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1
Experiment 1.1 shear force variation with for various loading conditions
THEORY
Shear stress is the force applied on per unit area of the member whereas the Shear force is the
resultant force acting on any one of the parts of the beam normal to the axis.
Shearing forces are unaligned forces pushing one part of a body in one specific direction, and
another part of the body in the opposite direction. When the forces are aligned into each other,
they are called compression forces. An example is a deck of cards being pushed one way on
the top, and the other at the bottom, causing the cards to slide. Another example is when wind
blows at the side of a peaked roof of a home - the side walls experience a force at their top
pushing in the direction of the wind, and their bottom in the opposite direction, from the
ground or foundation. William A. Nash defines shear force in terms of planes: "If a plane is
passed through a body, a force acting along this plane is called a shear force or shearing force.
OBJECTIVE
 This experiment examines how shear forces varies with an increasing point load.
2
APPARATUS
1. Load
2. Digital force display
3. Beam tester
3
PROCEDURES
1. The Digital Force Display meter was checked read zero with no load.
2. A hanger with a 100g mass was placed to the left of the ‘cut’.
3. The digital force display were recorded.
4. Using masses of 200g, 300g, 400g, and 500g were repeated.
5. The mass was converted into a load ( in N).
6. The theoretical shear force were calculated at the cut.
4
RESULT
Mass (g) Load (N) Experimental shear
Force (N)
Theoretical shear force (N)
0 0 0.0 0
100 0.98 0.6 0.67
200 1.96 1.1 1.34
300 2.94 1.7 2.0
400 3.92 2.2 2.68
500 4.90 2.7 3.34
5
CALCULATION
Theoretical shear force.
Formula =
𝑤𝑎
𝐿
For mass 100g
=
0.98 𝑥 0.26
0.2
= 0.62N
For mass 200g
=
1.96 𝑥 0.26
0.2
= 1.34N
For mass 300g
=
2.94 𝑥 0.26
0.2
= 2.0N
For mass 400g
=
3.92 𝑥 0.26
0.2
= 2.68N
For mass 500g
=
4.90 𝑥 0.26
0.2
= 4.90N
6
Shear force
For mass 100g
∑mb=0
FA(0.44) – 0.98(0.14) = 0
0.44FA=0.1372
FA=0.312N
∑Fy ↑ = ∑Fy↓
FB + 0.312 = 0.98
FB = 0.668N
FA= 0.312N
FA-w = 0.312N
Fw = 0.312 – 0.98
= -0.668N
FW-B = - 0.668N
FB = -0.668 + 0.668
= 0
For mass 200g
∑mb=0
FA(0.44) – 1.96(0.14) = 0
0.44FA=0.274
7
FA=0.624N
∑Fy ↑ = ∑Fy↓
FB + 0.624 = 1.96
FB = 1.336N
FA= 0.624N
FA-w = 0.624N
Fw = 0.624 – 1.96
= -1.336N
FW-B = - 1.336N
FB = -1.336 + 1.336
= 0
For mass 300g
∑mb=0
FA(0.44) – 2.94(0.14) = 0
0.44FA=0.412
FA=0.935N
∑Fy ↑ = ∑Fy↓
FB + 0.935 = 2.94
FB = 2.005N
FA= 0.935N
8
FA-w = 0.935N
Fw = 0.935 – 2.94
= -2.005N
FW-B = - 2.005N
FB = -2.005 + 2.005
= 0
For mass 400g
∑mb=0
FA(0.44) – 3.92(0.14) = 0
0.44FA=0.549
FA=1.247N
∑Fy ↑ = ∑Fy↓
FB + 1.247 = 3.92
FB = 2.673N
FA= 1.247N
FA-w = 1.247N
Fw = 1.247 – 3.92
= -2.673N
FW-B = - 2.673N
FB = -2.673 + 2.673
= 0
9
For mass 500g
∑mb=0
FA(0.44) – 4.90(0.14) = 0
0.44FA=0.686
FA=1.560N
∑Fy ↑ = ∑Fy↓
FB + 1.560 = 4.90
FB = 3.34N
FA= 1.560N
FA-w = 1.560N
Fw = 1.560 – 4.90
= -3.340N
FW-B = - 3.340N
FB = -3.340 + 3.340
= 0
10
OBSERVATION AND DISCUSSION
From the experiment;
a. Graph with compares your experimental results to theoretical.
0
0.3
0
0.28
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
mass 100g
Experimental theoretical
0
1.1
0
1.34
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
mass 200g
Experimental theoretical
11
b. Comment on the shape of the graph. What does it tell us about how shear force varies
due to increased load?
From the graph, the value of shear force increase when the load that has been place
increase. The gradient of the shear force also increase due than increased load.
0
2.2
0
2.68
0
0.5
1
1.5
2
2.5
3
mass 100g
Experimental theoretical
0
2.7
0
3.34
0
0.5
1
1.5
2
2.5
3
3.5
4
mass 100g
Experimental theoretical
12
c. Does the equation we used accurately predict the behaviour of the beam?
The value that we got from the equation determined the size of the beam so it
accurately predict the behaviour of the beam.
CONCLUSION
In conclusion, aim of this task was to study the effect of different forces on the shear force in
the beam and the result show that there is a linear relationship between shear force and
applied load. Experimental and theoretical shear force shows perfect linear relationship with
applied load with very little difference in the values of shear force.
13
Experiment 1.2: shear force variation with for Various loading conditions.
THEORY
The shearing force (SF) at any section of a beam represents the tendency for the portion of
the beam on one side of the section to slide or shear laterally relative to the other portion.
