Answer:
a) 159.07 MPa
b) 10.45 MPa
c) 79.535 MPa
Explanation:
Given data :
length of cantilever beam = 1.5m
outer width and height = 100 mm
wall thickness = 8mm
uniform load carried by beam along entire length= 6.5 kN/m
concentrated force at free end = 4kN
first we determine these values :
Mmax = ( 6.5 *(1.5) * (1.5/2) + 4 * 1.5 ) = 13312.5 N.m
Vmax = ( 6.5 * (1.5) + 4 ) = 13750 N
A) determine max bending stress
б =
=
= 159.07 MPa
B) Determine max transverse shear stress
attached below
ζ = 10.45 MPa
C) Determine max shear stress in the beam
This occurs at the top of the beam or at the centroidal axis
hence max stress in the beam = 159.07 / 2 = 79.535 MPa
attached below is the remaining solution
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Answer:
No
Explanation:
51 / 100 = 510 / 1000
Chance of getting a head is 1 / 2 of total throws
= 1 / 2 × 1000
= 500 is the probability
and the number of heads was just 10 more the the probability...if the was a greater gap, there would be evidence to say the coin is unfair
Answer:
Part a: The yield moment is 400 k.in.
Part b: The strain is 
Part c: The plastic moment is 600 ksi.
Explanation:
Part a:
As per bending equation

Here
- M is the moment which is to be calculated
- I is the moment of inertia given as

Here
- b is the breath given as 0.75"
- d is the depth which is given as 8"



The yield moment is 400 k.in.
Part b:
The strain is given as

The stress at the station 2" down from the top is estimated by ratio of triangles as

Now the steel has the elastic modulus of E=29000 ksi

So the strain is 
Part c:
For a rectangular shape the shape factor is given as 1.5.
Now the plastic moment is given as

The plastic moment is 600 ksi.
Take a look at the pictures that should help you out.