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Elements of Civil Engineering and Mechanics 18CIV14/24, , MODULE -03, (Session-05), , Example 3.10 Analyze the truss shown in figure by the method of joints., Tabulate the result and indicate the nature of force in the truss., , ΣFx = 0, RAx = 0, ΣFy = 0,  RAy + RD – 20 – 10 = 0, RAy + RD = 30 kN, , …(i), , ΣMA = 0,  –RD × 9 + 10 × 6 + 20 × 3 = 0,  RD × 9 = 120, RD = 13.333 kN, , Page 1

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Elements of Civil Engineering and Mechanics 18CIV14/24, , Substituting the value of RD in equation (i), we get,,  RAy = 30 – 13.333 = 16.667 kN, RAy = 16.667 kN, , Consider a joint A, , Applying conditions of equilibrium, we get,, ΣFy = 0,  16.667 + FAB sin 45° = 0, FAB = –23.570 KN (C),, compression, , member AB is under, , ΣFx = 0,  FAE + FAB cos 45° = 0, FAE = 16.667 kN (T), , Page 2

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Elements of Civil Engineering and Mechanics 18CIV14/24, , Consider joint E, , Applying conditions of equilibrium, ΣFx = 0, FEF = 16.667 kN (T), ΣFy = 0, FBE = 0, , Consider joint B, , Page 3

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Elements of Civil Engineering and Mechanics 18CIV14/24, , Consider joint F, , ΣFx = 0,  FFD – 16.667 + 4.714 cos 45° = 0, FFD = 13.333 kN (T), , Sl. no., , Member, , 1., , AB, , 2., 3., 4., 5., 6., 7., 8., 9., , AE, BC, BE, BF, CD, CF, DF, EF, , Magnitude of force (N), , Nature of force, , 23.570, 16.667, 13.333, 0, 4.714, 18.856, 3.333, 13.333, 16.667, , Compression, Tension, Compression, —, Compression, Compression, Tension, Tension, Tension, , Page 5

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Elements of Civil Engineering and Mechanics 18CIV14/24, , Reference Books:, 1)ELEMENTS OF CIVIL ENGINEERING AND MECHANICS, , by, , S.S.Bhavikatti., 2)ELEMENTS, , OF, , CIVIL, , ENGINEERING, , AND, , ENGINEERING, , MECHANICS by M.N.Shesha Prakash and Ganesh B. Mogaveer., 3)ELEMENTS, , OF, , CIVIL, , ENGINEERING, , AND, , ENGINEERING, , MECHANICS by R.V.Raikar., , Page 6