Problem 1 (28 points): The member shown below is subject to a distributed load. It is known that the member has a uniform cross section area 𝐴𝐴, second area moment 𝐼𝐼, and a shear shape factor 𝑓𝑓𝑠𝑠. The member is composed of a material whose Young’s modulus is 𝐸𝐸, and shear modulus 𝐺𝐺.

Using Castigliano’s theorem, determine the reaction forces in the beam. Please include the shear, normal and bending shear strain energies in your solution. Leave your answer in terms of 𝑤𝑤, 𝐿𝐿,𝐻𝐻,𝐸𝐸,𝐺𝐺,𝐴𝐴, 𝐼𝐼, and 𝑓𝑓𝑠𝑠.

Solution: FBD:

Σ𝐹𝐹𝑁𝑁 = 0 = 𝑁𝑁 + 𝐹𝐹𝐴𝐴 − 𝑤𝑤𝐻𝐻 (1)

Σ𝐹𝐹𝑉𝑉 = 0 = 𝑉𝑉 = 0 (2)

Σ𝑀𝑀𝑂𝑂 = 0 ⇒ 𝑀𝑀 + 𝐹𝐹𝐴𝐴𝐻𝐻 − 𝑤𝑤𝐻𝐻 ×𝐻𝐻2

= 0 (3)

There are two equations (1), and (3), and three unknowns 𝑁𝑁,𝐹𝐹𝐴𝐴, and 𝑀𝑀. The system is indeterminate of order one. It is best to start energy equations where there are least number of unknowns, namely 𝐹𝐹𝐴𝐴.

0 ≤ 𝑦𝑦 ≤ 𝐻𝐻

Σ𝐹𝐹𝑁𝑁 = 0 ⇒ 𝑁𝑁

Σ𝐹𝐹𝑉𝑉 = 0 = 𝑉𝑉 − 𝐹𝐹𝐴𝐴 + 𝑤𝑤𝑦𝑦 ⇒ 𝑉𝑉

Σ𝑀𝑀𝑂𝑂 = 0

⇒ 𝑀𝑀 = −𝐹𝐹𝐴𝐴𝑦𝑦 +𝑤𝑤2𝑦𝑦2

𝜕𝜕𝑁𝑁𝜕𝜕𝐹𝐹𝐴𝐴

= 0; 𝜕𝜕𝑉𝑉𝜕𝜕𝐹𝐹𝐴𝐴

= 1; 𝜕𝜕𝑀𝑀𝜕𝜕𝐹𝐹𝐴𝐴

= −𝑦𝑦

0 ≤ 𝑥𝑥 ≤ 𝐿𝐿

Σ𝐹𝐹𝑁𝑁 = 0 = 𝑁𝑁 + 𝐹𝐹𝐴𝐴 − 𝑤𝑤𝐻𝐻 ⇒ 𝑁𝑁 = 𝑤𝑤𝐻𝐻 − 𝐹𝐹𝐴𝐴 (7)

Σ𝐹𝐹𝑉𝑉 = 0 = 𝑉𝑉 (8)

Σ𝑀𝑀𝑂𝑂 = 0 = 𝑀𝑀 + 𝐹𝐹𝐴𝐴𝐻𝐻 −𝑤𝑤𝐻𝐻 ×𝐻𝐻2

(9)

