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ME 323: Mechanics of Materials Homework Set 7
Fall 2019 Due: Wednesday, October 16
Note: Students are free to use MATLAB /Maple /Mathematica to solve the final algebra and plot the deflection curves.
Problem 7.1 (10 points)
A rigid bar BEF is welded (fixed attachment) to an elastic cantilever beam ABCD (elastic modulus 𝐸 and second moment of area 𝐼) at the location B as shown in Fig. 7.1. A point load 𝑃 is applied at F. Using the second order integration method:
1) Calculate the deflection at the free end D 𝑣 in terms of 𝑃, 𝐿, 𝐸 and 𝐼. 2) Plot the deflection of the beam ABCD. (Non-dimensionalize the plot, i.e., 𝑥 becomes 𝑥/𝐿
and deflection 𝑣 becomes 𝑣/|𝑣 |, where |𝑣 | is the magnitude of deflection at free end D). Does the maximum deflection (absolute value) occur at free end D? (Answer Yes or No using the plot developed)
Fig. 7.1
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Solution
FBD:
Rigid Member BEF Elastic Beam ABD
Equilibrium
For member BEF:
Force Balance along Y:
Σ𝐹 = 0 => 𝐹 − 𝑃 = 0
=> 𝐹 = 𝑃
[1.1]
Moment balance about B:
Σ𝑀 = 0 => 𝑀 −𝑃𝐿
2= 0
=> 𝑀 =𝑃𝐿
2
[1.2]
For beam ABD:
Force Balance along Y:
Σ𝐹 = 0 => 𝑉 − 𝐹 = 0
=> 𝑉 = 𝐹 = 𝑃
[1.3]
Moment balance about B:
Σ𝑀 = 0 => 𝑀 + 𝑀 + 𝑃𝐿
2= 0
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=> 𝑀 = −𝑃𝐿
2−
𝑃𝐿
2= −𝑃𝐿
[1.4]
For the deflections we need section the beam into two parts.
Section 1: 𝟎 ≤ 𝒙 ≤ 𝑳/𝟐
Taking moment about O
ΣM = 0 => 𝑀 (𝑥) + 𝑃𝐿 − 𝑃𝑥 = 0
=> 𝑀 (𝑥) = −𝑃𝐿 + 𝑃𝑥
[1.5]
Integrating 𝑀 (𝑥) w.r.t x,
𝐸𝐼𝑣 (𝑥) = −𝑃𝐿𝑥 +1
2𝑃𝑥 + 𝐶1
[1.6] Integrating 𝐸𝐼𝑣 (𝑥) w.r.t x,
𝐸𝐼𝑣 (𝑥) = −𝑃𝐿𝑥
2+
𝑃𝑥
6+ 𝐶1𝑥 + 𝐶2
[1.7] The left end is fixed; hence the boundary conditions are 𝑣 (0) = 0 and 𝑣(0) = 0. Hence the values of constants C1 and C2 can be found to be:
𝐸𝐼𝑣 (𝑥 = 0) = 0
=> 𝐶1 = 0
[1.8]
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And
𝐸𝐼𝑣 (𝑥 = 0) = 0
=> 𝐶2 = 0
[1.9]
Section 2: 𝑳/𝟐 ≤ 𝒙 ≤ 𝑳
Taking moment about O
ΣM = 0 => 𝑀 (𝑥) + 𝑃𝐿 − 𝑃𝑥 + 𝐹 𝑥 −𝐿
2− 𝑀 = 0
Substituting from Eq. [1.1] and [1.2]
=> 𝑀 (𝑥) = −𝑃𝐿 − 𝑃𝑥 − 𝑃 𝑥 −𝐿
2−
𝑃𝐿
2= 0
𝑀 (𝑥) = 0
[1.10]
Integrating 𝑀 (𝑥) w.r.t x,
𝐸𝐼𝑣 (𝑥) = 𝐶3
[1.11]
Integrating 𝐸𝐼𝑣 (𝑥) w.r.t x,
𝐸𝐼𝑣 (𝑥) = 𝐶3 𝑥 + 𝐶4
[1.12]
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The values of the constants can be found using the compatibility conditions at 𝑥 = . Since the
beam is a single continuous member, the slope and displacement evaluated from both
expressions should be equal at 𝑥 = .
