Problems on support reaction.pdf
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Engineering mechanics subject Equilibrium chapter problems solved on BEAM REACTION.
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Problems on support reaction.pdf
- 1. Problems on Q1. Calculate the support reactions for the beam shown in fig ANS : Step 1: Consider Step 2: Consider Step 3: Consider 𝑅𝐴 + Step 4: Consider Prepared by: Prof. V.V. Nalawade 1 Problems on Support Reactions Calculate the support reactions for the beam shown in fig : Consider FBD of beam : Consider ƩFx Ʃ𝐹 = 0 As there are no horizontal forces : Consider ƩFy Ʃ𝐹 = 0 3 + 4 + 5 − 𝑅𝐴 − 𝑅𝐵 = 0 𝑅𝐴 + 𝑅𝐵 = 3 + 4 + 5 + 𝑅𝐵 = 12 𝐾𝑁 … … … … … … … … … … : Consider ƩM@A Prepared by: Prof. V.V. Nalawade Support Reactions Calculate the support reactions for the beam shown in fig. As there are no horizontal forces . 𝐸𝑞 1
- 2. (3 Put the value of RB in Eq Step 5: Final Answer Reaction at support A = Reaction at support B = Q2. Calculate the support reactions for the beam shown in fig ANS : Step 1: Consider Step 2: Consider Prepared by: Prof. V.V. Nalawade 2 ƩM = 0 (3 × 2) + (4 × 3) + (5 × 5) − RB × 6 (3 × 2) + (4 × 3) + (5 × 5) = RB × RB × 6 = 6 + 12 + 25 RB = 43 6 = 7.17 KN Put the value of RB in Eqn 1, we get RA = 12-7.17 = 4.83 KN Final Answer Reaction at support A = RA = 4.83 KN Reaction at support B = RB = 7.17 KN Calculate the support reactions for the beam shown in fig : Consider FBD of beam : Consider ƩFx Prepared by: Prof. V.V. Nalawade = 0 6 4.83 KN RB = 7.17 KN Calculate the support reactions for the beam shown in fig.
- 3. Prepared by: Prof. V.V. Nalawade 3 Ʃ𝐹 = 0 As there are no horizontal forces Step 3: Consider ƩFy Ʃ𝐹 = 0 𝑅𝐴 + 𝑅𝐵 = 4 + (2 × 1.5) + 1.5 𝑅𝐴 + 𝑅𝐵 = 8.5 𝐾𝑁 … … … … … … … … … … . 𝐸𝑞 1 Step 4: Consider ƩM@A ƩM = 0 (4 × 1.5) + (3 × 2.25) + (1.5 × 4.5) = RB × 6 RB × 6 = 6 + 6.75 + 6.75 RB = 19.5 6 = 3.25 KN Put the value of RB in Eqn 1, we get RA = 8.5 – 3.25 = 5.25 KN Step 5: Final Answer Reaction at support A = RA = 5.25 KN Reaction at support B = RB = 3.25 KN Q3. A simply supported beam AB of Span 4.5 m is loaded as shown in fig. Find the support reactions at A & B.
