1.) Volume of the top sheet = 5*0.08*0.005m^3 = 0.002 m^3
Volume of the both sheets = 2 * 0.002 m^3 = 0.004 m^3
Volume of vertical sheet = 10*10^(-3)*190*10^(-3)*5 = 0.0095m^3
Total volume of the top beam = Sum of the volumes two parallel sheets and vertical sheet
= (9.5+4)*10^(-3) m^3 = 13.5 * 10^(-3) m^3
2.) Now, In this part we would calculate the volume of the rectangular bars,
Height of the bar = 1.06 m
Breadth of the bar = 50 mm
width of the bar = 50 mm
, therefore the volume of this bar = 1.06*0.05*0.05 m^3
= 2.65 * 10^(-3) m^3
Since this bar is hollow from inside, we would calculate the volume of unoccupied space
= 1.06*0.04*0.04 m^3 ( the thickness is 5 mm)
= 1.69*10^(-3) m^3
The original volume occupied by HSLA steel = (2.65-1.69)*10^(-3) m^3
= 0.96*10^(-3) m^3
As the total no. of bars used in this gantry crane is 2, so multiplying the result by 2, we get
= 2*0.96 *10^(-3) m^3 = 1.92*10^(-3)m^3
3.) Here we will determine the amount of material used i.e. the volume of the four hollow legs used in this crane.
Height of the rectangular bar = 0.74m
Width of the bar = 50mm
breadth of the bar = 50mm
Volume = 0.74*0.05*0.05 m^3
= 1.85 * 10^(-3) m^3
Since the thickness of the steel is 5 mm, therefore in this case the breadth and width becomes 40 mm.
The volume of the space with these dimension = 0.74*0.04*0.04m^3
= 1.184*10^(-3) m^3
The volume of the steel used in this bar = (1.85-1.184)*10^(-3) m^3
= 0.666*10^(-3) m^3
Now, the number of these bar legs used = 4
The total volume of the HSLA steel used = 4*0.666*10(-3)m^3
=2.664*10^(-3) m^3
4.) Volume of the spheres used = 4*4/3*3.14*((0.15/2)^3) m^3
= 4*1.767*10^(-3) m^3
= 7.068*10^(-3) m^3
The material used for these spherical balls is volcanic rubber which would give the crane extra stability on uneven ground.
According to these calculations, the total volume of the HSLA steel used = 18.084*10^(-3) m^3
Volume of the both sheets = 2 * 0.002 m^3 = 0.004 m^3
Volume of vertical sheet = 10*10^(-3)*190*10^(-3)*5 = 0.0095m^3
Total volume of the top beam = Sum of the volumes two parallel sheets and vertical sheet
= (9.5+4)*10^(-3) m^3 = 13.5 * 10^(-3) m^3
2.) Now, In this part we would calculate the volume of the rectangular bars,
Height of the bar = 1.06 m
Breadth of the bar = 50 mm
width of the bar = 50 mm
, therefore the volume of this bar = 1.06*0.05*0.05 m^3
= 2.65 * 10^(-3) m^3
Since this bar is hollow from inside, we would calculate the volume of unoccupied space
= 1.06*0.04*0.04 m^3 ( the thickness is 5 mm)
= 1.69*10^(-3) m^3
The original volume occupied by HSLA steel = (2.65-1.69)*10^(-3) m^3
= 0.96*10^(-3) m^3
As the total no. of bars used in this gantry crane is 2, so multiplying the result by 2, we get
= 2*0.96 *10^(-3) m^3 = 1.92*10^(-3)m^3
3.) Here we will determine the amount of material used i.e. the volume of the four hollow legs used in this crane.
Height of the rectangular bar = 0.74m
Width of the bar = 50mm
breadth of the bar = 50mm
Volume = 0.74*0.05*0.05 m^3
= 1.85 * 10^(-3) m^3
Since the thickness of the steel is 5 mm, therefore in this case the breadth and width becomes 40 mm.
The volume of the space with these dimension = 0.74*0.04*0.04m^3
= 1.184*10^(-3) m^3
The volume of the steel used in this bar = (1.85-1.184)*10^(-3) m^3
= 0.666*10^(-3) m^3
Now, the number of these bar legs used = 4
The total volume of the HSLA steel used = 4*0.666*10(-3)m^3
=2.664*10^(-3) m^3
4.) Volume of the spheres used = 4*4/3*3.14*((0.15/2)^3) m^3
= 4*1.767*10^(-3) m^3
= 7.068*10^(-3) m^3
The material used for these spherical balls is volcanic rubber which would give the crane extra stability on uneven ground.
According to these calculations, the total volume of the HSLA steel used = 18.084*10^(-3) m^3
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