⇒ 𝑀𝑀 =𝑤𝑤𝐻𝐻2

2− 𝐹𝐹𝐴𝐴𝐻𝐻

𝜕𝜕𝑁𝑁𝜕𝜕𝐹𝐹𝐴𝐴

= −1; 𝜕𝜕𝑉𝑉𝜕𝜕𝐹𝐹𝐴𝐴

= 0; 𝜕𝜕𝑀𝑀𝜕𝜕𝐹𝐹𝐴𝐴

= −𝐻𝐻

From Castigliano’s theorem,

𝛿𝛿𝐹𝐹𝐴𝐴 = �𝑁𝑁𝐸𝐸𝐴𝐴

𝜕𝜕𝑁𝑁𝜕𝜕𝐹𝐹𝐴𝐴

𝐻𝐻

0𝑑𝑑𝑦𝑦 + �

𝑀𝑀𝐸𝐸𝐼𝐼

𝜕𝜕𝑀𝑀𝜕𝜕𝐹𝐹𝐴𝐴

𝐻𝐻

0𝑑𝑑𝑦𝑦 + �

𝑓𝑓𝑠𝑠𝑉𝑉𝐺𝐺𝐴𝐴

𝜕𝜕𝑉𝑉𝜕𝜕𝐹𝐹𝐴𝐴

𝐻𝐻

0𝑑𝑑𝑦𝑦 + �

𝑁𝑁𝐸𝐸𝐴𝐴

𝜕𝜕𝑁𝑁𝜕𝜕𝐹𝐹𝐴𝐴

𝐿𝐿

0𝑑𝑑𝑥𝑥 + �

𝑀𝑀𝐸𝐸𝐼𝐼

𝜕𝜕𝑀𝑀𝜕𝜕𝐹𝐹𝐴𝐴

𝐿𝐿

0𝑑𝑑𝑥𝑥

+ �𝑓𝑓𝑠𝑠𝑉𝑉𝐺𝐺𝐴𝐴

𝜕𝜕𝑉𝑉𝜕𝜕𝐹𝐹𝐴𝐴

𝐿𝐿

0𝑑𝑑𝑥𝑥 = 0 (10)

Non zero terms:

�𝑀𝑀𝐸𝐸𝐼𝐼

𝜕𝜕𝑀𝑀𝜕𝜕𝐹𝐹𝐴𝐴

𝐻𝐻

0𝑑𝑑𝑦𝑦 =

1𝐸𝐸𝐼𝐼� �−𝐹𝐹𝐴𝐴𝑦𝑦 +

𝑤𝑤2𝑦𝑦2� (−𝑦𝑦)

𝐻𝐻

0𝑑𝑑𝑦𝑦 =

𝐹𝐹𝐴𝐴𝐻𝐻3

3𝐸𝐸𝐼𝐼−𝑤𝑤𝐻𝐻4

8𝐸𝐸𝐼𝐼#(11)

�𝑓𝑓𝑠𝑠𝑉𝑉𝐺𝐺𝐴𝐴

𝜕𝜕𝑉𝑉𝜕𝜕𝐹𝐹𝐴𝐴

𝐻𝐻

0𝑑𝑑𝑦𝑦 =

𝑓𝑓𝑠𝑠𝐺𝐺𝐴𝐴

� {𝐹𝐹𝐴𝐴 − 𝑤𝑤𝑦𝑦}(1)𝐻𝐻

0𝑑𝑑𝑦𝑦 =

𝑓𝑓𝑠𝑠𝐹𝐹𝐴𝐴𝐻𝐻𝐺𝐺𝐴𝐴

−𝑓𝑓𝑠𝑠𝑤𝑤𝐻𝐻2

2𝐺𝐺𝐴𝐴#(12)

�𝑁𝑁𝐸𝐸𝐴𝐴

𝜕𝜕𝑁𝑁𝜕𝜕𝐹𝐹𝐴𝐴

𝐿𝐿

0𝑑𝑑𝑥𝑥 =

1𝐸𝐸𝐴𝐴

� (𝑤𝑤𝐻𝐻 − 𝐹𝐹𝐴𝐴)(−1)𝐿𝐿

0𝑑𝑑𝑥𝑥 =

𝐹𝐹𝐴𝐴𝐿𝐿𝐸𝐸𝐴𝐴

−𝑤𝑤𝐻𝐻𝐿𝐿𝐸𝐸𝐴𝐴

#(13)

�𝑀𝑀𝐸𝐸𝐼𝐼

𝜕𝜕𝑀𝑀𝜕𝜕𝐹𝐹𝐴𝐴

𝐿𝐿

0𝑑𝑑𝑥𝑥 =

1𝐸𝐸𝐼𝐼� �

𝑤𝑤𝐻𝐻2

2− 𝐹𝐹𝐴𝐴𝐻𝐻� (−𝐻𝐻)

