Here is the calculation of the **Reaction of Beam** for simply-supported-beam | Mechanics Solution | Ex-12.1.1

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- A simply supported beam AB of span 4 m is carrying a point loads of 5, 2 and 3 kN at 1, 2

and 3 m respectively from the support A. Calculate the reactions at the supports A and B.

[Ans. 5.5 kN and 4.5 kN] - A simply supported beam of span 6 m is carrying a uniformly distributed load of 2 kN/m

over a length of 3 m from the right end B. Calculate the support reactions.[Ans. RA = 1.5 kN, RB = 4.5 kN] - A simply supported beam AB of span 6 m is loaded as shown in Fig. 12.14. Determine the reactions at A and B. [Ans. 6.875 kN, 9.125 kN]
- A beam AB 6 m long rests on two supports 4 m apart, the right hand end is overhanging by 2 m. The beam carries a uniformly distributed load of 1 kN/m over the entire length of the beam. Determine the reactions at the two supports. [Ans. RA = 1.5 kN, RB = 4.5 kN]
- A beam ABCDEF of 7.5 m long and span 4.5 m is supported at B and E. The beam is loaded as shown in Fig. 12.15. Find graphically, or otherwise, the support reactions at the two supports. [Ans. RB = 29.33 kN, RE = 12.57 kN]
- A beam ABCDE hinged at A and supported on rollers at D, is loaded as shown in Fig. 12.16. Find the reactions at A and D. [Ans. RA = 5.94 kN, RD = 7.125 kN, θ = 61°]

ΣM_{A} = 0 (Clockwise+)

⇒(C)+(D)+(E)+(B)=0

⇒(5×1)+(2×2)+(3×3)+(-R_{B}x4)=0

⇒5+4+9-4R_{B}=0

⇒4R_{B}=18

⇒R_{B}=4.5 kN

ΣV = 0 (Upward+)

⇒(A)+(C)+(D)+(E)+(B)=0

⇒R_{A}-5-2-3+R_{B}=0

⇒R_{A}-10+4.5=0

⇒R_{A}=5.5kN

**The answer to question 1:** The reaction at the supports A=5.5kN, and B=4.5 kN

ΣM_{A} = 0 (Clockwise+)

⇒(2×3)x4.5-(R_{B}x6)=0

⇒R_{B}=4.5 kN

ΣV = 0 (Upward+)

⇒R_{A}-(2×3)+R_{B}=0

⇒R_{A}=1.5kN

**The answer to question 2:** The reaction at the supports A=1.5kN, and B=4.5 kN

ΣM_{A} = 0 (Clockwise+)

⇒(2×1.5)x0.75+(2×1.5)+(2×3)x4.5+(4×4.5)-(R_{B}x6)=0

⇒R_{B}=9.125 kN

ΣV = 0 (Upward+)

⇒R_{A}-(2×1.5)-2-(2×3)-5+R_{B}=0

⇒R_{A}=6.875kN

**The answer to question 3:** The reaction at the supports A=6.875kN, and B=9.125 kN

ΣM_{A} = 0 (Clockwise+)

⇒(1×6)x3-(R_{B}x4)=0

⇒R_{B}=4.5 kN

ΣV = 0 (Upward+)

⇒R_{A}-(1×6)+R_{B}=0

⇒R_{A}=1.5kN

**The answer to question 4:** The reaction at the supports A=1.5kN, and B=4.5 kN

Find Reaction of Overhang Beam

ΣM_{B} = 0 (Clockwise+)

⇒(9×3)x0+(3×4.5)x3.75+(5×2.7)-(R_{E}x4.5)=0

⇒R_{E}=14.25 kN

ΣV = 0 (Upward+)

⇒R_{B}-(9×3)-(3×4.5)-5+R_{E}=0

⇒R_{B}=31.25kN

**The answer to question 5:** The reaction at the supports B=31.25kN, and E=14.25 kN

Summation of the Anticlockwise moment at point A

=R_{D}x8

=8R_{D} _________________(i)

Summation of the Clockwise moment at point A

=6sin30x2+(4×1)x6+3×9

=57kN _______________(ii)

Now, equating the anticlockwise and clockwise moment equation (i) & (ii)

⇒8R_{D}=57

⇒R_{D}=7.125kN

Now the,

ΣV = 0 (Upward+)

⇒R_{A}-6sin30-(4×1)+R_{D}-3=0

⇒R_{A}=2.875kN

**The answer to question 6:** The reaction at the supports A=2.87kN, and D=7.125 kN