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Two solid cylindrical rods AB and BC are welded together at B and loaded as shown. Given: P = 15 kips.
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Section BreakDifficulty: Easy
1.
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Find the average normal stress at the midsection of rod AB.
The average normal stress at the midsection of rod AB is 22
Find the average normal stress at the midsection of rod AB.
The average normal stress at the midsection of rod AB is ksi.
Explanation:
Rod AB:
𝑃 = 15 kips + 12 kips = 27 kips
𝐴 =
𝜋
4
𝜎 𝐴𝐵 =
𝑑12 =
𝑃
𝐴
=
𝜋
4
2
(1.25 in.) = 1.22718 in 2
27 kips
1.22718 in 2
= 22.002 ksi
𝜎 𝐴𝐵 = 22 ksi
2.
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Find the average normal stress at the midsection of rod BC.
ksi.
The average normal stress at the midsection of rod BC is 33.94
ksi.
Find the average normal stress at the midsection of rod BC.
The average normal stress at the midsection of rod BC is ksi.
Explanation:
Rod BC:
𝑃 = 15 kips
𝐴 =
𝜋
4
𝜎𝐵𝐶 =
𝑑22 =
𝑃
𝐴
=
𝜋
4
2
(0.75 in.) = 0.44179 in 2
15 kips
0.44179 in 2
= 33.953 ksi
𝜎𝐵𝐶 = 33.95 ksi
Each of the four vertical links has an 8 × 36-mm uniform rectangular cross section, and each of the four pins has a 16mm diameter. Take P = 20 kN.
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Section BreakDifficulty: Medium
3.
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Determine the maximum value of the average normal stress in the links connecting points B and D.
The maximum value of the average normal stress in the links connecting points B and D is 101.6
Determine the maximum value of the average normal stress in the links connecting points B and D.
The maximum value of the average normal stress in the links connecting points B and D is MPa.
Explanation:
Use bar ABC as a free body.
MPa.
∑ 𝑀𝐶 = 0: (0.40 m)𝐹𝐵𝐷 − (0.25 m + 0.40 m)(20 × 10 3 N) = 0
𝐹 𝐵𝐷 = 32.5 × 10 3 N
Link BD is in tension.
Net area of one link for tension = 0.008 m0.036 m − 0.016 m = 160 × 10 −6 m 2
For two parallel links,
𝐴net = 320 × 10 −6 m 2
𝜎𝐵𝐷 =
𝐹𝐵𝐷
𝐴net
=
32.5 × 10 3 N
320 × 10 −6 m 2
= 101.6 × 10 6 Pa
𝜎𝐵𝐷 = 101.6 MPa
4.
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Determine the maximum value of the average normal stress in the links connecting points C and E. (Input the answer
with the appropriate sign.)
The maximum value of the average normal stress in the links connecting points C and E is -21.7
MPa.
Determine the maximum value of the average normal stress in the links connecting points C and E. (Input the answer
with the appropriate sign.)
The maximum value of the average normal stress in the links connecting points C and E is MPa.
Explanation:
Use bar ABC as a free body.
∑ 𝑀𝐵 = 0: − (0.40 m)𝐹𝐶𝐸 − (0.25 m)(20 × 10 3 N) = 0
𝐹𝐶𝐸 = -12.5 × 10 3 N
Link CE is in compression.
Area for one link in compression = 0.008 m0.036 m = 288 × 10 −6 m 2
For two parallel links,
𝐴net = 576 × 10 −6 m 2
𝜎𝐶𝐸 =
𝐹𝐶𝐸
𝐴net
=
-12.5 × 10 3 N
576 × 10 −6 m 2
= -21.7 × 10 6 Pa
𝜎𝐶𝐸 = -21.7 MPa
5.
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When the force P reached 1640 lb, the wooden specimen shown failed in shear along the surface indicated by the
dashed line. Determine the average shearing stress along that surface at the time of failure.
The average shearing stress is 911
psi.
When the force P reached 1640 lb, the wooden specimen shown failed in shear along the surface indicated by the
dashed line. Determine the average shearing stress along that surface at the time of failure.
The average shearing stress is psi.