𝐿𝐿

0𝑑𝑑𝑥𝑥 =

𝐹𝐹𝐴𝐴𝐻𝐻2𝐿𝐿𝐸𝐸𝐼𝐼

−𝑤𝑤𝐻𝐻3𝐿𝐿

2𝐸𝐸𝐼𝐼#(14)

Adding (11) -(14), and plugging into (10),

𝐹𝐹𝐴𝐴𝐻𝐻3

3𝐸𝐸𝐼𝐼−𝑤𝑤𝐻𝐻4

8𝐸𝐸𝐼𝐼+𝑓𝑓𝑠𝑠𝐹𝐹𝐴𝐴𝐻𝐻𝐺𝐺𝐴𝐴

−𝑓𝑓𝑠𝑠𝑤𝑤𝐻𝐻2

2𝐺𝐺𝐴𝐴+𝐹𝐹𝐴𝐴𝐿𝐿𝐸𝐸𝐴𝐴

−𝑤𝑤𝐻𝐻𝐿𝐿𝐸𝐸𝐴𝐴

+𝐹𝐹𝐴𝐴𝐻𝐻2𝐿𝐿𝐸𝐸𝐼𝐼

−𝑤𝑤𝐻𝐻3𝐿𝐿

2𝐸𝐸𝐼𝐼= 0

⇒ 𝐹𝐹𝐴𝐴 �𝐻𝐻3

3𝐸𝐸𝐼𝐼+𝑓𝑓𝑠𝑠𝐻𝐻𝐺𝐺𝐴𝐴

+𝐿𝐿𝐸𝐸𝐴𝐴

+𝐻𝐻2𝐿𝐿𝐸𝐸𝐼𝐼

� =𝑤𝑤𝐻𝐻4

8𝐸𝐸𝐼𝐼+𝑓𝑓𝑠𝑠𝑤𝑤𝐻𝐻2

2𝐺𝐺𝐴𝐴+𝑤𝑤𝐻𝐻𝐿𝐿𝐸𝐸𝐴𝐴

+𝑤𝑤𝐻𝐻3𝐿𝐿

2𝐸𝐸𝐼𝐼

⇒ 𝐹𝐹𝐴𝐴 =�𝑤𝑤𝐻𝐻

4

8𝐸𝐸𝐸𝐸+ 𝑓𝑓𝑠𝑠𝑤𝑤𝐻𝐻2

2𝐺𝐺𝐴𝐴+ 𝑤𝑤𝐻𝐻𝐿𝐿

𝐸𝐸𝐴𝐴+ 𝑤𝑤𝐻𝐻3𝐿𝐿

2𝐸𝐸𝐸𝐸�

�𝐻𝐻3

3𝐸𝐸𝐸𝐸+ 𝑓𝑓𝑠𝑠𝐻𝐻

𝐺𝐺𝐴𝐴+ 𝐿𝐿

𝐸𝐸𝐴𝐴+ 𝐻𝐻2𝐿𝐿

𝐸𝐸𝐸𝐸�

From (1),

𝑁𝑁 = −𝐹𝐹𝐴𝐴 + 𝑤𝑤𝐻𝐻 = 𝑤𝑤𝐻𝐻 −�𝑤𝑤𝐻𝐻

4

8𝐸𝐸𝐸𝐸+ 𝑓𝑓𝑠𝑠𝑤𝑤𝐻𝐻2

2𝐺𝐺𝐴𝐴+ 𝑤𝑤𝐻𝐻𝐿𝐿

𝐸𝐸𝐴𝐴+ 𝑤𝑤𝐻𝐻3𝐿𝐿

2𝐸𝐸𝐸𝐸�

�𝐻𝐻3

3𝐸𝐸𝐸𝐸+ 𝑓𝑓𝑠𝑠𝐻𝐻

𝐺𝐺𝐴𝐴+ 𝐿𝐿

𝐸𝐸𝐴𝐴+ 𝐻𝐻2𝐿𝐿

𝐸𝐸𝐸𝐸�

From (3),

𝑀𝑀 = −𝐹𝐹𝐴𝐴𝐻𝐻 +𝑤𝑤𝐻𝐻2

2=𝑤𝑤𝐻𝐻2

2−𝐻𝐻 �𝑤𝑤𝐻𝐻

4

8𝐸𝐸𝐸𝐸+ 𝑓𝑓𝑠𝑠𝑤𝑤𝐻𝐻2

2𝐺𝐺𝐴𝐴+ 𝑤𝑤𝐻𝐻𝐿𝐿

𝐸𝐸𝐴𝐴+ 𝑤𝑤𝐻𝐻3𝐿𝐿

2𝐸𝐸𝐸𝐸�

�𝐻𝐻3

3𝐸𝐸𝐸𝐸+ 𝑓𝑓𝑠𝑠𝐻𝐻

𝐺𝐺𝐴𝐴+ 𝐿𝐿

𝐸𝐸𝐴𝐴+ 𝐻𝐻2𝐿𝐿

𝐸𝐸𝐸𝐸�

Problem 2 (25 points) SOLUTION

The closed, thin-walled tank shown above has an inner radius of r and wall thickness t and contains a gas of pressure of p. The state of stress at point “a” on the tank is represented by stress components shown above for an unknown stress element rotation angle of θ . Note that

τ ′x ′y is in the direction shown; however, its magnitude is also

unknown. For this problem, you are asked for the following (in no particular order):

• Determine the principal components of stress, σ1 and σ 2 .

• Determine the magnitude of τ ′x ′y .

• Draw the Mohr’s circle for this state of stress. Show the location of the x '− axis in your Mohr’s circle. From this, determine the rotation angle θ .

• Determine the absolute maximum shear stress for this state of stress.

xa

y

x

y

12 ksi

stressstateatpointa

′x

′y

θ = ?

18 ksi

τ ′x ′y = ?

r

t

tankcrosssec-on

Problem 2 (25 points):

The closed, thin-walled tank shown above has an inner radius of r and wall thickness t and contains a gas of pressure of p. The state of stress at point “a” on the tank is represented by stress components shown above for an unknown stress element rotation angle of θ . Note that

