MASS DENSITY AND VOLUME

AREAS OF SHAPES RECTANGLE = L * B CIRCLE = ((π* D2)/4) TRIANGE =(0.5 B* H) B D H B

2) CIRCULAR ((π* D2)/4) * H) VOLUMES OF SHAPES 1) RECTANGULAR L*B*H 2) CIRCULAR ((π* D2)/4) * H) 3 TRIANGULAR (0.5 B* H * L)

THE CONCEPT OF MASS , DENSITY AND LOADING DENSITY IS THE MASS OF AN OBJECT / m3 MASS =VOLUME * DENSITY =Kg LOAD= KG * 9.81 = NEWTONS LOAD IN KN = NEWTONS /1000 =KN

DENSITY DIFFERENT MATERIALS HAVE DIFFERENT DENSITIES AND DIFFERENT MASSES / VOLUME WATER =1000 Kg / m3 CONCRETE =2400Kg / m3 STEEL =6800Kg / m3 WOOD =1650 Kg / m3

Rectangular shapes

SECTIONAL VIEWS OF A BLOCK

A SIMPLE EXAMPLE VOLUME =L*B*H 1*1*2.5 = 2.5m3 MASS = V* DENSITY WORK OUT THE FORCE THAT THE BLOCK SHOWN BELOW EXERTS DUE TO ITS MASS AND THE GRAVITATIONAL PULLOF THE EARTH IF IT HAS THE SAME DENSITY AS WATER VOLUME =L*B*H 1*1*2.5 = 2.5m3 MASS = V* DENSITY 2.5m3 *1000kgs =2500kgs LOAD = MASS*9.81 (2500* 9.81)/1000 =24.53KN

WORK OUT THE PRESSURE EXERTED ON THE GROUND BY THE SOLID CONCRETE BLOCK

rectangular Area = L*B 10*5 = 50m2 Volume L*B*H 10*5 *2 =100m3 Mass = Volume * Density 100*2400 = 240000kgs Load = (Mass *9.81 )/1000=Kn (240000*9.81)/1000= 2354.4 Kn Pressure = F/A 2354.4/ 50= 47.088 Kp

WORK OUT THE PRESSURE EXERTED ON THE GROUND BY THE CONCRETE TANK

Tank Volume L*B*H concrete (10*5 *2)- ( 9*4*1.5)=46m3 Mass concrete 46 *2400 = 110400 kgs Load (110400*9.81)/1000=1083.02 kn Pressure 1083.02/ 50 =21.66 kpa

WORK OUT THE PRESSURE EXERTED ON THE GROUND BY THE CONCRETE TANK FILLED WITH WATER

Tank filled with water Volume L*B*H concrete Mass concrete 46 *2400 = 110400 kgs Volume water ( 9*4*1.5)=54 m3 Mass of water 54*1000=54000 kgs Load ((110400+54000)*9.81)/1000=1612.76 Pressure = 1612.76/ 50 = 32.26

FORCES THAT STRUCTURES EXERT ON THE GROUND FROM THE DRAWING OF A CONCRETE FOUNDATION WORK OUT THE FOLLOWING (DENSITY = 2400Kg/m3) ITS VOLUME ITS MASS ITS LOAD

MODEL ANSWER Vol = (1.5* 0.4*0.6) – (1 *0.6 *0.2)+ ( 0.4*0.6*0.25)= 0.3m3 Mass = 0.3 *2400= 720 kgs Load = ( 720*9.81)/ 1000 = 7.06 kn

Cylinders

SECTIONAL VIEWS OF A CYCLINDER THREE DIMENSIONAL SECTIONAL PLAN

Work out the load Of the structure due To gravity ( P 2300) 0.5m hole ((π* 6.22)/4)- ((π* 52)/4) *0.5 Volume =5.3m3 Mass = 5.3*2300=12190 kg Load = (12190*9.81)/1000 = 120 Kn

