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ME 323: Mechanics of Materials Homework Set 1

Fall 2019 Due: Wednesday, August 28

Problem 1.1 (10 points)

Rod ABC has a circular cross section with diameter 𝒅 and is fixed to a wall at A as shown in Fig. 1.1. The rod has radius of curvature 𝑹, where 𝑹/𝒅 > 10. Two forces 𝑷 and 𝑸 are applied at the end C of the rod. Determine the internal resultant forces and moments as a function of 𝜽 acting at the centroid (which is coincident with the mean radius) on the cross section of the bar at B.

Fig. 1.1

Solution

Free Body Diagram (FBD)

Making a cut at B and inspecting the lower section of the beam,ie, BC, the FBD can be constructed as shown in Fig. 1A.

From FBD the unknowns are: 𝑵(𝜽), 𝑽(𝜽), 𝑴(𝜽)

Drawing a normal to the line ODC through B, we have two right triangles OBD and BDC. Then the lengths:

𝐿 = 𝑅𝑐𝑜𝑠(𝜃)

𝐿 = 𝑅𝑠𝑖𝑛(𝜃)

𝐿 = 𝐿 − 𝐿 = 𝑅 − 𝑅𝑐𝑜𝑠(𝜃)

Fig. 1A:FBD of cut section

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Equilibrium equations

Force balance in the tangential direction 𝑡:

Σ𝐹 = 0 => 𝑁(𝜃 ) − 𝑃 cos(𝜃) − 𝑄𝑠𝑖𝑛(𝜃) = 0

=> 𝑁(𝜃 ) = 𝑃 cos(𝜃) + 𝑄𝑠𝑖𝑛(𝜃)

[1.1]

Force balance in the radial direction 𝑟 :

Σ𝐹 = 0 => 𝑉(𝜃 ) − 𝑃 sin(𝜃) + 𝑄𝑐𝑜𝑠(𝜃) = 0

=> 𝑉(𝜃 ) = 𝑃 sin(𝜃) − 𝑄𝑐𝑜𝑠(𝜃)

[1.2]

Moment balance about B:

Σ𝑀 = 0 => −𝑀(𝜃 ) − (𝑅 − 𝑅 cos(𝜃)) 𝑃 + 𝑅𝑠𝑖𝑛(𝜃)𝑄 = 0

=> 𝑀(𝜃 ) = −𝑃𝑅 1 − 𝑐𝑜𝑠(𝜃) + 𝑄𝑅 sin(𝜃) [1.3]

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Problem 1.2 (10 points)

Fig. 1.2 represents a frame ABCD with forces 𝑷 and 𝑾 applied at the cross section D. Calculate the reaction forces and moments developed at the wall A for the structure.

Fig. 1.2

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Solution

Free Body Diagram (FBD)

Figure 2A

From the FBD, the forces acting at the point D can be written as 𝑭 = 𝑃 ̂ − 𝑊 𝒌

The position vector 𝒓𝑨𝑫 = 2𝐿 ̂ + 𝐿 ̂ − 𝐿 𝒌

Equilibrium equations

Force balance (vector approach)

Σ𝐹 = 0 => 𝐹 ̂ + 𝐹 ̂ + 𝐹 𝒌 + 𝑭 = 0

=> 𝐹 ̂ + 𝐹 ̂ + 𝐹 𝒌 + 𝑃 ̂ − 𝑊 𝒌 = 0

Equating the components,

[2.1]

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𝐹 = −𝑃

𝐹 = 0

𝐹 = 𝑊

Moment balance about the wall A

𝛴𝑴 = 0 => 𝑀 ̂ + 𝑀 ̂ + 𝑀 𝒌 + 𝒓𝑨𝑫 x 𝑭 = 0

=> 𝑀 ̂ + 𝑀 ̂ + 𝑀 𝒌 + 2𝐿 ̂ + 𝐿 ̂ − 𝐿 𝒌 x 𝑃 ̂ − 𝑊 𝒌 = 0

𝑀 ̂ + 𝑀 ̂ + 𝑀 𝒌 + −𝑊𝐿 ̂ + (2𝑊𝐿 − 𝑃𝐿) ̂ − 𝑃𝐿 𝒌 = 0

Equating the components,

𝑀 = 𝑊𝐿

𝑀 = 𝑃𝐿 − 2𝑊𝐿

𝑀 = 𝑃𝐿

[2.2]

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Problem 1.3 (10 points)

A bar ABC is loaded by a compressive force 𝑷 acting at an angle 𝜽 as shown in Fig. 1.3. The bar ABC and the link BD have the same cross sectional area 𝑨. If the link BD can only support a maximum axial stress of 𝝈, determine the magnitude of the largest allowable force 𝑷 when:

1. 𝜃 = 0 2. 𝜃 = 90 3. 𝜃 = 30

Fig. 1.3

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Solution

Free Body Diagram (FBD)

Member BD is supported by pinned joints at both ends and hence is a 2 force member. Hence the FBDs can be drawn as

Fig. 3A: FBD of link BD Fig. 3B: FBD of link ABC

Unknowns: 𝑅 (x and y components), and 𝐹

Since we are only interested in the force 𝐹 , this can be calculated with moment balance about the point A, thus eliminating the need for calculating 𝑅 .

