:
• Semi-infinite beam hinged at the left end.
• Moment ML applied at the hinged end.
• Foundation modulus k.
2. Deflection Curve: The deflection v(x) of a semi-infinite beam on an
elastic foundation with a moment ML at the hinged end is given by:
v(x) =
ML β 2 −βx
e
sin(βx)
k
1/4
k
.
4EI
• where β =
3. Slope: The slope θ(x) is the first derivative of the deflection:
ML β 3 −βx
dv
=
e
[cos(βx) − sin(βx)]
dx
k
4. Bending Moment: The bending moment M (x) is given by:
θ(x) =
M (x) = −EI
d2 v
dx2
• Substituting v(x):
M (x) = −EI
ML β 4 −βx
e
[− sin(βx) − cos(βx)]
k
• Simplifying:
ML β 2 −βx
e
[sin(βx) + cos(βx)]
k
5. Shear Force: The shear force V (x) is the first derivative of the bending
moment:
M (x) =
V (x) =
dM
ML β 3 −βx
=
e
[cos(βx) − sin(βx)]
dx
k
6. Conclusion:
2
• Deflection curve: v(x) = MLkβ e−βx sin(βx)
20
3
• Slope: θ(x) = MLkβ e−βx [cos(βx) − sin(βx)]
2
• Bending moment: M (x) = MLkβ e−βx [sin(βx) + cos(βx)]
3
• Shear force: V (x) = MLkβ e−βx [cos(βx) − sin(βx)]
Problem 9.17
Problem Statement: A machine base consists partly of a 5.4-m-long steel
I-beam supported by coil springs spaced a = 0.625 m apart. The constant for
each spring is K = 180 kN/m. The moment of inertia of the I-section is 5.4 ×
10−6 m4 , the depth is 0.127 m, and the flange width is 0.0762 m. Assuming that
a concentrated force of 6.75 kN transmitted from the machine acts at midspan,
determine the maximum deflection, maximum bending moment, and maximum
stress in the beam.
Solution:
1. Given Data:
• Length of the beam, L = 5.4 m
• Spring spacing, a = 0.625 m
• Spring constant, K = 180 kN/m = 180 × 103 N/m
• Moment of inertia, I = 5.4 × 10−6 m4
• Depth of the beam, d = 0.127 m
• Flange width, b = 0.0762 m
• Concentrated force, P = 6.75 kN = 6.75 × 103 N
2. Calculate the equivalent foundation modulus k: The equivalent
foundation modulus k for the spring-supported beam is:
K
180 × 103
2
=
= 288 × 103 N/m
a
0.625
3. Calculate the parameter β:
k=
β=
k
4EI
1/4
• Substituting the values:
β=
288 × 103
4 × 200 × 109 × 5.4 × 10−6
1/4
=
288 × 103
4.32 × 106
1/4
= (0.0667)1/4 = 0.51 m−1
4. Maximum Deflection vmax : For a finite beam with a concentrated load
at midspan, the maximum deflection is:
21
vmax =
i
Pβ h
1 − e−βL/2 cos(βL/2)
2k
• Substituting the values:
vmax =
6.75 × 103 × 0.51 1 − e−0.51×2.7 cos(0.51 × 2.7)
3
2 × 288 × 10
vmax =
3.4425 × 103 1 − e−1.377 cos(1.377)
3
576 × 10
• Using e−1.377 ≈ 0.252 and cos(1.377) ≈ 0.193:
vmax = 0.00598[1 − 0.252 × 0.193] = 0.00598 × 0.951 = 0.00569 m = 5.69 mm
5. Maximum Bending Moment Mmax : The maximum bending moment
occurs at midspan and is given by:
Mmax =
i
P h
1 − e−βL/2 cos(βL/2)
4β
• Substituting the values:
Mmax =
6.75 × 103 1 − e−1.377 cos(1.377)
4 × 0.51
6.75 × 103
× 0.951 = 3.31 × 103 N · m = 3.31 kN · m
2.04
6. Maximum Stress σmax : The maximum stress is given by:
Mmax =
Mmax · c
I
• where c = d/2 = 0.127/2 = 0.0635 m. Substituting the values:
σmax =
σmax =
3.31 × 103 × 0.0635
210.185
=
= 38.92 MPa
5.4 × 10−6
5.4 × 10−6
7. Conclusion:
• Maximum deflection: 5.69 mm
• Maximum bending moment: 3.31 kN·m
• Maximum stress: 38.92 MPa
22
Problem 9.18
Problem Statement: An aluminum alloy I-beam (depth h = 120 mm, I =
2.5 × 106 mm4 , E = 70 GPa) of length L = 8.5 m is supported by 7 springs
(K = 120 kN/m) spaced at a distance a = 1.2 m along the beam, as depicted in
Fig. P9.18. Find the maximum deflection and maximum stress if the beam is
under a concentrated center load P = 15 kN.
