Answer:
P=361.91 KN
Explanation:
given data:
brackets and head of the screw are made of material with T_fail=120 Mpa
safety factor is F.S=2.5
maximum value of force P=??
<em>solution:</em>
to find the shear stress
T_allow=T_fail/F.S
=120 Mpa/2.5
=48 Mpa
we know that,
V=P
<u>Area for shear head:</u>
A(head)=π×d×t
=π×0.04×0.075
=0.003×πm^2
<u>Area for plate:</u>
A(plate)=π×d×t
=π×0.08×0.03
=0.0024×πm^2
now we have to find shear stress for both head and plate
<u>For head:</u>
T_allow=V/A(head)
48 Mpa=P/0.003×π ..(V=P)
P =48 Mpa×0.003×π
=452.16 KN
<u>For plate:</u>
T_allow=V/A(plate)
48 Mpa=P/0.0024×π ..(V=P)
P =48 Mpa×0.0024×π
=361.91 KN
the boundary load is obtained as the minimum value of force P for all three cases. so the solution is
P=361.91 KN
note:
find the attached pic
Answer:
a) 0.3
b) 3.6 mm
Explanation:
Given
Length of the pads, l = 200 mm = 0.2 m
Width of the pads, b = 150 mm = 0.15 m
Thickness of the pads, t = 12 mm = 0.012 m
Force on the rubber, P = 15 kN
Shear modulus on the rubber, G = 830 GPa
The average shear strain can be gotten by
τ(average) = (P/2) / bl
τ(average) = (15/2) / (0.15 * 0.2)
τ(average) = 7.5 / 0.03
τ(average) = 250 kPa
γ(average) = τ(average) / G
γ(average) = 250 kPa / 830 kPa
γ(average) = 0.3
horizontal displacement,
δ = γ(average) * t
δ = 0.3 * 12
δ = 3.6 mm
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Answer:
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