From slope compatibility,
𝐸𝐼𝑣 𝑥 =𝐿
2= 𝐸𝐼𝑣 𝑥 =
𝐿
2
where
𝐸𝐼𝑣 𝑥 =𝐿
2= −
𝑃𝐿𝐿2
2+
𝑃𝐿26
= −3𝑃𝐿
8
and
𝐸𝐼𝑣 𝑥 =𝐿
2= 𝐶3
Which gives
𝐶3 = −3𝑃𝐿
8
[1.13]
and
𝐸𝐼𝑣 (𝑥 = 𝐿/2) = 𝐸𝐼𝑣 (𝑥 = 𝐿/2)
where
𝐸𝐼𝑣 𝑥 =𝐿
2= −
𝑃𝐿𝐿2
2+
𝑃𝐿26
= −5𝑃𝐿
48
and
𝐸𝐼𝑣 𝑥 =𝐿
2= 𝐶3
𝐿
2+ 𝐶4
Which gives
[1.14]
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𝐶4 ∶=𝑃𝐿
12
The final deflection function is:
𝑣(𝑥) =
⎩⎨
⎧ −𝑃𝐿𝑥
2𝐸𝐼+
𝑃𝑥
6𝐸𝐼, 𝑓𝑜𝑟 0 ≤ 𝑥 ≤
𝐿
2
−3𝑃𝐿 𝑥
8𝐸𝐼 +
𝑃𝐿
12𝐸𝐼, 𝑓𝑜𝑟,
𝐿
2≤ 𝑥 ≤ 𝐿
[1.15]
a) The deflection at the free end is
𝑣(𝑥 = 𝐿) = 𝑣 (𝑥 = 𝐿) = −3𝑃𝐿
8𝐸𝐼 +
𝑃𝐿
12𝐸𝐼
−> 𝑣(𝑥 = 𝐿) = −7𝑃𝐿
24𝐸𝐼
[1.16]
b) The deflection of the beam is
The maximum bending moment does occur at the free end D(𝑥 = 𝐿), as evident from the deflection plot.
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Problem 7.2 (10 points)
A cantilever beam ABC made from a material with elastic modulus 𝐸 and second moment of area 𝐼 is loaded as shown in Fig. 7.2. Using the second order integration method:
1) Determine the deflection at the free end 𝑣 in terms of 𝑃, 𝐿, 𝐸 and 𝐼. 2) Plot the deflection of the beam ABCD. (Non-dimensionalize the plot, i.e., 𝑥 becomes 𝑥/𝐿,
deflection 𝑣 becomes 𝑣/|𝑣 | where |𝑣 | is the magnitude of deflection at free end A). Does the maximum deflection (absolute value) occur at free end A? (Answer Yes or No using the plot developed)
For calculations, use 𝑞 =
Fig. 7.2
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Solution
FBD:
Equilibrium
Force Balance along Y:
Σ𝐹 = 0 => −𝑃 + 𝑅 − 𝑞(4𝐿) + 𝑅 = 0
[2.1]
Moment balance about B:
Σ𝑀 = 0 => 𝑃𝐿 − 𝑞(4𝐿)(2𝐿 − 𝐿) + 𝑅 (3𝐿) = 0
=> 𝑅 = −𝑃
3+
4𝑞𝐿
3
[2.2] Substituting into Eq. [2.1], we get
𝑅 =4𝑃
3+
8𝑞𝐿
3
[2.3]
For the deflections we need section the beam into two parts.
