Problem 2 (10pts): Compute for the net area of the plate with staggered fasteners considering the failure line shown in the figure below. ►| 2″ |◄ 3″′ ►le 3″′ ►¶ 3"″ ► ¶ 3″ ►¶ 3″►le—–— 2" + 4 이하 2" O O O 3/4-in.-diameter bolts -PL ¾ x12
Problem 2 (10pts): Compute for the net area of the plate with staggered fasteners considering the failure line shown in the figure below. ►| 2″ |◄ 3″′ ►le 3″′ ►¶ 3"″ ► ¶ 3″ ►¶ 3″►le—–— 2" + 4 이하 2" O O O 3/4-in.-diameter bolts -PL ¾ x12
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
100%
Answer Problem 2
Transcribed Image Text:8:57
CE497_R3 FE
Show complete solution and draw if necessary. Correct answer with incorrect or incomplete
solution will not be given points.
Problem 1 (20pts): A simply supported rectangular wooden beam has a span of 6 m. and carries a
total uniform load of 25 kN/m including its own weight. The beam is made up of 50% stress grade
Agoho. (Refer to table for the allowable stresses)
a. If the beam width is 200mm and beam depth is 400 mm. What is the maximum flexural stress
of the beam? Is it safe?
b. What should be the area of the beam that will not exceed the allowable shearing stress?
Species
1. High Strength Group
Agoho
Liusin
Malabayabas
Manggachapui
Molave
Narig
Sasalit
Yakal
2"
Table TMBR-2: Properties of Philippine Wood at 50% Grade
Bending and
Modulus of
Tension Parallel Elasticity in
Compression
Parallel to
to Grain
Bending
Grain
O
MPa
16.40
15.60
17.90
16.10
15.00
13.60
19.60
15.30
GPa
O
5.14
5.85
5.19
6.02
4.09
5.20
6.08
3.11
MPa
9.06
9.76
9.90
10.0
9.60
8.59
13-5
9.55
3/4-in.-diameter bolts
Compression
Perpendicular
to Grain
MPa
O
ASTM
Designation
5G 53
3.69
2.69
5-44
3-77
3-96
3.11
6.40
3-92
Problem 2 (10pts): Compute for the net area of the plate with staggered fasteners considering the
failure line shown in the figure below.
−| 2"″ |¶ 3″ ►¶ 3"″ ►◄ 3″ |¶ 3″ ►¶ 3″ ►| — |
Commit to the LORD whatever you do, and your plans will succeed.
Proverbs 16:3
God bless. Ⓒ
Shear Parallel
to Grain
MPa
1.84
1.65
1.89
1.74
1.80
1.63
2.12
Fy Fu
Yield Tensile
Stress Stress
(ksi) (ksi)
1.55
-PL ¾ x12
Problem 3 (20pts): Compute the nominal compressive strength of the member shown. Use the
theoretical value for K (Refer to table C-A-7.1).
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