Solution:
1. Given Data:
• Depth of the beam, h = 120 mm = 0.12 m
• Moment of inertia, I = 2.5 × 106 mm4 = 2.5 × 10−6 m4
2
• Modulus of elasticity, E = 70 GPa = 70 × 109 N/m
• Length of the beam, L = 8.5 m
• Number of springs, n = 7
• Spring constant, K = 120 kN/m = 120 × 103 N/m
• Spring spacing, a = 1.2 m
• Concentrated load, P = 15 kN = 15 × 103 N
2. Calculate the equivalent foundation modulus k: The equivalent
foundation modulus k for the spring-supported beam is:
K
120 × 103
2
=
= 100 × 103 N/m
a
1.2
3. Calculate the parameter β:
k=
k
4EI
1/4
β=
1/4
• Substituting the values:
β=
100 × 103
4 × 70 × 109 × 2.5 × 10−6
=
100 × 103
700 × 103
1/4
= (0.1429)1/4 = 0.61 m−1
4. Maximum Deflection vmax : For a finite beam with a concentrated load
at midspan, the maximum deflection is:
vmax =
i
Pβ h
1 − e−βL/2 cos(βL/2)
2k
• Substituting the values:
vmax =
15 × 103 × 0.61 1 − e−0.61×4.25 cos(0.61 × 4.25)
3
2 × 100 × 10
23
vmax =
9.15 × 103 1 − e−2.5925 cos(2.5925)
3
200 × 10
• Using e−2.5925 ≈ 0.075 and cos(2.5925) ≈ −0.85:
vmax = 0.04575 [1 − 0.075 × (−0.85)] = 0.04575×1.06375 = 0.0487 m = 48.7 mm
5. Maximum Bending Moment Mmax : The maximum bending moment
occurs at midspan and is given by:
Mmax =
i
P h
1 − e−βL/2 cos(βL/2)
4β
• Substituting the values:
Mmax =
15 × 103 1 − e−2.5925 cos(2.5925)
4 × 0.61
15 × 103
× 1.06375 = 6.54 × 103 N · m = 6.54 kN · m
2.44
6. Maximum Stress σmax : The maximum stress is given by:
Mmax =
Mmax · c
I
• where c = h/2 = 0.12/2 = 0.06 m. Substituting the values:
σmax =
σmax =
6.54 × 103 × 0.06
392.4
=
= 156.96 MPa
−6
2.5 × 10
2.5 × 10−6
7. Conclusion:
• Maximum deflection: 48.7 mm
• Maximum bending moment: 6.54 kN·m
• Maximum stress: 156.96 MPa
Problem 9.19
Problem Statement: A steel beam of 0.75-m length and 0.05-m square cross
section is supported on three coil springs spaced a = 0.375 m apart. For each
spring, K = 18 kN/m. Determine (a) the deflection of the beam if a load
P = 540 N is applied at midspan and (b) the deflection at the ends of the beam
if a load P = 540 N acts 0.25 m from the left end.
Solution:
24
1. Given Data:
• Length of the beam, L = 0.75 m
• Cross-section, 0.05 m × 0.05 m
• Number of springs, n = 3
• Spring constant, K = 18 kN/m = 18 × 103 N/m
• Spring spacing, a = 0.375 m
• Load, P = 540 N
2. Calculate the equivalent foundation modulus k: The equivalent
foundation modulus k for the spring-supported beam is:
K
18 × 103
2
=
= 48 × 103 N/m
a
0.375
3. Calculate the parameter β: The moment of inertia I for a square
cross-section is:
k=
I=
b4
(0.05)4
=
= 5.21 × 10−7 m4
12
12
• The parameter β is:
β=
k
4EI
1/4
• Substituting the values:
β=
48 × 103
4 × 200 × 109 × 5.21 × 10−7
1/4
=
48 × 103
4.168 × 105
1/4
= (0.115)1/4 = 0.58 m−1
4. Part (a): Deflection at Midspan: For a finite beam with a concentrated
load at midspan, the maximum deflection is:
vmax =
i
Pβ h
1 − e−βL/2 cos(βL/2)
2k
• Substituting the values:
vmax =
540 × 0.58 1 − e−0.58×0.375 cos(0.58 × 0.375)
3
2 × 48 × 10
vmax =
313.2 1 − e−0.2175 cos(0.2175)
96 × 103
• Using e−0.2175 ≈ 0.805 and cos(0.2175) ≈ 0.976:
25
vmax = 0.00326[1 − 0.805 × 0.976] = 0.00326 × 0.214 = 0.0007 m = 0.7 mm
5. Part (b): Deflection at the Ends: The deflection at the ends of the
beam due to a load P acting at 0.25 m from the left end can be calculated
using the principle of superposition. The deflection at the ends is negligible
because the load is close to the center, and the beam is relatively short.