Section 1: 𝟎 ≤ 𝒙 ≤ 𝑳
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Taking moment about O
ΣM = 0 => 𝑀 (𝑥) + 𝑃𝑥 + 𝑞𝑥𝑥
2= 0
=> 𝑀 (𝑥) = −𝑃𝑥 −𝑞𝑥
2
[2.4]
Integrating 𝑀 (𝑥) w.r.t x,
𝐸𝐼𝑣 (𝑥) = −𝑃𝑥
2−
𝑞𝑥
6+ 𝐶1
[2.5]
Integrating 𝐸𝐼𝑣 (𝑥) w.r.t x,
𝐸𝐼𝑣 (𝑥) = −𝑃𝑥
6−
𝑞𝑥
24+ 𝐶1 𝑥 + 𝐶2
[2.6]
Section 1: 𝑳 ≤ 𝒙 ≤ 𝟒𝑳
Taking moment about O
ΣM = 0 => 𝑀 (𝑥) + 𝑃𝑥 + 𝑞𝑥𝑥
2− 𝑅 (𝑥 − 𝐿) = 0
=> 𝑀 (𝑥) = −𝑃𝑥 −𝑞𝑥
2+ 𝑅 (𝑥 − 𝐿)
[2.7]
Integrating 𝑀 (𝑥) w.r.t x,
𝐸𝐼𝑣 (𝑥) = −𝑃𝑥
2−
𝑞𝑥
6+ 𝑅
𝑥
2− 𝐿𝑥 + 𝐶3
[2.8]
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Integrating 𝐸𝐼𝑣 (𝑥) w.r.t x,
𝐸𝐼𝑣 (𝑥) = −𝑃𝑥
6−
𝑞𝑥
24+ 𝑅
𝑥
6−
𝐿𝑥
2+ 𝐶3𝑥 + 𝐶4
[2.9]
We have 4 unknowns (𝐶1, 𝐶2, 𝐶3, 𝐶4) which needs to be solved for. The four equations needed can be developed from the boundary and compatibility conditions:
Boundary Conditions
As point B is a roller joint
𝑣(𝑥 = 𝐿) = 0
Which gives
𝐸𝐼𝑣(𝑥 = 𝐿) = 𝐸𝐼𝑣 (𝑥 = 𝐿) = 𝐸𝐼𝑣 (𝑥 = 𝐿) = 0
Thus, 𝐸𝐼𝑣 (𝑥 = 𝐿) = 0 gives
𝐸𝐼𝑣 (𝑥 = 𝐿) = −𝑃𝐿
6−
𝑞𝐿
24+ 𝐶1 𝐿 + 𝐶2 = 0
Similarly, 𝐸𝐼𝑣 (𝑥 = 𝐿) = 0 gives
𝐸𝐼𝑣 (𝑥 = 𝐿) = −𝑃𝐿
6−
𝑞𝐿
24−
𝑅 𝐿
3+ 𝐶3 (𝐿) + 𝐶4 = 0
[2.10]
[2.11]
[2.12]
At the point D, the joint is a pinned joint, so
𝑣(𝑥 = 4𝐿) = 𝑣 (𝑥 = 4𝐿) = 0
Which gives
𝐸𝐼𝑣 (𝑥 = 4𝐿) = −32𝑃𝐿
3−
32𝑞𝐿
3+
8𝑅 𝐿
3+ 𝐶3 (4𝐿) + 𝐶4 = 0
[2.13]
We get 3 equations from the boundary conditions. The final equation comes from compatibility.
Compatibility Conditions
The slope at point B should be the same as the beam is a single continuous member
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𝑣 (𝑥 = 𝐿) = 𝑣 (𝑥 = 𝐿)
Which gives
−𝑃𝐿
2−
𝑞𝐿
6+ 𝐶1 = −
𝑃𝐿
2−
𝑞𝐿
6−
𝑅 𝐿
2+ 𝐶3
[2.14]
Which is the final equation needed. Note that the displacement compatibility condition is already accounted for in the displacement boundary condition at B (Eq. [2.10]).
Solving the Eqs. [2.11] to [2.14], we have the value of the constants:
𝐶1 =85
24𝑞𝐿 +
7
2𝑃𝐿 −
3
2𝑅 𝐿
𝐶2 = −7
2𝑞𝐿 −
10
3𝑃𝐿 +
3
2𝑅 𝐿
𝐶3 =85
24𝑞𝐿 +
7
2𝑃𝐿 − 𝑅 𝐿
𝐶4 = −7
2𝑞𝐿 −
10
3𝑃𝐿 +
4
3𝑅 𝐿
Using Eq.[2.3] and substiuting for 𝑅 we get
