<p>17.1.1 Assignment No. 1</p><p>TASK 1 a) Explain briefly, the effects of gravitational force on the structural elements (walls, floors, roof, beams and columns) of a building. </p><p>Answer:</p><p>Every object is attracted towards the Earth due to the gravitational pull. The mass of an object can be determined from its density and volume, but for structural calculations we need to find the weight of the object, as part of the design load. As explained in chapter 3 of the book, the weight of an object is the force with which it is attracted towards the Earth. A structural element exerts dead load and imposed load on the building and both of them are due to the gravitational pull (or force). There are some exceptions like wind load, the effect of which is different. </p><p>Any object which is situated above the ground surface has to be supported otherwise it will fall due to the gravitational pull. For example, the roof of a building is supported on load- bearing walls. Similarly, the floors are supported on the walls, and the walls are supported on the foundation. Finally the foundation is supported on subsoil which is also known as the natural foundation.</p><p>The floor of a building would deflect under the action of dead load and imposed load, and as a consequence of the deflection the joists are subjected to compression and tension. The joists are also subjected to shear force due to the combined load (dead + imposed load). The roof trusses, which are the structural part of the roof structure, are subjected to tension and compression due to the combined load. The dead load of the walls creates compression in bricks, limiting the height of the walls; similarly in concrete columns compressive stress in concrete is created. b) Explain the effect of gravitational forces on the shape of structural elements like walls, columns and beams.</p><p>Answer:</p><p>Beams deflect under the combined effect of dead load and imposed load. Simply supported beams sag whereas cantilevers hog under the effect of forces. For more details refer to </p><p>Chapter 10 of the book - Figures 10.3 and 10.5.</p><p>The forces acting on walls and columns cause compression in these elements. Compression may cause the shortening of walls and columns. c) Explain how the structural elements of a building may fail as a result of overloading.</p><p>Answer:</p><p>Overloading in timber joists and reinforced concrete beams may cause failure due to excessive deflection and excessive shear force. This has been explained in Chapter 10, section 10.3. Overloading will create stress in a material which is more than its elastic limit, therefore, there is a permanent change in the shape. This may cause distortion in the shape of the building and could result in failure.</p><p>Overloading in walls may cause crushing of the bricks or blocks. Aerated concrete blocks are more vulnerable to failure as their strength is far less than that of bricks. d) A 6.0 m long concrete beam which is 200 mm × 400 mm in cross-section, is resting on two walls. Calculate the dead load of the beam if the density of concrete is 2400 kg/m3.</p><p>Answer:</p><p>Cross-sectional area of the beam = 0.2 m × 0.4 m = 0.08 m2</p><p>Volume = 6.0 m × 0.08 = 0.48 m2 </p><p>Mass = Density × Volume = 2400 × 0.48 = 1152 kg</p><p>Weight or dead load of the beam = Mass × g = 1152 × 9.81 = 11 301.12 N e) Calculate reactions R1 and R2 for the beam shown in Figure 17.1</p><p>Figure 17.1</p><p>Answer:</p><p>Calculation of reaction R1</p><p>R1 will cause clockwise moment, and both 3 kN and 5 kN forces will cause anti-</p><p> clockwise moments. </p><p>Clockwise moment = R1 × 6 m </p><p>Anti-clockwise moment due to the 3 kN force</p><p>= 3 kN × 4.5 m = 13.5 kNm</p><p>Anti-clockwise moment due to the 5 kN force</p><p>= 5 kN × 3 m = 15 kNm</p><p>Total anti-clockwise moment = 13.5 + 15 = 28.5 kNm</p><p>Moment due to R2 is zero as it acts at point B (see example 11.2 for explanation).</p><p>Clockwise moments must be equal to the anti-clockwise moments to maintain the </p><p> stability of the beam at point B:</p><p>R1 × 6 m = 28.5 kNm</p><p> or R1 = = 4.75 kN</p><p>R2 = (3 + 5) – 4.75 = 3.25 kN</p>