Answer: the allowable load P is 242.7877 kips
Explanation:
Given that;
diameter bolts d = 1.83 in
ultimate shear strength of the bolts = 60 ksi
we know that
shear area = 2×(π/4)d²
= 2×(π/4)×(1.83)² = 5.2604 in²
so
p/3(5.2604) = 60000/3.9
p/15.7812 = 15384.6153
p = 15.7812 × 15384.6153
p = 242787.691 lb
p = 242.7877 kips
therefore the allowable load P is 242.7877 kips
Answer: you can watch a video on how to solve this question on you tube
Answer: I don't agree with the colleague. Because the lab technician may be conducting the shear strength experiment properly.
The difference between the shear strength of the sample in the lab and the field may be due to all or any of the following:
1) Sample size.
2) Sample shape.
3) Sample proportion.
4) Multiple test run on the same sample.
Explanation:
1) The sample size is very important in determining the shear strength of a sample, because the cohesion of a sample will vary with respect to the size of the sample, if the proportion remains the same. Therefore the sample size used in the lab and the field may not be equal.
2) The sample shape tells the arrangements of the particle lattice in the sample, which determines how cohesive a sample can be. Therefore if the shape of the samples are different, the stress properties won't be the same.
3) The cohesion of the sample will increase as the proportion of the sample increases. Therefore if the sample proportion are not the same, then the stress properties of the Mohr circle won't be the same.
4) If at all exactly the same sample used in the lab was the one used in the field, this will reduce the shear strength of the sample, because a repeative experiment on shear strength of a sample will reduce the cohesion of the sample. Therefore the cohesion of the sample during the lab experiment will be more than the cohesion of the sample during the field experiment.
Answer:
i would throw it in a 90 degree angle to gain speed
Explanation: