Question

please solve this problem clearly specially the shear-moment diagram , and clear photoType or paste question here

The roof is supported by joists that rest on two girders. Each joist can be considered simply supported, and the front girder can be considered attached to the three columns by a pin at A and rollers at B and C. Assume the roof will be made from 75 mm-thick cinder concrete, and each joist has a weight of 2.5 kN. According to code the roof will be subjected to a snow loading of 1.2 kN/m². The joists have a length of 8 m. Draw the shear and moment diagrams for the girder. Assume the supporting columns are rigid. | 1 m 1 m 1 m 1 m 1 m 1 m 1 m 1 m

Answer 1

Schution Given data, Thickness of slab (ts) = 75mm ; joist (W;)= 2.5 kN; weight of Snow load on roof = 1,2 kN/m; unit wt of the cinder Concrete = 200/m²; The length of joist = Bin; no. of joist on Girder =9; al Snow load on Girder por metor length . : (Ws) = 1,2x8 = 9.6 kN lm The load ... due to joist on girder per meter length = (9x22 / 0) wj .= 1.41 kN lm

. The Self weight of roof which is made with Cinder concrete = wr 0.075x8x. 1820 = 120/m. = 0.012. +.9.6+ 1.41 = 11.022 kulmi

- 11.0 2262/m l=3m . I de 50 12.399 kn-m 134.44 kw-m maximum bending moment in the span. AB = 11.022 *3 = 12.4 km-m 8 = 12.4 kw-m maximum bending moment in the span | Bọ - lios - 34-49 kem m Area of Bim diagram in span AB : a = 2/3 * 1243 99 *3 = 24.798 kn / m C.G of ai from end A = 3/2= 1.5m. Aorca Of Bom diagram in Span BC. - Q₂ = 2%*34.4445 = 114.8 kN/m

CLAPEYRON'S The olem The equation is; Made + 2mg (det de) + med + Garmin Garten 20 level and the same flexural rigidity (Ellia E2 I) . The beam is simply supported at A and a bence mA=M..=0; 2MB (le tl2) + Gain GQqk * Ime (3+5) +(6x sey 24.798 * 1.5) + 6x 114.8 x 25 mga 26.16kw -m. To find reactions, take moment about "B towards right hand side, MB = 11.022*5*5 - Rexs Rc = 22-32 kN.

To find reaction at. A take moment about B, toward left hand side, mg = 11-022*3* 3 - RAX3 RA = 7.813 kw RB = Total load -RA-Rc Rp = (11.022*8) - 7.8 13~ 28:32.. RB = 58.043 kn et B.m at mid point in span AB. = 12.4 – 28;16 = -0.68 kw-m. 2. . 34.8 - 26.16 - = 21.72 kw-m.

llo 022 kw lm - 3m X - --- too - 12.4 kom 34.8 korm 2G116 kw.m i . Bom diagram. . . [25.253 kn 122.39 |22.32 KN . 7.813w .: . 32.79 kn : Shear force diagram