For the moment balance, consider the triangles ABD and ACE as shown Figure 2B.

Using the similar triangles approach

𝐿

𝐿=

𝐿

𝐿=

𝐿

𝐿

which gives

𝐿 =𝐿

𝐿 𝐿 =

3𝐿

5𝐿 7𝐿 =

21

5𝐿

[3.1]

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𝐿 =𝐿

𝐿 𝐿 =

4𝐿

5𝐿 7𝐿 =

28

5𝐿

Moment balance about the A gives

𝛴𝑀 = 0 => −𝐿 𝐹 − 𝐿 𝑃𝑠𝑖𝑛(𝜃) + 𝐿 𝑃𝑐𝑜𝑠(𝜃) = 0 [3.2]

=> (4𝐿)𝐹 +28

5𝐿 𝑃𝑠𝑖𝑛(𝜃) −

21

5𝐿 𝑃𝑐𝑜𝑠(𝜃) = 0

[3.3]

Since the bar BD has a cross sectional area 𝐴 and maximum axial stress of 𝜎, the maximum force it can withstand is

=> 𝐹 = 𝜎𝐴 [3.4]

Using equations [3.3] and [3.4] the force 𝑃 can be expressed as

𝑃 =20 𝜎𝐴

21 𝑐𝑜𝑠(𝜃) − 28 sin(𝜃)

[3.5]

Case 1: 𝜃 = 0

𝑃 =20

21 𝜎𝐴 = 0.9524 𝜎𝐴

The bar experiences a tensile load.

[3.6]

Case 2: 𝜃 = 90

𝑃 = −20

28 𝜎𝐴 = −

5

7𝜎𝐴 = − 0.7143 𝜎𝐴

Now the load is compressive!

[3.7]

Case 3: 𝜃 = 30

𝑃 =20

21√32

− 2812

𝜎𝐴 = 4.7772 𝜎𝐴

Interestingly, the force P that can be applied now is significantly higher!

[3.8]

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Problem 1.4 (10 points)

Circular bar 1 is attached to the wall by brackets 2 and 3 as shown in Fig 1.4. Bracket 3 rests on a pinned joint and has negligible thickness. Bracket 2 is attached to bracket 3 by bolt assemblies. Bar 1 is made of a material which can withstand a maximum normal stress 𝜎 = 10 𝑀𝑃𝑎 and the bolted connections are made of material which can withstand a maximum shear stress 𝜏 =160 𝑀𝑃𝑎. To avoid the failure of the structure, determine:

1. The smallest diameter of the bar AB. 2. The smallest diameters of the bolts at the joint C and D.

Fig. 1.4

Solution

Free Body Diagram (FBD)

Making a cut at pin joint F, there is only one unknown force 𝐹 , so the assembly is determinate.

Fig. 4A:FBD of assembly

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ΣF = 0 => −5000 − 4000 + 3000 + 𝐹 = 0

𝐹 = 6000 𝑁

[4.1]

Part 1: Smallest diameter of the bar AB

Consider the cut 𝑋𝑋’ in the bar AB

Fig. 4B: FBD of AB at cut XX’

From the figure, the maximum stress the bar can withstand is

𝜎 =5000

𝜋𝑑 4

𝑀𝑃𝑎

Given maximum normal stress 𝜎 = 10 𝑀𝑃𝑎, the limiting condition is

𝜎 =5000

𝜋𝑑 4

= 𝜎 = 10 𝑀𝑃𝑎

Which gives 𝑑 = 25.2313 𝑚𝑚

To be safe, we round off to a higher number hence

𝑑 = 25.24 𝑚𝑚

Part 2: Smallest diameters of the bolts at the joint C and D

Consider the cut YY on member 3

Fig. 4C: FBD of link 3 at cut YY’

The member 3 is symmetric about the horizontal axis. Hence forces acitng on the cut sections are equal (=𝐹 ). Using force balance

ΣF = 0 => −2𝐹 + 3000 + 𝐹 = 0

which gives

𝐹 =4500 N

Consider the pin joint at C. The joint is in single shear due to the geometry of the design.

Hence the effective shear stress is

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𝜏 =𝑉

𝐴=

𝐹

𝜋𝑑4

Since the max shear cannot exceed 𝜏 = 160 𝑀𝑃𝑎, the minimum diameter of the bolt at C should be

𝑑 =4 x 4500

160𝜋= 5.9841 𝑚𝑚

To be safe in design, we round off to the higher number,ie, 𝑑 = 5.99 𝑚𝑚.

Now for the pin joint at D. The joint is in double shear.

Hence the effective shear stress is

𝜏 =𝑉

2𝐴=

𝐹

𝜋𝑑2

Since the max shear cannot exceed 𝜏 = 160 𝑀𝑃𝑎, the minimum diameter of the bolt at D should be

𝑑 =2 x 4500

160𝜋= 4.2314 𝑚𝑚

To be safe we round off to the higher number,ie, 𝑑 = 4.24 𝑚𝑚.