Analyse the beam:
Analyse the beam illustrated in Figure and draw the SFD, BND & thrust diagram. Situated the point of contraflexure, if any.
Figure
Solution
Vertical component of 4 kN at C, = 4 × cos 60^{o} = 2 kN ( ↓ )
Horizontal component of 4 kN at C, = 4 × sin 60^{o} = 3.464 kN ( ← )
Vertical component of 2 kN at D, = 2 × cos 30^{o} = 1.732 kN ( ↓ )
Horizontal component of 2 kN at D, = 2 × sin 30^{o} = 1 kN ( → )
Taking moment around A,
R _{B} × 8 - (3 × 10) - 2 × 4 × (4 + (4/2)) - (1.732 × 3) - (2 × 1) = 0
R_{B} = 10.6495 kN
R_{A} = 2 + 1.732 + (2 × 4) + 3 - R_{B} = 14.732 - 10.6495 = 4.0825 kN
Shear Force (beginning from the Left End A)
SF at A, F_{A} = + 4.0825 kN
SF just left of C, F_{C} = + 4.0825 kN
SF just right of C, F_{C} = + 4.0825 - 2 = + 2.0825 kN
SF just left of D, F_{D} = + 2.0825 kN
SF just right of D, F_{D} = + 2.0825 - 1.732 = + 0.3505 kN
SF at E, F_{E} = + 0.3505 kN
SF just left of B, F_{B} = + 0.3505 - (2 × 4) = - 7.6495 kN
SF just right of B, F_{B} = - 7.6495 + 10.6495 = + 3 kN
SF just left of F, F_{F} = + 3 kN = load at the end F.
Bending Moment (beginning from the F)
BM at F, M_{F} = 0
BM at B, M _{B} = - (3 × 2) = - 6 kN-m
BM at E, M _{E} =+ (10.46 95 × 4) - (3 × 6) - ( 2 × 4 ×( 4/2)) =+ 8.599 kN-m
BM at D, BM at C,
M_{ D} = + (4.0825 × 3) - (2 × 2) = + 8.2475 kN-m
M _{C} = + (4.0825 × 1) = + 4.0825 kN-m
BM at A, M_{A} = 0
Maximum Bending Moment
Maximum bending moment shall occur at B and among B and E. Consider a section XX at a distance x from the end F.
F_{x} = - 10.6495 + 3 + 2 ( x - 2)
For maximum bending moment, F_{x} must be equal to zero.
- 10.6495 + 3 + 2x - 4 = 0
x = 5.82475 m ≈ 5.825 m
BM at section XX,
∴ M x = + 10.6495 ( x - 2) - 3x - 2 ( x - 2) × (( x - 2)/2)
= 10.6495 ( x - 2) - 3x - ( x - 2)^{2}
M _{max} = 10.6495 (5.825 - 2) - 3 (5.825) - (5.825 - 2)^{2}
=+ 8.629 kN-m
Maximum positive bending moment = + 8.629 kN-m
Maximum negative bending moment = - 6 kN-m.
Point of Contraflexure
Equating the BM at section XX, to zero.
10.6495 ( x - 2) - 3x - ( x - 2)^{2} = 0
or 10.6495 x - 21.299 - 3x - x^{2} + 4 x - 4 = 0
or x^{2} + 11.6495x - 25.299 = 0
or x^{2} - 11.6495x + 25.299 = 0
Solving out by trial and error, we obtain x = 2.9 m.
Point of contraflexure is at a distance of 2.9 m from the end F.
Thrust Diagram
Horizontal reaction at A,
+ H _{A} + 3.464 - 1 = 0
+ H _{A} + 2.464 = 0
∴ H _{A} =- 2.464 kN
('-' denote that the reaction is towards right)
The portion AC is subjected to a compressive force of 2.464 kN. The part CD is subjected to a tensile force of 1 kN (that means3.464 - 2.464 = 1).