I. Problem Solving: Answer what is being required. Show you complete solution. 1. Compression test was applied to a certain type of wood. Compute the stress based on the following dimensions. The grains are parallel with the length. а. Compression Test Dimension (mm) Load (N) Compressive stress (MPa) L = 50 B= 75 W = 75 Parallel to the grain 130,000 L = 50 Perpendicular to the Grain В %3 75 W = 75 35,000 b. What would be the shear stress for a single shear for the same data in a.? c. What is the modulus of rupture if the width of the specimen is 50mm, depth is 50mm and the effective length is 30cm. The specimen failed at 24KN

Structural Analysis
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ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
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I.
Problem Solving: Answer what is being required. Show you complete solution.
1. Compression test was applied to a certain type of wood. Compute the stress based on the
following dimensions. The grains are parallel with the length.
а.
Dimension (mm)
Load (N)
Compressive stress
(MPa)
Compression Test
L= 50
Parallel to the grain
В 3 75
130,000
W = 75
L = 50
Perpendicular to the
Grain
В - 75
W = 75
35,000
b. What would be the shear stress for a single shear for the same data in a.?
c. What is the modulus of rupture if the width of the specimen is 50mm, depth is
50mm and the effective length is 30cm. The specimen failed at 24KN
2. A moisture test was conducted and the following weights were acquired before and after
the wood was placed inside the oven for 24 hours.
Initial weight
40g
Final Weight
37g
Moisture content
Transcribed Image Text:I. Problem Solving: Answer what is being required. Show you complete solution. 1. Compression test was applied to a certain type of wood. Compute the stress based on the following dimensions. The grains are parallel with the length. а. Dimension (mm) Load (N) Compressive stress (MPa) Compression Test L= 50 Parallel to the grain В 3 75 130,000 W = 75 L = 50 Perpendicular to the Grain В - 75 W = 75 35,000 b. What would be the shear stress for a single shear for the same data in a.? c. What is the modulus of rupture if the width of the specimen is 50mm, depth is 50mm and the effective length is 30cm. The specimen failed at 24KN 2. A moisture test was conducted and the following weights were acquired before and after the wood was placed inside the oven for 24 hours. Initial weight 40g Final Weight 37g Moisture content
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