Answer:
a) Δd(change in wood diameter) = 5%
b) The wood would swell since the moisture content is increasing which will also led to increase in the wood's diameter
C) new diameter (D2) = 10.5 in
Explanation:
Wood pole diameter = 10 inches
moisture content = 5%
FSP = 30%
A) The percentage change in the wood's diameter
note : moisture fluctuations from 5% to 30% causes dimensional changes in the wood but above 30% up to 55% causes no change. hence this formula can be used to calculate percentage change in the wood's diameter
Δd/d = 1/5(30 - 5)
Δd/d = 5%
Δd = 5%
B) would the wood swell or shrink
The wood would swell since the moisture content is increasing which will also led to increase in the wood's diameter
C) The new diameter of the wood
D2 = D + D( 
)
D = initial diameter= 10 in , M1 = initial moisture content = 5%
therefore D2 = 10 + 10( 5/100 )
new diameter (D2) = 10.5 in
4.75cm * 5.22cm is 24.795cm²
but let's do this with m² instead:
0.0475m * 0.0522m = 0.0024795
now we can compare it with the 49780 much easier.
devide. that's the <u>roughly</u> 20,000,000fold of the plot, but it's still squared, so let's take the root
so the representative fraction is 1:4,480.
the exact value is in the screens
have a nice day:)
Answer:
13.6mm
Explanation:
We consider diameter to be a chord that runs through the center point of the circle. It is considered as the longest possible chord of any circle. The center of a circle is the midpoint of its diameter. That is, it divides it into two equal parts, each of which is a radius of the circle. The radius is half the diameter.
See attachment for the step by step solution of the problem
Answer:
σ =5.39Mpa
Explanation:
step one:
The flexure strength is defined as the tendency with which unreinforced concrete yield to bending forces
Flexural strength test Flexural strength is calculated using the equation:
σ = FL/ (bd^2 )----------1
Where
σ = Flexural strength of concrete in Mpa
F= Failure load (in N).
L= Effective span of the beam
b= Breadth of the beam
step two:
Given data
F=40.45 kN= 40450N
b=0.15m
d=0.15m
L=0.45m
step three:
substituting into the expression we have
σ = 40450*0.45/ (0.15*0.15^2 )
σ =18202.5/ (0.15*0.15^2 )
σ =18202.5/ (0.15*0.0225 )
σ =18202.5/0.003375
σ =5393333.3
σ =5393333.3/1000000
σ =5.39Mpa
Therefore the flexure strength of the concrete is 5.39Mpa