τ ′x ′y is in the direction shown; however, its magnitude is also

unknown. For this problem, you are asked for the following (in no particular order):

• Determine the principal components of stress, σ1 and σ 2 .

• Determine the magnitude of τ ′x ′y .

• Draw the Mohr’s circle for this state of stress. Show the location of the x '− axis in your Mohr’s circle. From this, determine the rotation angle θ .

• Determine the absolute maximum shear stress for this state of stress.

xa

y

x

y

12 ksi

stressstateatpointa

′x

′y

θ = ?

18 ksi

τ ′x ′y = ?

r

t

tankcrosssec-on

2θ = 180°− tan−1 4

15−12⎛⎝⎜

⎞⎠⎟= 126.9° ⇒ θ = 63.4°

AlsofromMohr’scircle:

τ

max,abs=σmax −σmin

2= 20

2= 10 ksi

σ1 = 20 =σmaxσ ave = 1512σ 2 = 10σmin = 0

2θ

x '− axis

4

σ

τ

Problem 4 (10 points):

Five points (i.e., A, B, C, D, and E) are marked on the structures shown above. The state of stress of each of these points can be represented by a Mohr’s circle and the Mohr’s circle can be used to determine principal stresses σ1 > σ2 > σ3.

Circle the state(s) of stress represented by the Mohr’s circle depicted below.

Page 11 of 11

Problem 4.5 (4 Points): The cantilever beam is loaded as shown below. Which one of the Mohr's circle shown here represents the stress element at point A and B?

Stress element at A

a) b) c) d)

Stress element at B

a) b) c) d)

Problem 4.6 (2 Points): The beam is loaded as shown below. Which one of the Mohr's circle shown here represents the stress element at point A?

a) b) c) d)

ME 323 Examination # 3 Name ___________________________________ (Print) (Last) (First) _____ ___, 2016 Instructor _______________________________

PART C - 8 points:

Beam (i) – Steel Beam (ii) – Aluminum

Beams (i) and (ii) shown above are identical, except that beam (i) is made up of steel and beam (ii) is made up of aluminum. Note that 𝐸𝐸𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 ≥ 𝐸𝐸𝑎𝑎𝑠𝑠𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎.