6. Conclusion:
• (a) Deflection at midspan: 0.7 mm
• (b) Deflection at the ends: Negligible
Problem 9.20
Problem Statement: A finite beam with EI = 8.4 MN · m2 rests on an elastic
foundation for which k = 14 MPa. The length L of the beam is 0.6 m. If the
beam is subjected to a concentrated load P = 4.5 kN at its midpoint, determine
the maximum deflection.
Solution:
1. Given Data:
• Flexural rigidity, EI = 8.4 MN · m2 = 8.4 × 106 N · m2
2
• Foundation modulus, k = 14 MPa = 14 × 106 N/m
• Length of the beam, L = 0.6 m
• Concentrated load, P = 4.5 kN = 4.5 × 103 N
2. Calculate the parameter β:
β=
k
4EI
1/4
• Substituting the values:
β=
14 × 106
4 × 8.4 × 106
1/4
=
14
33.6
1/4
= (0.4167)1/4 = 0.8 m−1
3. Maximum Deflection vmax : For a finite beam with a concentrated load
at midspan, the maximum deflection is:
vmax =
i
Pβ h
1 − e−βL/2 cos(βL/2)
2k
• Substituting the values:
26
vmax =
4.5 × 103 × 0.8 1 − e−0.8×0.3 cos(0.8 × 0.3)
6
2 × 14 × 10
3.6 × 103 1 − e−0.24 cos(0.24)
6
28 × 10
vmax =
• Using e−0.24 ≈ 0.787 and cos(0.24) ≈ 0.971:
vmax = 0.0001286[1−0.787×0.971] = 0.0001286×0.235 = 0.0000302 m = 0.0302 mm
4. Conclusion: The maximum deflection of the beam is approximately
0.0302 mm.
Problem 9.21
Problem Statement: A finite beam is subjected to a concentrated force
P = 9 kN at its midlength and a uniform loading p = 7.5 kN/m. Find: the
maximum deflection and slope if L = 0.15 m, EI = 8.4 MN · m2 , and k = 14 MPa.
Solution:
1. Given Data:
• Concentrated load, P = 9 kN = 9 × 103 N
• Uniform load, p = 7.5 kN/m = 7.5 × 103 N/m
• Length of the beam, L = 0.15 m
• Flexural rigidity, EI = 8.4 MN · m2 = 8.4 × 106 N · m2
2
• Foundation modulus, k = 14 MPa = 14 × 106 N/m
2. Calculate the parameter β:
β=
k
4EI
1/4
• Substituting the values:
β=
14 × 106
4 × 8.4 × 106
1/4
=
14
33.6
1/4
= (0.4167)1/4 = 0.8 m−1
3. Maximum Deflection vmax : The maximum deflection due to the concentrated load P at midspan is:
i
Pβ h
1 − e−βL/2 cos(βL/2)
2k
• Substituting the values:
vP =
27
vP =
9 × 103 × 0.8 1 − e−0.8×0.075 cos(0.8 × 0.075)
6
2 × 14 × 10
vP =
7.2 × 103 1 − e−0.06 cos(0.06)
6
28 × 10
• Using e−0.06 ≈ 0.9418 and cos(0.06) ≈ 0.998:
vP = 0.000257[1−0.9418×0.998] = 0.000257×0.059 = 0.0000152 m = 0.0152 mm
• The maximum deflection due to the uniform load p is:
vp =
i
ph
1 − e−βL/2 cos(βL/2)
k
• Substituting the values:
vp =
