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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

Page 2 of 23

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:

Page 12 of 23

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

Page 14 of 23

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

Page 16 of 23

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

Page 17 of 23

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

Page 18 of 23

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

Page 19 of 23

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

Page 20 of 23

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

Page 21 of 23

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

Page 22 of 23

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]

Page 23 of 23

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