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

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Question: Exercise 21.1

1. 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]
2. 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]
3. 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]
4. 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]
5. 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]
6. 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°]

The solution of reaction of Beam of 1 to 6

Question 01: [Solution]

ΣM_A = 0 (Clockwise+)
⇒(C)+(D)+(E)+(B)=0
⇒(5×1)+(2×2)+(3×3)+(-R_Bx4)=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

Question 02: [Solution]

ΣM_A = 0 (Clockwise+)
⇒(2×3)x4.5-(R_Bx6)=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

Question 03: [Solution]

ΣM_A = 0 (Clockwise+)
⇒(2×1.5)x0.75+(2×1.5)+(2×3)x4.5+(4×4.5)-(R_Bx6)=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

Question 04: [Solution]

ΣM_A = 0 (Clockwise+)
⇒(1×6)x3-(R_Bx4)=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

Question 05: [Solution]

ΣM_B = 0 (Clockwise+)
⇒(9×3)x0+(3×4.5)x3.75+(5×2.7)-(R_Ex4.5)=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

Question 06: [Solution]

Summation of the Anticlockwise moment at point A

=R_Dx8
=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