Diameter 3.5m

Total vol concrete = 84.9- 65.34 = 19.56m3 A circular tank is to be filled with sand density 1234 find the pressure the tank exerts on the ground A) when empty B) when fill of sand xsa cylinder = 28.3m2 vol = 28.3*3=84.9 xsa hole = 23.76m2 vol = 23.76*2.75=65.34 Total vol concrete = 84.9- 65.34 = 19.56m3 Mass = 19.56 * 2400 = 46944 kgs Load =( 46944*9.81)/ 1000= 460.52 kn Pressure = 460.52 / 28.3 = 16.27 kpa Mass sand = 65.34 *1234 = 80629.56 Load = (80629.56*9.81)/1000=790.97 kn Pressure = (460.52+790.97)/ 28.3=44.22kpa

What is the mass of The structure What is the pressure On the ground. Density of concrete = 2400kg/m3 Empty When cylinders half filled with sand P = 1340 kg/m3

Rec V= 15 *8 *1 = 120 m 3 CIR V= (((π * 3 2 /4 )-(π * 1.5 2 /)4 ))*2 ) * 2) = 21.21 m 3 MASS CON = 141.21* 2400 = 338904 kgs LOAD = (338904 *9.81)/1000= 3324.65 kn PRESSURE = 3324.65 /(8*15 ) = 27.71 SAND V = (π * 1.5 2 /)4 )*2 = 3.53m 3 LOAD = (3.53 *1340 * 9.81 )/1000= 46.4 KN TOTAL PRESSURE = (3324.65+46.4) /(8*15 )= 28.1

Plan What upward load can the anchor Resist What pressure does the structure Exert on the ground

Sectional view Density soil =1859 kg / m3 Density con =2358 kg / m3 Natural ground

Volume of base = (π* 182)/4)*2)-(π* 142)/4)*1.5)= 278.03 Model answer Volume of base = (π* 182)/4)*2)-(π* 142)/4)*1.5)= 278.03 Volume of column = (6*6*2 ) – (4*3*2) = 48 Volume of sand = (4*3*2) =24 Total volume = 278.03+48 = 326.03 Mass empty = 326.03 *2358 =768778.74 Mass sand = 24 * 1859 = 44616 Total mass = 813394.74 kgs Load = (813394.74 *9.81) /1000 = 7979.40 kn Pressure = 7979.40)/(π* 142)/4)=51.83

Examples A concrete footing consists of a circular base and a hollow square column. The base has a diameter of 3,0 m and is 500 mm thick. The external dimensions of the 1,5 m high column are 500 mm x 500 mm with a wall thickness of 100 mm. Concrete has a density of 2 400 kg/m3 and the footing is loaded as shown. Calculate the pressure exerted by the concrete footing on the ground below.

3d view

QUESTION ONE (12 marks) Vbase = x 32 x 0,5 = 3,534 m3 (1 mark) Vcol = [(0,5 x 0,5) – (0,3 x 0,3)] x 1,5 = 0,24 m3 (1 mark) Vtot = 3,774 m3 Wconc = 2400 x 3,774 x 9,81 / 1000 = 88,856 kN (2 marks) Total downward load = 88,856 + 10 sin43 + 8 = 103,676 kN (2 marks) Area = x 32 = 7,069 m2 (2 marks) Pressure = Force / Area = 103,676 / 7,069 = 14,666 kPa (2 marks)

Example 2 The drawing BELOW shows the plan view and sectional view of a circular concrete tank if the density of concrete and water is 2345 kg/ m3 and 1000kg/ m3 respectively calculate the following. The resultant load of the structure the pressure exerted by the tank on the ground when it is filled with water

4.2m 4.2m

Vol cylinder = (( π* 4.62)/4 )*2.6)) = 43.21 Vol hole = (( π* 4.22)/4 )*2.5)) = 34.61 Vol concrete = 43.21 – 34.64 = 8.57 Load =( 8.57 * 2345 * 9.81 )/1000 = 197. 15 kn Vol w = 34.64 Load = 34.64 *9.81 = 339.82 Tot load = 197.15 + 339.82 = 536.97 Pressure 536.97 /(( π* 4.62)/4 )= 16.62 kpa