ME323 F19 HW7 sol v3 - Purdue University · 2019. 10. 21. · w p ó } ( î ï 3ureohp srlqwv $...

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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:


Σ𝐹 = 0 => 𝑉 − 𝐹 = 0

=> 𝑉 = 𝐹 = 𝑃

[1.3]


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


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]


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


2= −

𝑃𝐿𝐿2

2+

𝑃𝐿26

= −3𝑃𝐿

8

and


2= 𝐶3

Which gives

𝐶3 = −3𝑃𝐿

8

[1.13]

and

𝐸𝐼𝑣 (𝑥 = 𝐿/2) = 𝐸𝐼𝑣 (𝑥 = 𝐿/2)

where


2= −

𝑃𝐿𝐿2

2+

𝑃𝐿26

= −5𝑃𝐿

48

and


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.

Page 7 of 23


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

Page 8 of 23

Solution

FBD:

Equilibrium


Σ𝐹 = 0 => −𝑃 + 𝑅 − 𝑞(4𝐿) + 𝑅 = 0

[2.1]


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

Page 9 of 23


ΣM = 0 => 𝑀 (𝑥) + 𝑃𝑥 + 𝑞𝑥𝑥

2= 0

=> 𝑀 (𝑥) = −𝑃𝑥 −𝑞𝑥

2

[2.4]


𝐸𝐼𝑣 (𝑥) = −𝑃𝑥

2−

𝑞𝑥

6+ 𝐶1

[2.5]



6−

𝑞𝑥

24+ 𝐶1 𝑥 + 𝐶2

[2.6]

Section 1: 𝑳 ≤ 𝒙 ≤ 𝟒𝑳


ΣM = 0 => 𝑀 (𝑥) + 𝑃𝑥 + 𝑞𝑥𝑥

2− 𝑅 (𝑥 − 𝐿) = 0

=> 𝑀 (𝑥) = −𝑃𝑥 −𝑞𝑥

2+ 𝑅 (𝑥 − 𝐿)

[2.7]



2−

𝑞𝑥

6+ 𝑅

𝑥

2− 𝐿𝑥 + 𝐶3

[2.8]

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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

Page 13 of 23


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]


𝑀(𝑥) = −𝑞𝑥

6𝐿−

𝑞𝑥

2+ 𝐶1 𝑥 + 𝐶2

[3.3]


𝐸𝐼𝑣′(𝑥) = −𝑞𝑥

24𝐿−

𝑞𝑥

6+

𝐶1𝑥

2+ 𝐶2 𝑥 + 𝐶3

[3.4]


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

Page 15 of 23

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


𝑀(𝑥 = 0) = −7

60 𝑞𝐿 = −0.1167 𝑞𝐿

[3.15]

Location 3: the right end boundary

𝑥 = 𝐿


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


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


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


ΣM = 0 => 𝑀 (𝑥) − 𝑀 − 𝑉 𝑥 = 0

=> 𝑀 (𝑥) = 𝑀 + 𝑉 𝑥

[4.3]


𝐸𝐼𝑣 (𝑥) = 𝑀 𝑥 +𝑉 𝑥

2+ 𝐶1

[4.4]


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


2

[4.8]


2+

𝑉 𝑥

6

[4.9]

Section 2: 𝟐 𝒇𝒕 ≤ 𝒙 ≤ 𝟖 𝒇𝒕

Page 21 of 23


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]



2−

50(𝑥 − 2)

3+ 𝐶3

[4.11]



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


ΣM = 0 => 𝑀 (𝑥) − 𝑀 − 𝑉 𝑥 + 100𝑙𝑏

𝑓𝑡(6 𝑓𝑡)(𝑥 − 5) = 0

=> 𝑀 (𝑥) = 𝑀 + 𝑉 𝑥 − 600𝑥 + 3000

[4.15]



2− 300𝑥 + 3000𝑥 + 𝐶5

[4.16]



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

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