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Structural Mechanics 5 REACTIONS, SFD,BMD – with multiple loads 20 kN 1.50m2.00m RARA RBRB 30 kN10 kN 0.50m1.00m.

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Presentation on theme: "Structural Mechanics 5 REACTIONS, SFD,BMD – with multiple loads 20 kN 1.50m2.00m RARA RBRB 30 kN10 kN 0.50m1.00m."— Presentation transcript:

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2 Structural Mechanics 5 REACTIONS, SFD,BMD – with multiple loads 20 kN 1.50m2.00m RARA RBRB 30 kN10 kN 0.50m1.00m

3 Sum of moments about R A = 0 R B x ( ) = 10 x (1.50) + 20 x( ) + 30x( ) R B x (5.0) = 10 x (1.5) + 20x(2.0) + 30x(4) R B = {10 x (1.5) + 20x(2.0) +30x(4)} / 5.0 R B = { } / 5.0 R B = {175} / 5.0 R B = 35kN 20 kN RARA RBRB 30 kN10 kN 1.50m2.00m0.50m1.00m + 20 x( )10 x (1.50)+ 30x( )- R B x ( )= 0

4 Sum of moments about R B = 0 R A x ( ) - 10 x ( ) - 20 x( ) - 30x( 1.0) R A x (5.0) = 10 x (3.5) + 20x(3.0) + 30x(1.0) R A = { } / 5.0 R A = {125} / 5.0 R A = 25kN

5 CHECK VERTICAL FORCES UPWARD REACTION = R B + R A = = 60 kN DOWNWARD FORCE = =60 kN So likely that calculations are correct!

6 SHEAR FORCE DIAGRAM Space diagram SHEAR FORCE DIAGRAM 25kN 35kN 0 25kN =0 20 kN R A = R B = 30 kN10 kN = -5kN = -35kN =15kN

7 BENDING MOMENT DIAGRAM Space diagram 45.0kNm 0kNm 20 kN30 kN10 kN R A =R B =25kN35kN 1.50m2.00m0.50m1.00m 37.5kNm 35.0kNm Section 1 Section 2Section 3

8 BMD CALCULATIONS Section 1 25kN [1.5]m = 37.5 kNm Section 2 25kN [ ]m -10kN[0.5]m = 45.0kNm (NB max BM where shear force crosses the zero axis) Section 3 25kN [ ]m -10kN [ ]m-20kN[2.0]m = 35.0kNm Check: (taking BM from the RHS) 35kN [1.0]m = 35.0kNm


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