𝐶1 = −11
24𝑞𝐿 +
3
2𝑃𝐿
𝐶2 = −1
2𝑞𝐿 −
4
3𝑃𝐿
𝐶3 =7
8𝑞𝐿 +
13
6𝑃𝐿
𝐶4 = −1
18𝑞𝐿 −
14
9𝑃𝐿
[2.15] The final deflection equation is:
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𝑣(𝑥) =
⎩⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎧ −
𝑃𝑥
6𝐸𝐼−
𝑞𝑥
24𝐸𝐼−
11𝑞𝐿 𝑥
24𝐸𝐼+
3𝑃𝐿 𝑥
2𝐸𝐼
+𝑞𝐿
2𝐸𝐼−
4𝑃𝐿
3𝐸𝐼, 𝑓𝑜𝑟 0 ≤ 𝑥 ≤ 𝐿
𝑃𝑥
18𝐸𝐼−
𝑞𝑥
24𝐸𝐼+
4𝑞𝐿𝑥
9𝐸𝐼−
4𝑞𝐿 𝑥
3𝐸𝐼−
2𝑃𝐿𝑥
3𝐸𝐼+
7𝑞𝐿 𝑥
8𝐸𝐼+
13𝑃𝐿 𝑥
6𝐸 𝐼
+𝑞𝐿
18𝐸𝐼±
14𝑃𝐿
9𝐸𝐼, 𝑓𝑜𝑟, 𝐿 ≤ 𝑥 ≤ 4𝐿
[2.16]
a) The deflection at the free end is
𝑣(𝑥 = 0) = 𝑣 (𝑥 = 0) = −𝑞𝐿
18𝐸𝐼 +
14𝑃𝐿
9𝐸𝐼
Given 𝑞 = , the defelction at the free end is
−> 𝑣 = 𝑣(𝑥 = 0) =2𝑃𝐿
3𝐸𝐼
[2.17]
b) The deflection of the beam is
c) Note that the maximum deflection does not occur at the free end A, which is obvious from the displacement plot
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Problem 7.3 (10 points)
A fixed beam AB (elastic modulus 𝐸 and second moment of area 𝐼) is subject to a trapezoidal load that varies from 𝑞 at 𝑥 = 0 to 2𝑞 at 𝑥 = 𝐿, as shown in the figure. Using second or fourth order integration method. Determine:
1) The maximum bending moment magnitude in terms of 𝑞, 𝐿, 𝐸 and 𝐼, and the corresponding 𝑥- coordinate location.
2) Calculate |𝑣 / | (the deflection magnitude at 𝑥 = 𝐿/2) in terms of 𝑃, 𝐿, 𝐸 and 𝐼.
3) Plot the deflection of the beam AB. (Non-dimensionalize the plot, i.e., 𝑥 becomes 𝑥/𝐿 and deflection 𝑣 becomes 𝑣/|𝑣 / |)
Fig. 7.3
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Solution
Using the 4th order integration method, the distributed load can be written as
𝑝(𝑥) = −𝑞𝑥
𝐿− 𝑞
[3.1]
Integrating w.r.t. x
𝑉(𝑥) = −𝑞𝑥
2𝐿− 𝑞𝑥 + 𝐶1
[3.2]
Integrating w.r.t. x
𝑀(𝑥) = −𝑞𝑥
6𝐿−
𝑞𝑥
2+ 𝐶1 𝑥 + 𝐶2
[3.3]
Integrating w.r.t. x
𝐸𝐼𝑣′(𝑥) = −𝑞𝑥
24𝐿−
𝑞𝑥
6+
𝐶1𝑥
2+ 𝐶2 𝑥 + 𝐶3
[3.4]
Integrating w.r.t. x
𝐸𝐼𝑣(𝑥) = −𝑞𝑥
120𝐿−
𝑞𝑥
24+
𝐶1 𝑥
6+
𝐶2 𝑥
2+ 𝐶3 𝑥 + 𝐶4
[3.5]
We have 4 unknown constants of integration, namely 𝐶1, 𝐶2, 𝐶3 and 𝐶4.
Hence we need 4 equations to solve the problem, which can be obtained from boundary conditions.
