B.M And S.F Diagram Simply Supported Beam with Three Points Load
B.M And S.F Diagram Simply Supported Beam |Simply Supported Beam Three Points Load
B.M And S.F Diagram Simply Supported Beam with Three Points Load
Today I will discuss the topic of bending moment and shear force. what is the bending moment | What is the Shear Force |B.M And S.F Diagram Simply Supported Beam with Three Points Load| B.M And S.F Diagram Simply Supported Beam |Simply Supported Beam Three Points Load.
What is the bending moment?
the bending moment may be a force commonly measured during a force x length (e.g. kNm). Bending moments occur once a force is applied at a given distance off from a degree of reference; inflicting a bending result. within the simplest terms, a bending moment is essentially a force that causes one thing to bend. If the thing isn’t well-restrained the bending force can cause the thing to rotate for a few sure purposes. it would even be value noting you’ll experiment and check out our free calculator to calculate the shear and bending moment diagrams of a beam.
What is the Shear Force?
Shear force is the force performing on a substance in a direction perpendicular to the extension of the substance, acting in a direction to a flat cross-section of a body.
Shear force is that the summation of the result of shear stress over a surface and infrequently leads to shear strain. Bending moment and shear force calculations ar essential whereas planning any structural members
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EXAMPLE
We have the one simply supported beam with the three-point load F1=10kN, F2=30kN, and F3=10kN. The distance between two Reaction Force and span =10m and distance between F1 to F2= 2m and F2 to F3=1m.calculte the bending moment and shear force diagram and calculate the Reaction of RA And R.B.
Given data
F1=C1=10KN
F2=C2=30KN
F3=C3=10KN
The span of the beam =10m
Solution
We know
The sum of upward force = The sum of downward force
R.A+R.B = F1+F2+F3
R.A+R.B = 10+30+10
R.A+R.B = 50KN———————- 01
Sum of bending moment at A = 0
R.B x 10-10×6.5-30×5.5-10×3.5=0
10R.B-65-165-35=0
10R.B=65+195+35
10R.B = 265
10 divided by both sides for calculate the value the Rb
10R.B= 265/10
R.B = 26.5 KN
From equation 01
R.A+R.B = 50KN
R.A+26.5 =50
R.A= 50-26.5
R.A = 23.5 KN
Now Fist we calculate the shear force
SFB = -26.5 KN
SFc3 = -26.5+10 = -16.5KN
SFc2 = -16.5+30 = 13.5 KN
SFc1 = 13.5+10 = 23.5KN
SFA= 23.5 KN
Now we calculate the bending moment
BM.b = 0 KN.M
BM.C3 = RB X distance (from RB to C3)
BM.C3 = 26.5 x 3.5 = 92.75KN.M
BM.C2 = RB X distance RB to C2 – F3 X distance F3 to C2
BM.C2 =26.5 X 4.5 -10 X 1= 119.25 – 109.25= KN.M
BM.C1 = RB X distance RB to C1– F3 X distance F3 to C1- F2 X distance F2 to C1
BM.C1 =26.5 X 6.5 – 10 X 3 – 30 X 2 = 159 – 30 – 60 = 69 KN.M
BM.A = RB X distance RB to A– F3 X distance F3 to A- F2 X distance F2 to A –F1 X distance F1 to Point A
BM.A = 26.5 X 10 – 10 X 6.5 – 30 X 5.5 – 10 X 3.5 = 265 – 65 -165 – 35= 0
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