Simply supported beam - bending moment, Mechanical Engineering

Simply supported beam -  Bending Moment:

Sketch the SFD and BMD for a simply supported beam of 15 m span loaded as illustrated in Figure

1782_Simply supported beam -  Bending Moment.png

Figure

Solution

Taking moment about A,

R B   × 15 - 1.5 × 6 × (9 + (6/2)) - (7.5 × 8) + 9 - ((½) × 6 × 3 × (2/3) × 6 ) = 0

RB  = 13 kN

R A   = (( ½) × 6 × 3?)+ 7.5 + (1.5 × 6) - RB = 25.5 - 13 = + 12.5 kN

Shear Force (Beginning from the Left End A)

SF at A,        FA  = + 12.5 kN

SF at C, F C   =+ 12.5 - ((½) × 6 × 3 ) = + 3.5 kN

SF just left of E,         FE  =+ 3.5 kN

SF just right of E,      FE  =+ 3.5 - 7.5 = - 4 kN

SF at F,           FF  =- 4 kN

SF just left of B,     FB  =- 4 - (1.5 × 6) = - 13 kN = Reaction at B.

 Bending Moment (Beginning from Right End B)

BM at B,         MB = 0

BM at F,

M F      =+ (13 × 6) - (1.5 × 6 × (6/2)   =+ 51 kN-m

BM at E,      M E         =+ (13 × 7) - 1.5 × 6 × (1 + (6 /2))  =+ 55 kN-m

BM just right of D, M D =+ (13 × 8) -1.5 × 6 × (2 + (6/2)) - 7.5 =+ 51.5 kN-m

BM just left of D, M D  = + 51.5 + 9 = + 60.5 kN-m

BM at C, M C  =+ (12.5 × 6) - (½) × 3 × 6 × ((1/3) × 6 )  =+ 57 kN-m

Posted Date: 1/22/2013 1:10:11 AM | Location : United States







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