BC1: Slope at left end, 𝑣 (𝑥 = 0) = 0 gives
𝐶3 = 0
[3.6] BC2: Deflection at left end, 𝑣 (𝑥 = 0) = 0 gives
𝐶4 = 0
[3.7] BC3: Slope at right end, 𝑣 (𝑥 = 𝐿) = 0 gives
𝐸𝐼𝑣′(𝑥 = 𝐿) = −𝑞𝐿
24−
𝑞𝐿
6+
𝐶1𝐿
2+ 𝐶2 𝐿 = 0
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Dividing by 𝐿,
−𝑞𝐿
24−
𝑞𝐿
6+
𝐶1 𝐿
2+ 𝐶2 = 0
=> −5𝑞𝐿
24+
𝐶1 𝐿
2+ 𝐶2 = 0
[3.8]
BC4: Deflection at right end, 𝑣 (𝑥 = 𝐿) = 0 gives
𝐸𝐼𝑣(𝑥 = 𝐿) = −𝑞𝐿
120−
𝑞𝐿
24+
𝐶1 𝐿
6+
𝐶2 𝐿
2= 0
Dividing by 𝐿
−𝑞𝐿
120−
𝑞𝐿
24+
𝐶1 𝐿
6+
𝐶2
2= 0
=> −𝑞𝐿
20+
𝐶1 𝐿
6+
𝐶2
2= 0
[3.9]
Solving Eqs. [3.8] and [3.9],
𝐶1 =13𝑞𝐿
20
[3.10]
and
𝐶2 = −7𝑞𝐿
60
[3.11]
The final deflection curve is
𝑣(𝑥) =1
𝐸𝐼−
𝑞𝑥
120𝐿−
𝑞𝑥
24+
13𝑞𝐿𝑥
120−
7𝑞𝐿 𝑥
120
[3.12]
a) The maximum bending moment magnitude
Since we have continuous load 𝑝(𝑥) over 0 ≤ 𝑥 ≤ 𝐿, we need to check 3 locations
Location 1: where = 𝑉(𝑥) = 0
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Using Eqs. [3.2] and [3.10], we have
𝑉(𝑥) = −𝑞𝑥
2𝐿− 𝑞𝑥 +
13𝑞𝐿
20= 0
[3.13]
The solution of the quadriatc equation is
𝑥 = 0.5166 𝐿, −2.5166𝐿
Since negative solution for 𝑥 is not possible,
𝑥 = 0.5166 𝐿
Which gives bending moment magnitude
𝑀(𝑥 = 0.5166 𝐿) = 0.0627 𝑞𝐿
[3.14]
Location 2: the left end boundary
𝑥 = 0
Which gives bending moment magnitude
𝑀(𝑥 = 0) = −7
60 𝑞𝐿 = −0.1167 𝑞𝐿
[3.15]
Location 3: the right end boundary
𝑥 = 𝐿
Which gives bending moment magnitude
𝑀(𝑥 = 𝐿) = −2
15 𝑞𝐿 = −0.1334 𝑞𝐿
[3.16]
So the maximum bending moment magnitude is
|𝑴|𝒎𝒂𝒙 = −0.1334 𝑞𝐿 𝑎𝑡 𝑥 = 𝐿
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This can be verified by plotting the bending moment
b) Deflection 𝒗𝑳
𝟐
at 𝒙 =𝑳
𝟐
From Eq [3.12]
𝑣 𝑥 =𝐿
2= −
𝑞𝐿
256 𝐸𝐼
[3.17]
c)The deflection of the beam is
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Problem 7.4 (10 points)
A steel (𝐸 = 30,000 𝑘𝑠𝑖) square beam ABCD with a side length 𝑎 = 6" is subject to loading as shown in Fig. 7.4. Using second order integration method:
1) Determine the reactions at the supports at ends A and D. 2) Plot the deflection of the beam ABCD.
Fig. 7.4
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Solution
FBD:
Equilibrium
Force Balance along Y:
Σ𝐹 = 0 => 𝑉 + 𝑉 − 100𝑙𝑏
𝑓𝑡(6 𝑓𝑡) = 0
𝑉 + 𝑉 = 600 𝑙𝑏
[4.1] Moment balance about D:
Σ𝑀 = 0 => 𝑀 + (10𝑓𝑡)𝑉 − 100𝑙𝑏
𝑓𝑡(6 𝑓𝑡)(5𝑓𝑡) = 0
=> 𝑀 + 10 𝑉 = 3000 𝑙𝑏 𝑓𝑡
[4.2] We have three unknowns: 𝑉 , 𝑉 and 𝑀 , and only two unknowns. So, we need to make cuts and use the deflection compatibility equations.