Explanation:
Area being sheared: 𝐴 = 3 in. × 0.6 in. = 1.8 in 2
Force: 𝑃 = 1640 lb
Shearing stress:
𝜏 =
𝑃
𝐴
=
1640 lb
1.8 in 2
= 911 psi
𝜏 = 911 psi
A 32-kN axial load P is applied to a short wooden post that is supported by a square concrete footing resting on
undisturbed soil.
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Section BreakDifficulty: Easy
6.
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Determine the maximum bearing stress on the concrete footing.
The maximum bearing stress on the concrete footing is 2.67
MPa.
Determine the maximum bearing stress on the concrete footing.
The maximum bearing stress on the concrete footing is MPa.
Explanation:
Bearing stress on concrete footing:
𝑃 = 32 kN = 32 × 10 3 N
𝐴 = (100 mm)(120 mm) = 12 × 10 3 mm 2 = 12 × 10 −3 m 2
𝜎 =
𝑃
𝐴
=
32 × 10 3 N
12 × 10 −3 m 2
= 2.6667 × 10 6 Pa
𝜎 = 2.67 MPa
7.
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Determine the size of the footing for which the average bearing stress in the soil is 145 kPa.
The size of the footing is 469
mm.
Determine the size of the footing for which the average bearing stress in the soil is 145 kPa.
The size of the footing is mm.
Explanation:
Footing area:
𝑃 = 32 × 10 3 N
𝜎 = 145 kPa = 145 × 10 3 Pa
𝜎 =
𝐴 =
𝑃
𝐴
𝑃
𝜍
=
32 × 10 3 N
145 × 10 3 Pa
= 0.22069 m 2
Since the area is square,
𝐴 = 𝑏2
𝑏 =
=
√𝐴
√0.22069
m2
= 0.47 m
𝑏 = 470 mm
A force P of magnitude 950 N is applied to the pedal shown.
NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part.
References
Section BreakDifficulty: Medium
8.
Award: 10 out of 10.00 points
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Determine the diameter of the pin at C for which the average shearing stress in the pin is 40 MPa.
The diameter of the pin at C is 6.27
mm.
Determine the diameter of the pin at C for which the average shearing stress in the pin is 40 MPa.
The diameter of the pin at C is mm.
Explanation:
Since BCD is a three-force member, the reaction at C is directed toward E, the intersection of the lines of action of the
other two forces.
Using geometry,
𝐶𝐸 = √300 mm 2 + 125 mm 2 = 325 mm
Using the free-body diagram of BCD,
+↑𝛴𝐹𝑦 = 0:
𝜏pin =
1
𝐶
2
𝐴𝑃
125 mm
325 mm
=
1
𝐶
2
𝜋 2
𝑑
4
𝐶 − 𝑃 = 0 ⇒ 𝐶 = 2.6𝑃 = 2.6(950 N) = 2470 N
=
2𝐶
2
𝜋𝑑
⇒𝑑=
2𝐶
√ 𝜋𝜏pin
=
2(2470 N)
√ 𝜋(40×10 6 Pa)
= 6.2699 × 10 −3 m = 6.2699 mm
9.
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Determine the corresponding bearing stress in the pedal at C.
The corresponding bearing stress in the pedal at C is 43.77
MPa.
Determine the corresponding bearing stress in the pedal at C.
The corresponding bearing stress in the pedal at C is MPa.
Explanation:
𝜎𝑏 =
𝐶
𝐴𝑏
=
𝐶
𝑑𝑡
=
(2470 N)
(6.2699×10 −3 m)(9×10 −3 m)
= 43.7720 × 10 6 Pa = 43.7720 MPa
10.
Award: 10 out of 10.00 points
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Determine the corresponding bearing stress in each support bracket at C.
The corresponding bearing stress in each support bracket at C is 39.39
MPa.
Determine the corresponding bearing stress in each support bracket at C.
The corresponding bearing stress in each support bracket at C is MPa.
Explanation:
𝜎𝑏 =
1
𝐶
2
𝐴𝑏
=
𝐶
2𝑑𝑡𝑏
=
(2470 N)
2(6.2699×10 −3 m)(5×10 −3 m)
= 39.3948 × 10 6 Pa = 39.3948 MPa