Shear Force and Bending Moment Diagrams
Dr Jonathan Leaver
Structural Joints
Source: Wei Loo
Structural Joints (cont.)
Source: Wei Loo
Determining Reactions on a Beam An object that is not accelerating (e.g. stationary or moving at constant
velocity) must have zero net force acting on it. e.g. ΣFx = 0, ΣFx = 0, ΣFx = 0
If an object is not rotating it must have zero net moment acting on it. Consider the following beam: Loads (downwards) are forces imposed on the beam Reactions (upwards) result from the imposed loads.
Determining Reactions on a Beam (cont.) To determine the reactions. Follow these steps: 1. Draw free body diagram (i.e. replace supports with forces and moments as applicable) 2. Replace any UDL’s with a single force acting at the midpoint of the UDL. 3. Find the right hand reaction by taking moments about the point of action of the left hand reaction. 4. Find the left hand reaction by adding all the forces together for the x and y directions and putting these equal to zero.
Example 1
Find the reactions.
Calculate reactions Sum moment about A:
6kN*3m + 10kN*5m – RD*10m = 0 therefore RD = 6.8 kN
Sum forces:
RA – 6kN -10kN +6.8 = 0 therefore RD = 9.2 kN
Example 2
1) Determine reactions ΣMA = 0 (Clockwise +ve) 2kN*1m - RB*3m + (2kN/m*1m)*2.5m + 4kN*2m = 0 RB = 5kN ΣF = 0 gives RA – 2kN + 5kN – 4kN – 2kN/m*1m = 0 RA = 3 kN
Examples (a) (a)
(b)
(c)
Find the reactions.
Solutions Reactions: (a) (a)
(b)
2.857 kN (left support), 7.143 kN . No moments.
50 kN (left support), 150 kN. No moments.
150 kN, -410 kN-m
(c)
Example 45o
RA
RB
Find the reactions RA and RB. RAX = 4.24 kN, RAY = -0.621 kN, RB = 7.86 kN
Solution
Source: Wei Loo
Example Find the reactions and moment at A.
3 kN/m
4 kN-m
A
1.5 m
2m
1m
Solution 1) Draw free body diagram RAY MA
3 kN/m
RAx
1.5 m
2m
2) ΣMA = 0 MA +3 kN/m*2m*(1.5+1) – 4 kN-m = 0
4 kN-m
1m
Notice how the distance of the 4 kN-m moment from A does not enter the equation. Hence the influence of the moment is independent of location.
MA = - 11 kN-m
3) ΣFAY = 0 RAY – 3kN/m*2m = 0 RAY = 6 kN 4) ΣFAx = 0 RAX = 0
Problem Calculate the reactions at the supports. For equilibrium: • Sum of forces in x and y directions = 0 • sum of moments =0
Answers: RA = 75 kN RB= 60 kN
DE4101 Students can ignore slides past this point. ENGG5038 students should study all slides.
Finding Reactions Shear and Bending Moments 1. Draw Free Body Diagram •
Draw free body diagrams (i.e. replace supports with forces and moments as applicable)
2. Find Reactions •
Take moments about one of the supports to determine one reaction.
•
Sum forces in x and y directions to determine the other reaction.
•
Draw deflected shape
+ve+veshear shear
3. Find Shear Forces •
Determine SFs by taking cuts just before and just after each applied load and reaction.
•
Draw the SF diagram remembering +ve shear convention (see diagram).
Finding Reactions Shear and Bending Moments 1. Find Bending Moments a) Theoretical method •
Determine BMs by taking cuts at each applied load. In civil engineering a positive BM is usually drawn on the tension side of the beam. (As long as you get the shape of the BM correct no marks will be deducted for drawing on the compression side of the beam as mechanical engineers use this method)
b) Simplified method •
BM is the area under the shear force diagram. (SF is the gradient of the BM diagram.)
a. SF = 0 when the BM is a maximum or minimum. b. BM=0 at points of contraflexure.
