Question

3. Details of a one-way slab are illustrated in Figure 3.9(1). Based on the load combination formula, ultimate load 1.2g + 1.59, compute the uniformly distributed live load (9) that may be carried by the slab. Take / = 25 MPa and pw = 24 kN/m². (Hint: take a typical strip 1000 mm wide and analyse it as a simply supported beam.) Simple support a 5500 A N16 @ 200 mm O 125 100 1 O 200 Section a-a

Answer 1

Given data! Ultimate load fé = 1.2 g +1.59 = 25 mpa Pw = 24 kN/m² length 5500mm 160 mm d 125 mm. N16 @ 200 mm spacing. say fy=420 mpa. Step! Calculation of Area of steel per meter length. T/4x16² X100 As= 200 As= 1005.3 mm 2 Step 2 To calculate Location of neutral axis (C) if CET As fy = 0.85.feb.a Asfy a Cheight of stress block) 0.85€6.b (1005:31) x 4p 20 a = 0.85 X 25 XIDO 19.86 m {B, = 0.85 for fa=25mpal a = a NiA (C)= 0.85 C- 19.86 0.85 c = 23.37 mm Step 3: Mn = Calculation of T (d-1/2) o Asty (d-/) 0.9% 1005.3 y 420 (125- 23.73 ?) 42.99x106 N-mm こ 2 42.99 KN-m Step 4: calculation of ultimate load. As slab es Simply supported, wa Lultimate load) C mumma C ME Well (for simly support) 5500mm 8 42.99= Wax 5.52 8 We = 11.37 KN m steps: Wself (9) - self weight a SW x DXD = 24(Km2) x 0.16X1 9= 3.84 (ICN m2) 4 UDL (9) calculation Step 6 Given Wu= 1.29 + 1.59 11.37=11.243.84) + 1.59 1.59 = 6.762 q=4.51 KN/m2 carried slab by (9) 2 4.51 UDL me