- 4. ANS : Step 1: Consider Step 2: Consider Step 3: Consider 𝑅𝐴 + Step 4: Consider Put the value of RB in Eq Step 5: Final Answer Reaction at support A = Reaction at support B = Prepared by: Prof. V.V. Nalawade 4 : Consider FBD of beam : Consider ƩFx Ʃ𝐹 = 0 As there are no horizontal forces : Consider ƩFy Ʃ𝐹 = 0 𝑅𝐴 + 𝑅𝐵 = 4.5 + 2.25 + 𝑅𝐵 = 6.75 𝐾𝑁 … … … … … … … … … … : Consider ƩM@A ƩM = 0 (4.5 × 2.25) + (2.25 × 3) = RB × 4. RB = 16.88 4.5 = 3.75 KN Put the value of RB in Eqn 1, we get RA = 6.75 – 3.75 = 3 KN Final Answer Reaction at support A = RA = 3 Reaction at support B = RB = 3.75 Prepared by: Prof. V.V. Nalawade there are no horizontal forces … . 𝐸𝑞 1 .5 KN 3.75 KN
- 5. Q4. Calculate the support reactions for the beam shown in fig ANS : Step 1: Consider Step 2: Consider As there are no horizontal Step 3: Consider 𝑉𝐴 + Step 4: Consider (120 × 3 Prepared by: Prof. V.V. Nalawade 5 Calculate the support reactions for the beam shown in fig : Consider FBD of beam : Consider ƩFx Ʃ𝐹 = 0 𝐻𝐴 = 0 As there are no horizontal forces acting on beam : Consider ƩFy Ʃ𝐹 = 0 𝑉𝐴 + 𝑅𝐵 = 120 + 30 + 90 + 𝑅𝐵 = 240 𝐾𝑁 … … … … … … … … … … : Consider ƩM@A ƩM = 0 3) + (30 × 6) + (40) + (90 × 8.67) = Prepared by: Prof. V.V. Nalawade Calculate the support reactions for the beam shown in fig. forces acting on beam … . 𝐸𝑞 1 ) = RB × 10
- 6. Put the value of RB in Eq Step 5: Final Answer Horizontal Reaction at support A = H Vertical Reaction at support A = V Reaction at support B = Q5. Find analytically the support reactions at B and the load P, for the beam shown in fig. If the reaction of support A is Zero ANS : Step 1: Consider Step 2: Consider Prepared by: Prof. V.V. Nalawade 6 Put the value of RB in Eqn 1, we get VA = 240-136.03 = 103.97 KN Final Answer Horizontal Reaction at support A = H Vertical Reaction at support A = VA = 103.97 Reaction at support B = RB = 136.03 Find analytically the support reactions at B and the load P, for shown in fig. If the reaction of support A is Zero : Consider FBD of beam : Consider ƩFx Prepared by: Prof. V.V. Nalawade Horizontal Reaction at support A = HA = 0 103.97 KN 136.03 KN Find analytically the support reactions at B and the load P, for shown in fig. If the reaction of support A is Zero
- 7. Prepared by: Prof. V.V. Nalawade 7 Ʃ𝐹 = 0 𝐻𝐴 = 0 As there are no horizontal forces acting on beam Step 3: Consider ƩFy Ʃ𝐹 = 0 𝑉𝐴 + 𝑅𝐵 = 10 + 36 + 𝑃 0 + 𝑅𝐵 − 𝑃 = 46 𝐾𝑁 … … … … … … … … … … 𝑎𝑠 𝑉𝐴 = 0 𝑅𝐵 − 𝑃 = 46 𝐾𝑁 … … … … … … … … … … . 𝐸𝑞 1 Step 4: Consider ƩM@A ƩM = 0 (10 × 2) − 20 + (36 × 5) − (P × 7) = RB × 6 6RB − 7P = 220… … … … … … … … … … … … 𝐸𝑞 2 Solving Eqn 1 & 2, we get Putin Eqn 1, we get Step 5: Final Answer Reaction at support B = RB = 102 KN Load = P = 56 KN