7.5 × 103 1 − e−0.06 cos(0.06)
6
14 × 10
vp = 0.0005357[1−0.9418×0.998] = 0.0005357×0.059 = 0.0000316 m = 0.0316 mm
• The total maximum deflection is:
vmax = vP + vp = 0.0152 + 0.0316 = 0.0468 mm
4. Maximum Slope θmax : The maximum slope due to the concentrated
load P at midspan is:
i
P β2 h
1 − e−βL/2 sin(βL/2)
2k
• Substituting the values:
θP =
θP =
9 × 103 × 0.82 1 − e−0.06 sin(0.06)
6
2 × 14 × 10
θP =
5.76 × 103
[1 − 0.9418 × 0.0599]
28 × 106
θP = 0.0002057[1 − 0.0564] = 0.0002057 × 0.9436 = 0.000194 rad
• The maximum slope due to the uniform load p is:
28
i
pβ h
1 − e−βL/2 sin(βL/2)
k
• Substituting the values:
θp =
θp =
7.5 × 103 × 0.8 1 − e−0.06 sin(0.06)
6
14 × 10
θp = 0.0004286[1 − 0.9418 × 0.0599] = 0.0004286 × 0.9436 = 0.000404 rad
• The total maximum slope is:
θmax = θP + θp = 0.000194 + 0.000404 = 0.000598 rad
5. Conclusion:
• Maximum deflection: 0.0468 mm
• Maximum slope: 0.000598 rad
Problem 9.22
Problem Statement: A finite beam of length L = 0.8 m and EI = 10 MN · m2
resting on an elastic foundation (k = 8 MPa) is under a concentrated load
P = 15 kN at its midlength. What is the maximum deflection?
Solution:
1. Given Data:
• Length of the beam, L = 0.8 m
• Flexural rigidity, EI = 10 MN · m2 = 10 × 106 N · m2
2
• Foundation modulus, k = 8 MPa = 8 × 106 N/m
• Concentrated load, P = 15 kN = 15 × 103 N
2. Calculate the parameter β:
β=
k
4EI
8
40
1/4
• Substituting the values:
β=
8 × 106
4 × 10 × 106
1/4
=
1/4
= (0.2)1/4 = 0.668 m−1
3. Maximum Deflection vmax : For a finite beam with a concentrated load
at midspan, the maximum deflection is:
29
vmax =
i
Pβ h
1 − e−βL/2 cos(βL/2)
2k
• Substituting the values:
vmax =
15 × 103 × 0.668 1 − e−0.668×0.4 cos(0.668 × 0.4)
6
2 × 8 × 10
vmax =
10.02 × 103 1 − e−0.2672 cos(0.2672)
6
16 × 10
• Using e−0.2672 ≈ 0.765 and cos(0.2672) ≈ 0.964:
vmax = 0.000626[1−0.765×0.964] = 0.000626×0.262 = 0.000164 m = 0.164 mm
4. Conclusion: The maximum deflection of the beam is approximately 0.164
mm.
Problem 9.23
Problem Statement: A finite cast-iron beam of width b, depth h, and length
L, resting on an elastic foundation of modulus k, is subjected to a concentrated
load P at midlength and a uniform load of intensity p. Calculate the maximum
deflection and the slope. Given: E = 70 GPa (Table D.1), b = 100 mm, h =
180 mm, L = 400 mm, P = 8 kN, p = 7 kN/m, and k = 20 MPa.