Let |𝜎𝜎|𝑎𝑎𝑎𝑎𝑚𝑚,(𝑎𝑎) and |𝜎𝜎|𝑎𝑎𝑎𝑎𝑚𝑚,(𝑎𝑎𝑎𝑎) represent the maximum magnitude of flexural stress in beams (i) and (ii), respectively. Circle the correct relationship between these two stresses:

a) |𝜎𝜎|𝑎𝑎𝑎𝑎𝑚𝑚,(𝑎𝑎) > |𝜎𝜎|𝑎𝑎𝑎𝑎𝑚𝑚,(𝑎𝑎𝑎𝑎)

b) |𝜎𝜎|𝑎𝑎𝑎𝑎𝑚𝑚,(𝑎𝑎) = |𝜎𝜎|𝑎𝑎𝑎𝑎𝑚𝑚,(𝑎𝑎𝑎𝑎)

c) |𝜎𝜎|𝑎𝑎𝑎𝑎𝑚𝑚,(𝑎𝑎) < |𝜎𝜎|𝑎𝑎𝑎𝑎𝑚𝑚,(𝑎𝑎𝑎𝑎)

Let |𝜏𝜏|𝑎𝑎𝑎𝑎𝑚𝑚,(𝑎𝑎) and |𝜏𝜏|𝑎𝑎𝑎𝑎𝑚𝑚,(𝑎𝑎𝑎𝑎) represent the maximum magnitude of the xy-component of shear stress in beams (i) and (ii), respectively. Circle the correct relationship between these two stresses:

a) |𝜏𝜏|𝑎𝑎𝑎𝑎𝑚𝑚,(𝑎𝑎) > |𝜏𝜏|𝑎𝑎𝑎𝑎𝑚𝑚,(𝑎𝑎𝑎𝑎)

b) |𝜏𝜏|𝑎𝑎𝑎𝑎𝑚𝑚,(𝑎𝑎) = |𝜏𝜏|𝑎𝑎𝑎𝑎𝑚𝑚,(𝑎𝑎𝑎𝑎)

c) |𝜏𝜏|𝑎𝑎𝑎𝑎𝑚𝑚,(𝑎𝑎) < |𝜏𝜏|𝑎𝑎𝑎𝑎𝑚𝑚,(𝑎𝑎𝑎𝑎)

Let |𝛿𝛿|𝑎𝑎𝑎𝑎𝑚𝑚,(𝑎𝑎) and |𝛿𝛿|𝑎𝑎𝑎𝑎𝑚𝑚,(𝑎𝑎𝑎𝑎) represent the maximum magnitude of deflection in beams (i) and (ii), respectively. Circle the correct relationship between these two stresses:

a) |𝛿𝛿|𝑎𝑎𝑎𝑎𝑚𝑚,(𝑎𝑎) > |𝛿𝛿|𝑎𝑎𝑎𝑎𝑚𝑚,(𝑎𝑎𝑎𝑎)

b) |𝛿𝛿|𝑎𝑎𝑎𝑎𝑚𝑚,(𝑎𝑎) = |𝛿𝛿|𝑎𝑎𝑎𝑎𝑚𝑚,(𝑎𝑎𝑎𝑎)

c) |𝛿𝛿|𝑎𝑎𝑎𝑎𝑚𝑚,(𝑎𝑎) < |𝛿𝛿|𝑎𝑎𝑎𝑎𝑚𝑚,(𝑎𝑎𝑎𝑎)

Let |𝐵𝐵𝑦𝑦|(𝑎𝑎) and |𝐵𝐵𝑦𝑦|(𝑎𝑎𝑎𝑎) represent the vertical reaction at B in beams (i) and (ii), respectively. Circle the correct relationship between these two stresses:

a) |𝐵𝐵𝑦𝑦|(𝑎𝑎) > |𝐵𝐵𝑦𝑦|(𝑎𝑎𝑎𝑎)

b) |𝐵𝐵𝑦𝑦|(𝑎𝑎) = |𝐵𝐵𝑦𝑦|(𝑎𝑎𝑎𝑎)

c) |𝐵𝐵𝑦𝑦|(𝑎𝑎) < |𝐵𝐵𝑦𝑦|(𝑎𝑎𝑎𝑎)