Section 1: 𝟎 ≤ 𝒙 ≤ 𝟐 𝒇𝒕
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Taking moment about O
ΣM = 0 => 𝑀 (𝑥) − 𝑀 − 𝑉 𝑥 = 0
=> 𝑀 (𝑥) = 𝑀 + 𝑉 𝑥
[4.3]
Integrating 𝑀 (𝑥) w.r.t x,
𝐸𝐼𝑣 (𝑥) = 𝑀 𝑥 +𝑉 𝑥
2+ 𝐶1
[4.4]
Integrating 𝐸𝐼𝑣 (𝑥) w.r.t x,
𝐸𝐼𝑣 (𝑥) =𝑀 𝑥
2+
𝑉 𝑥
6+ 𝐶1 𝑥 + 𝐶2
[4.5]
BC1: Slope at left end, 𝑣 (𝑥 = 0) = 0 gives
𝐶1 = 0
[4.6] BC2: Deflection at left end, 𝑣 (𝑥 = 0) = 0 gives
𝐶2 = 0
[4.7] So the slope and deflection for 0 ≤ 𝑥 ≤ 2 𝑓𝑡 is
𝐸𝐼𝑣 (𝑥) = 𝑀 𝑥 +𝑉 𝑥
2
[4.8]
𝐸𝐼𝑣 (𝑥) =𝑀 𝑥
2+
𝑉 𝑥
6
[4.9]
Section 2: 𝟐 𝒇𝒕 ≤ 𝒙 ≤ 𝟖 𝒇𝒕
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Taking moment about O
ΣM = 0 => 𝑀 (𝑥) − 𝑀 − 𝑉 𝑥 + 100𝑙𝑏
𝑓𝑡(𝑥 − 2)
𝑥 − 2
2 = 0
=> 𝑀 (𝑥) = 𝑀 + 𝑉 𝑥 − 50(𝑥 − 2)
[4.10]
Integrating 𝑀 (𝑥) w.r.t x,
𝐸𝐼𝑣 (𝑥) = 𝑀 𝑥 +𝑉 𝑥
2−
50(𝑥 − 2)
3+ 𝐶3
[4.11]
Integrating 𝐸𝐼𝑣 (𝑥) w.r.t x,
𝐸𝐼𝑣 (𝑥) =𝑀 𝑥
2+
𝑉 𝑥
6−
25
6(𝑥 − 2) + 𝐶3 𝑥 + 𝐶4
[4.12]
Using slope compatibility at B, i.e., 𝑥 = 2 𝑓𝑡
𝐸𝐼𝑣 (𝑥 = 2) = 𝐸𝐼𝑣 (𝑥 = 2)
=> 2𝑉𝐴 + 2𝑀𝐴 = 2𝑀𝐴 + 2𝑉𝐴 + 𝐶3
=> 𝐶3 = 0
[4.13] Using deflection compatibility at B, i.e., 𝑥 = 2 𝑓𝑡
𝐸𝐼𝑣 (𝑥) = 𝐸𝐼𝑣 (𝑥)
=> 2𝑀 +4𝑉
3= 2𝑀 +
4𝑉
3+ 𝐶4
=> 𝐶4 = 0
[4.14] Section 2: 𝟖 𝒇𝒕 ≤ 𝒙 ≤ 𝟏𝟎 𝒇𝒕
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Taking moment about O
ΣM = 0 => 𝑀 (𝑥) − 𝑀 − 𝑉 𝑥 + 100𝑙𝑏
𝑓𝑡(6 𝑓𝑡)(𝑥 − 5) = 0
=> 𝑀 (𝑥) = 𝑀 + 𝑉 𝑥 − 600𝑥 + 3000
[4.15]
Integrating 𝑀 (𝑥) w.r.t x,
𝐸𝐼𝑣 (𝑥) = 𝑀 𝑥 +𝑉 𝑥
2− 300𝑥 + 3000𝑥 + 𝐶5
[4.16]
Integrating 𝐸𝐼𝑣 (𝑥) w.r.t x,
𝐸𝐼𝑣 (𝑥) =𝑀 𝑥
2+
𝑉 𝑥
6− 100𝑥 + 1500𝑥 + 𝐶5 𝑥 + 𝐶6
[4.17]
Using slope compatibility at C, i.e., 𝑥 = 8 𝑓𝑡
𝐸𝐼𝑣 (𝑥 = 8) = 𝐸𝐼𝑣 (𝑥 = 8)
=> 32𝑉𝐴 + 8𝑀𝐴 − 3600 = 8𝑀𝐴 + 32𝑉𝐴 + 4800 + 𝐶5
=> 𝐶5 = −8400
[4.18] Using deflection compatibility at C, i.e., 𝑥 = 8 𝑓𝑡
𝐸𝐼𝑣 (𝑥) = 𝐸𝐼𝑣 (𝑥)
=> 32𝑀 +256𝑉
3− 5400 = 2𝑀 +
4𝑉
3− 44800 + 𝐶6 = 0
=> 𝐶6 = 17000
[4.19]
a) Reaction forces and moments at the ends A and D
BC3: Deflection at right end, 𝑣 (𝑥 = 10) = 0 gives
500𝑉
3+ 50𝑀 − 17000 = 0
[4.20]
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Solving the above equation with the equilibrium equations [4.1] and [4.2] we have
𝑀 = −990 𝑙𝑏 𝑓𝑡, 𝑉 = 399 𝑙𝑏 𝑎𝑛𝑑 𝑉 = 201 𝑙𝑏
[4.21]
b) Deflection of the beam