Draw SF and B.M diagrum for a beam shown in fig
50 KN
30 KR LOKN
A
7
A
T
ktnk
am ke
mundna
B
с
E
kom
Silm
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Question :
Draw the SF and BM diagram for the beam as shown in fig Also indicate the principal values on the diagrams.
Answer :
Let reaction at support A be RAV RAH and Manti clock wise First find the support reaction For that ∑V = 0 RAV –2 –3 –3 = 0 RAV = 8 ...1Taking moment about point A ∑MA = 0 –M + 2 X 1 + 3 X 3 + 3 X 5 = 0 M = 26KNm ...2 ∑H = 0 RAH = 0 ...3Calculation for the Shear force Diagram Draw the section line here total 3 section line which break the load RAV and 2KNBetween Point A and B 2KN and 3KNBetween Point B and C 3KN and 3KN Between Point C and D. Consider left portion of the beam Consider section 1-1Force on left of section 1-1 is RAV SF1-1 = 8KN constant valueConstant value means value of shear force at both nearest point of the section is equal i.e.SFA = SFB = 8KN ...4 Consider section 2-2 Forces on left of section 2-2 is RAV & 2KN SF2-2 = 8 – 2 = 6KN constant value Constant value means value of shear force at both nearest point of the section is equal i.e. SFB = SFC = 6KN ...5 Consider section 3-3 Forces on left of section 3-3 is RA 2KN 3KN SF3-3 = 8 – 2 – 3 = 3KN constant value Constant value means value of shear force at both nearest point of the section is equal i.e.SFC = SFD = 3KN ...6 Calculation for the Bending moment Diagram Let Distance of section 1-1 from point A is X1 Distance of section 2-2 from point A is X2 Distance of section 3-3 from point A is X3 Consider section 1-1 taking moment about section 1-1 BM1-1 = 8.X1 It is Equation of straight line Y = mX + C inclined linearInclined linear means value of bending moment at both nearest point of the section is varies with X1 = 0to X1 = 1 At X1 = 0 BMA = 0 ...8 At X1 = 1 BMB = 8 ...9 i.e. inclined line 0 to 8 Consider section 2-2 taking moment about section 2-2BM2-2 = 8.X2 – 2.X2 – 1 = 6.X2 + 2It is Equation of straight line Y = mX + C inclined linear.Inclined linear means value of Bending moment at both nearest point of the section is varies with X2 = 1 to X2 = 3At X2 = 1 BMB = 8 ...10 At X2 = 3 BMC = 20 ...11 i.e. inclined line 8 to 20Consider section 3-3 taking moment about section 3-3BM3-3 = 8.X3 – 2.X3 – 1 – 3.X3 – 3= 3.X3 + 11 It is Equation of straight line Y = mX + C inclined linear. Inclined linear means value of Bending moment at both nearest point of the section is varies withX3 = 3 to X3 = 5 At X3 = 3 BMC = 20 ...12 At X3 = 5 BMD = 26 ...13 i.e. inclined line 20 to 26 Plot the BMD with the help of above bending moment values. The SFD and BMD is shown in fig