- 8. Q6. Find the support reactions at A and B for the beam loaded as shown in fig. ANS : Step 1: Consider Step 2: Consider 𝐻𝐴 Step 3: Consider 𝑉𝐴 + 𝑅𝐵 Step 4: Consider Prepared by: Prof. V.V. Nalawade 8 Find the support reactions at A and B for the beam loaded as : Consider FBD of beam : Consider ƩFx Ʃ𝐹 = 0 𝐻𝐴 − 𝑅𝐵 𝐶𝑂𝑆 60 = 0 … … … … … … 𝐸𝑞 : Consider ƩFy Ʃ𝐹 = 0 𝑉𝐴 + 𝑅𝐵𝑠𝑖𝑛 60 = 120 + 90 + 80 𝑅𝐵 𝑠𝑖𝑛60 = 290 𝐾𝑁 … … … … … … … … … : Consider ƩM@A ƩM = 0 Prepared by: Prof. V.V. Nalawade Find the support reactions at A and B for the beam loaded as 𝐸𝑞 1 … … . 𝐸𝑞 2
- 9. (120 × Put in Eqn 1 & 2 Step 5: Final Answer Horizontal Reaction at support A = H Vertical Reaction at support A = V Reaction at support B = Q7. Find the support reactions at A and F for the beam loaded as shown in fig. ANS : Step 1: Consider FBD of beam AC FBD of beam DF Prepared by: Prof. V.V. Nalawade 9 × 5) + (90 × 6) + (80 × 13) = RB sin & 2, we get Final Answer Horizontal Reaction at support A = HA = Vertical Reaction at support A = VA = Reaction at support B = RB = 251.72 Find the support reactions at A and F for the beam loaded as : Consider FBD of beam FBD of beam AC DF Prepared by: Prof. V.V. Nalawade sin60 × 10 A = 125.86 KN A = 72 KN 251.72 KN Find the support reactions at A and F for the beam loaded as
- 10. Consider FBD of beam DF first, Step 2: Consider Step 3: Consider 𝑉𝐹 Step 4: Consider Put in Eqn 1, we get Now Consider Step 5: Consider Prepared by: Prof. V.V. Nalawade 10 Consider FBD of beam DF first, : Consider ƩFx Ʃ𝐹 = 0 𝐻𝐹 = 0 : Consider ƩFy Ʃ𝐹 = 0 + 𝑅𝐷 = 120 … … … … … … … … … … . 𝐸𝑞 : Consider ƩM@D ƩM = 0 (120 × 3) − (VF × 4) = 0 1, we get Consider the FBD of beam AC, : Consider ƩFx Ʃ𝐹 = 0 𝐻𝐴 + 2 cos 59.04 = 0 Prepared by: Prof. V.V. Nalawade 𝐸𝑞 1
- 11. Prepared by: Prof. V.V. Nalawade 11 𝐻𝐴 = −1.03 𝐾𝑁 i.e. our assumed direction is wrong, Therefore 𝐻𝐴 = 1.03 𝐾𝑁 (←) Step 6: Consider ƩFy Ʃ𝐹 = 0 𝑉𝐴 − 𝑅𝐶 = 2 sin 59.04 𝑉𝐴 = 1.71 + 30 = 31.41 𝐾𝑁 (↑) Step 7: Consider ƩM@A −M − 2 sin 59.04 × 1 − (30 × 2) = 0 Step 8: Final Answer Horizontal Reaction at support A = HA = 1.03 KN Vertical Reaction at support A = VA = 31.41 KN Moment at support A = MA = 61.71 KN.m Horizontal Reaction at support F = HF = 0 KN Vertical Reaction at support F = VF = 90 KN Reaction at support D = RD = 30 KN Q8. Two beams AB & CD are arranged as shown in fig. Find the support reactions at D.
- 12. ANS : Step 1: Consider FBD of beam AB FBD of beam Prepared by: Prof. V.V. Nalawade 12 : Consider FBD of beam FBD of beam AB FBD of beam CD Prepared by: Prof. V.V. Nalawade
- 13. Step 2: Consider ( Step 3: Consider Prepared by: Prof. V.V. Nalawade 13 : Consider FBD of AB & Consider ƩM@A ƩM = 0 (600 × 4) + 2400 = RB sin 53.13 × 12 Consider FBD of CD & Consider ƩM@C ƩM = 0 RD sin 36.87 × 10 − 500 × 4.2 = 0 Prepared by: Prof. V.V. Nalawade M@A 12 M@C 0
- 14. Q9. Prepared by: Prof. V.V. Nalawade 14 Prepared by: Prof. V.V. Nalawade
- 15. Prepared by: Prof. V.V. Nalawade 15 Q10. Q11.