Bending Shear and Deflection 1) Determine shear forces (assume SF at cut is +ve upwards) From A Cut just left of B: SFB- +4.4 = -4.4kN Cut just right of B: SFB+ +4.4kN – 12kN = +7.6 kN Cut just right of C: SFc + 4.4kN – 12kN +17kN = -9.4kN Cut just right of D: SFc + 4.4kN – 12kN +17kN -8kN = -1.4kN Cut just right of E: SFc + 4.4kN – 12kN +17kN -8kN – 12 kN= +10.6kN 2) Determine moments (assume moment at cut is +ve clockwise) From Left: Cut at B: MB + 4.4kN*2 = 0
MB = -8.8kN-m
Cut at C: Mc + 4.4kN*5m - 12kN*3m = 0
Mc = +14 kN-m
Cut at D: MD + 4.4kN*8m - 12kN*6m +17 kN*3m = 0MD= -14.2kN-m Cut at E: ME + 4.4kN*9.5m - 12kN*7.5m +17 kN*4.5m -8kN*1.5m = 0 ME = -16.3kN-m At F: MF = 0
Bending Shear and Deflection (Simplified Method) 1) Determine shear forces (assume SF at cut is +ve upwards) From A Cut just left of B: SFB- +4.4 = -4.4kN Cut just right of B: SFB+ +4.4kN – 12kN = +7.6 kN Cut just right of C: SFc + 4.4kN – 12kN +17kN = -9.4kN Cut just right of D: SFc + 4.4kN – 12kN +17kN -8kN = -1.4kN Cut just right of E: SFc + 4.4kN – 12kN +17kN -8kN – 12 kN= +10.6kN 2) Determine moments Moments are the area under the SF diagram. From Left: At B: MB = -4.4kN*2 = 0
MB = -8.8kN-m
At C: Mc = -8.8 kN-m + 3m * 7.6 kN
Mc =+14 kN-m
At D: MD = 14 kN-m + 3m * (-9.4) kN =
MD = -14.2kN-m
At E: ME = -14.2 kN-m + 1.5m*(-1.4) kN
ME = -16.3kN-m
At F: MF = 0
Common Results of Beam Analysis
Note: W(N) = w(N/m)*L(m) in right side figure Ref: Steel Designers Manual CSRDC 1972
Exercises
Exercises
Exercises
Exercises
Solution
Source: Wei Loo
Example Calculate reactions Sum moment about A:
6kN*3m + 10kN*5m – RD*10m = 0 therefore RD = 6.8 kN
Sum forces:
RA – 6kN -10kN +6.8 = 0 therefore RD = 9.2 kN
Point A+ B+ C+
SF (Sum forces to the left) kN = 9.2 = 9.2-6 = 3.2 = 9.2-6 -10 = -6.8
BM (Sum BM to left) kN-m =0 = 9.2*3 = 27.6 =9.2*5-6*2 = 34
Exercises i.
Draw the deflected shape.
ii.
Find reactions. Take moments at one support and then sum forces to find other reaction.
iii.
Find BMs. Take cuts at bend and where a force acts. Draw the BM on the compression side of the beam.
iv.
Find SFs. Take cuts from an end and remember +ve shear rule. The shear force is the slope of the BM diagram Source: Wei Loo
Solution
Source: Wei Loo
Solving for Shear Forces H
E
B
D C
F
G
A
Find SF Start from an end say C: Take a cut at D: SFD – 5 = 0; SFD = 5kN. Either +5kN or – 5kN is acceptable.
The shear force at D from D-C becomes an axial force in member DE . Hence there is no shear force in DE but the axial force is 5kN. You do not need to show the axial force.
Solving for Shear Forces Take a cut just above A: SFA- – 5kN = 0
SFA- = 5kN
5 kN
E 2m A
SFA
Take a cut just to the right of A: Summing all the forces in the vertical direction on the structure to the left SFA+ + 4.375 = 0 SFA+ = -4.375kN
Solving for Shear Forces
Solving for Bending Moments
Source: Wei Loo
Solving for Bending Moments
Note: The BM here is drawn on the tension side of the beam. Drawing on the compression side means that the shear force represents the slope of the BM with the correct sign. (Either method is correct)
Source: Wei Loo
Exercises (i) Draw the deflected shape. (ii) Find the reactions (iii) Draw SF and BM diagrams E
A
C
D
B
Solution: Deflected shape -3.203,-0.01,0
2m
1m -3.203,-2.579,0
y
0,0,0 x
3m
0,0,0
Solution 1) Determine reactions ΣMA = 0 (Clockwise +ve) 2kN*1m - RB*3m + (2kN/m*1m)*2.5m + 4kN*2m = 0 RB = 5kN ΣF = 0 gives RA – 2kN + 5kN – 4kN – 2kN/m*1m = 0 RA = 3 kN 2) Determine moments From Left: Cut at C: Mc + 3kN*1m = 0 Mc = -3 kN-m Cut at B: MB + 3kN*3 – 2kN*2m = 0 MB = -5kN-m At D since the distance to forces is the same as for B then MD = MB = -5kN-m At E: ME = 0 3) Determine shear forces From A Cut just left of C: SF = 3kN Cut just right of C: SF = 3kN – 2kN = 1 kN Cut just left of B: SF = same as above = 1kN From E Cut just right of E: SF = - 1kN Cut just left of D: SF = -4kN Just right of E SF = - 4 kN, At D SF = 4 + (2 kN/m*1m) = - 6 kN SF in D-B = 0 (but the axial force = 6 kN)
Solution
-3.203,-0.01,0
2m
1m -3.203,-2.579,0
y
0,0,0 x
3m
0,0,0
Exercises • Find the shear force and bending moment diagrams. • Calculate the magnitude of the SF and BM at point ‘K’.
Source: Materials and Structures: Whitlow 1991
Solutions
Source: Materials and Structures: Whitlow 1991