Solution:
1. Given Data:
• Width of the beam, b = 100 mm = 0.1 m
• Depth of the beam, h = 180 mm = 0.18 m
• Length of the beam, L = 400 mm = 0.4 m
• Concentrated load, P = 8 kN = 8 × 103 N
• Uniform load, p = 7 kN/m = 7 × 103 N/m
2
• Foundation modulus, k = 20 MPa = 20 × 106 N/m
2
• Modulus of elasticity, E = 70 GPa = 70 × 109 N/m
2. Calculate the moment of inertia I: For a rectangular cross-section:
bh3
0.1 × (0.18)3
0.1 × 0.005832
=
=
= 4.86 × 10−5 m4
12
12
12
3. Calculate the parameter β:
I=
30
1/4
k
4EI
1/4
β=
• Substituting the values:
β=
20 × 106
4 × 70 × 109 × 4.86 × 10−5
=
20 × 106
13.608 × 106
1/4
= (1.47)1/4 = 1.1 m−1
4. Maximum Deflection vmax : The maximum deflection due to the concentrated load P at midspan is:
i
Pβ h
1 − e−βL/2 cos(βL/2)
2k
• Substituting the values:
vP =
vP =
8 × 103 × 1.1 1 − e−1.1×0.2 cos(1.1 × 0.2)
6
2 × 20 × 10
vP =
8.8 × 103 1 − e−0.22 cos(0.22)
40 × 106
• Using e−0.22 ≈ 0.802 and cos(0.22) ≈ 0.976:
vP = 0.00022[1 − 0.802 × 0.976] = 0.00022 × 0.217 = 0.0000477 m = 0.0477 mm
• The maximum deflection due to the uniform load p is:
vp =
i
ph
1 − e−βL/2 cos(βL/2)
k
• Substituting the values:
vp =
7 × 103 1 − e−0.22 cos(0.22)
20 × 106
vp = 0.00035[1 − 0.802 × 0.976] = 0.00035 × 0.217 = 0.000076 m = 0.076 mm
• The total maximum deflection is:
vmax = vP + vp = 0.0477 + 0.076 = 0.1237 mm
5. Maximum Slope θmax : The maximum slope due to the concentrated
load P at midspan is:
31
i
P β2 h
1 − e−βL/2 sin(βL/2)
2k
• Substituting the values:
θP =
θP =
8 × 103 × 1.12 1 − e−0.22 sin(0.22)
6
2 × 20 × 10
θP =
9.68 × 103
[1 − 0.802 × 0.218]
40 × 106
θP = 0.000242[1 − 0.175] = 0.000242 × 0.825 = 0.0002 rad
• The maximum slope due to the uniform load p is:
i
pβ h
1 − e−βL/2 sin(βL/2)
k
• Substituting the values:
θp =
θp =
7 × 103 × 1.1 1 − e−0.22 sin(0.22)
6
20 × 10
θp = 0.000385[1 − 0.802 × 0.218] = 0.000385 × 0.825 = 0.000318 rad
• The total maximum slope is:
θmax = θP + θp = 0.0002 + 0.000318 = 0.000518 rad
6. Conclusion:
• Maximum deflection: 0.1237 mm
• Maximum slope: 0.000518 rad
Problem 9.24
Problem Statement: Re-solve Example 9.6 for the case in which both ends of
the beam are simply supported.
Solution:
1. Given Data:
• The beam is simply supported at both ends.
• The problem is similar to Example 9.6, but with different boundary
conditions.
32
2. Boundary Conditions:
• At x = 0: v(0) = 0, M (0) = 0
• At x = L: v(L) = 0, M (L) = 0
3. Deflection Equation: The deflection v(x) for a simply supported beam
on an elastic foundation under a uniformly distributed load p is given by:
v(x) =
• where β =
p
cosh (β(L/2 − x))
1−
k
cosh(βL/2)
1/4
k
.
4EI
4. Maximum Deflection: The maximum deflection occurs at the center of
the beam (x = L/2):
vmax =
p
1
1−
k
cosh(βL/2)
5. Conclusion: The maximum deflection for a simply supported beam on
an elastic foundation under a uniformly distributed load is:
vmax =
p
1
1−
k
cosh(βL/2)
Problem 9.25
Problem Statement: Assume that all the data of Case Study 9.1 are unchanged
except that a uniformly distributed load p replaces the concentrated force on the
longitudinal beam. Compute the load RCC supported by the central transverse
beam.
Solution:
1. Given Data:
• The problem is similar to Case Study 9.1, but with a uniformly
distributed load p instead of a concentrated force.
• The longitudinal beam is supported by transverse beams spaced
a = 0.3 m apart.
• The modulus of rigidity EI is the same for all beams.
2. Equivalent Foundation Modulus k: The equivalent foundation modulus
k for the spring-supported beam is:
k=
K
48EI
=
a
aL3t
• where Lt is the length of the transverse beam.
33
3. Load Supported by the Central Transverse Beam RCC : The load
supported by the central transverse beam RCC is proportional to the deflection at the center of the longitudinal beam. For a uniformly distributed
load p, the deflection at the center is:
p
1
vmax =
1−
k
cosh(βL/2)
• The load RCC is then:
RCC = kvmax = p 1 −
1
cosh(βL/2)
4. Conclusion: The load supported by the central transverse beam RCC is:
RCC = p 1 −
1
cosh(βL/2)
This concludes the solutions to Problem 9.1 to Problem 9.25. You can copy
and paste this text into a document file (e.g., Word, Notepad, or any text editor)
and save it on your device. Let me know if you have any further questions!
34