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1
3
2
8
Link BD consists of a single bar 1 in. wide and in. thick. Each pin has a -in. diameter and P = 8 kips.
References
Section BreakDifficulty: Medium
1.
Award: 10 out of 10.00 points
You did not receive full credit for this question in a previous attempt
Determine the maximum value of the average normal stress in link BD if θ = 0.
The maximum magnitude of the average normal stress is 22.17
ksi. This stress is in tension
Determine the maximum value of the average normal stress in link BD if θ = 0.
The maximum magnitude of the average normal stress is ksi. This stress is in tension
Explanation:
Use bar ABC as a free body.
𝜃 = 0
+↺ ∑ 𝑀 𝐴 = 0: (18 sin30°)(8 kips) − (12 cos30°)𝐹 𝐵𝐷 = 0
.
.
𝐹 𝐵𝐷 = 6.9282 kips (Tension)
Area for tension loading:
𝐴 = (𝑏 − 𝑑)𝑡 = (1 −
3
8
1
)( ) in 2 = 0.31250 in 2
2
Stress:
𝜎 =
𝐹𝐵𝐷
𝐴
=
6.9282 kips
0.31250 in 2
𝜎 = 22.17 ksi
This stress is in tension.
2.
Award: 10 out of 10.00 points
You received credit for this question in a previous attempt
Determine the maximum value of the average normal stress in link BD if θ = 90°.
The maximum magnitude of the average normal stress is 24
ksi. This stress is in compression
Determine the maximum value of the average normal stress in link BD if θ = 90°.
The maximum magnitude of the average normal stress is ksi. This stress is in compression
Explanation:
Use bar ABC as a free body.
𝜃 = 90°
+↺ ∑ 𝑀 𝐴 = 0: − (18 cos30°)(8 kips) − (12 cos30°)𝐹 𝐵𝐷 = 0
𝐹 𝐵𝐷 = -12 kips (Compression)
Area for compression loading:
1
𝐴 = 𝑏𝑡 = (1)( ) in 2 = 0.5 in 2
2
Stress:
.
.
𝜎 =
𝐹𝐵𝐷
𝐴
=
-12 kips
0.5 in 2
𝜎 = 24 ksi
This stress is in compression.
3.
Award: 10 out of 10.00 points
You received credit for this question in a previous attempt
The 1.6-kip load P is supported by two wooden members of uniform cross section that are joined by the simple glued scarf splice
shown. Determine the normal and shearing stresses in the glued splice.
The normal stress is 80
The shearing stress is 46.2
psi.
psi.
The 1.6-kip load P is supported by two wooden members of uniform cross section that are joined by the simple glued scarf splice
shown. Determine the normal and shearing stresses in the glued splice.
The normal stress is psi.
The shearing stress is psi.
Explanation:
𝑃 = 1600 lb
𝜃 = 90° − 60° = 30°
𝐴0 = (5.0 in.)(3.0 in.) = 15 in 2
𝜎 =
𝑃 cos 2 𝜃
𝐴0
=
(1600 lb)cos 2 30°
15 in 2
𝜎 = 80 psi
𝜏 =
𝑃 sin 2𝜃
2𝐴0
=
(1600 lb)(sin 60°)
(2)(15 in 2)
𝜏 = 46.2 psi
Two wooden members of uniform cross section are joined by the simple scarf splice shown. The maximum allowable tensile
stress in the glued splice is 83 psi.
NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part.
References
Section BreakDifficulty: Easy
4.
Award: 10 out of 10.00 points
You did not receive full credit for this question in a previous attempt
Determine the largest load P that can be safely supported.
The largest load P that can be safely supported is 1.66
Determine the largest load P that can be safely supported.
The largest load P that can be safely supported is kips.
Explanation:
kips.
𝐴0 = (5.0 in.)(3.0 in.) = 15 in 2
𝜃 = 90° − 60° = 30°
𝜎 =
𝑃 =
𝑃 cos 2 𝜃
𝐴0
𝜍𝐴0
cos 2 𝜃
=
(83 psi)(15 in 2)
cos 2 30°
= 1660 lb
𝑃 = 1.66 kips
5.
Award: 10 out of 10.00 points
You received credit for this question in a previous attempt
Determine the corresponding shearing stress in the splice.
The corresponding shearing stress in the splice is 47.9
Determine the corresponding shearing stress in the splice.
The corresponding shearing stress in the splice is psi.
Explanation:
𝜏 =
𝑃 sin 2𝜃
2𝐴0
𝜏 = 47.9psi
=
(1660 lb)(sin 60°)
(2)(15 in 2)
psi.