The diagram shows a beam carrying loads . It is simply supported at two
points where the reactions are . Assume that the beam is divided into two parts
by a section XX. The resultant of the loads and reaction acting on the left of AA is F
vertically upwards, and since the whole beam is in equilibrium, the resultant force to the right
of AA must be F downwards. F is called the Shearing Force at the section AA. It may be
defined as follows:-
The shearing force at any section of a beam is the algebraic sum of the lateral components of
the forces acting on either side of the section.
Where forces are neither in the lateral or axial direction they must be resolved in the usual
way and only the lateral components are used to calculate the shear force.
1. Vertical equilibrium (total force up = total force down)
2. Horizontal equilibrium (total force right = total force left)
3. Moment equilibrium (total clockwise moment = total anticlockwise moment)
14
OBJECTIVES
 This experiment examines how shear forces varies at the cut position of the beam for
various.
15
APPARATUS
7. Load
16
8. Digital force display
9. Beam tester
PROCEDURES
Figure 1 force diagram
W1 RA
140mm
cut
RB
17
W1=3.92 N (400g)
Figure 2 force diagram
W1= 1.96 N (200g)
W2= 3.92 N (400g)
W2
RA
cut
RB
W1
220mm
260mm
18
Figure 3 force diagram
W1 = 4.91N (500g)
W2 = 3.92N (400g)
1. The digital force display meter was checked read zero with no load.
2. Carefully load the beam with the hangers in the position shown in figure 1.
3. The digital force display reading were recorded.
4. The support reactions RA and RB were calculated and the theoretical shear force
were calculated at the cut.
5. The procedures with the beam loaded as in figure 2 and 3 were repeated.
RA
cut
RB
240mm
400mm
W1 W2
19
RESULT
Figure W1 (N) W2 (N) Force
(N)
Experimental
bending
moment
(Nm)
RA
(N)
RB (N) Theoretical
bending
moment
(Nm)
1 3.92 1.3 0.16 5.167 -1.247 0.37
2 1.96 3.92 3.2 0.40 2.584 3.296 0.64
3 4.91 3.92 3.5 0.44 2.588 6.242 0.68
20
OBSERVATION AND CALCULATION
Experimental bending moment
For mass 400g
=1.3 x 0.125
=0.16Nm
For mass 200g and 400g
=3.2 x 0.125
=0.40Nm
For mass 500g and 400g
=3.5 x 0.125
=0.44Nm
Bending moment
For mass 400g
∑mb = 0
RA(0.44) – 3.92(0.58) = 0
0.44RA = 2.274
RA = 5.167N
∑Fy ↑ = ∑Fy↓
21
5.167 + RB = 3.92
RB = -1.247N
RA = 5.167N
RA-W = 5.167N
RW = 5.167 – 3.92
= 1.247N
RW-B = 1.247N
RB = 1.247 – 1.247
= 0
For mass 200g and 400g
∑mb = 0
RA(0.44) – 1.96(0.22) – 3.92(0.18) = 0
0.44RA= 1.137
RA= 2.584N
∑Fy ↑ = ∑Fy↓
1.584 + RB = 1.96 + 3.92
RB = 3.296
RA = 2.584N
RA-W1 = 2.584N
22
RW1 = 2.584 – 1.96
= 0.624N
RW1-W2 = 0.624N
RW2 = 0.624 – 3.92
= -3.296 N
RW2-B = -3.296N
RB = -3.296 + 3.296
= O
For mass 500g and 400g
∑mb = 0
RA(0.44) – 4.91(0.2) – 3.92(0.04) = 0
0.44RA = 0.681
RA = 2.558N
∑Fy ↑ = ∑Fy↓
2.588 + RB = 4.91 + 3.92
RB = 6.242N
RA = 2.558N
RA-W1 = 2.558N
RW1 = 2.322 – 4.91
= -2.322N
RW1-W2 = -2.322N
RW2 = -2.322 – 3.92
23
= -6.242 N
RW2-B = -6.242N
RB = -6.242 + 6.242
= O
0
0.3
0
0.28
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
mass 100g
Experimental theoretical
24
0
0.4
0
0.64
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
mass 200g and 400g
Experimental theoretical
25
0
0.44
0
0.68
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
mass 500g and 400g
Experimental theoretical
26
DISCUSSION
1. Comment on how the result of the experiment compare with those calculated using
the theory.
The result of the experiment we get is the value from the experiment is less than the
value from theoretical.
2. Factors that effected in both experiment and theory.
a. The way we handle the apparatus when doing the experiment.
b. The way we calculated the theoretical equation.
3. The importance of shear force in civil engineering.
a. We can use a shear force to analyse the beam.
b. The shear force indicates the shear force resisted by the beam section along the
length of the beam.
27
CONCLUSION
From experiment, we can use the theoretical value to determine the beam, we can learn more
details about the beam and load base on the graph and we also can analyse the type of beam
and how the beam response when the load apply by look at shear force diagram.
28
REFERENCES
1. Braja M.Das (2011). Principles of structure engineering (9nd ed). British
2. Robert D.Holtz (2010). A Introduction to structure Engineering (2nd
ed). British.
3. Robert W.Day (2009). structure engineers handbook (2nd
ed). Us

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LAB REPORT SHEAR FORCE IN A BEAM

  • 1. 1 Experiment 1.1 shear force variation with for various loading conditions THEORY Shear stress is the force applied on per unit area of the member whereas the Shear force is the resultant force acting on any one of the parts of the beam normal to the axis. Shearing forces are unaligned forces pushing one part of a body in one specific direction, and another part of the body in the opposite direction. When the forces are aligned into each other, they are called compression forces. An example is a deck of cards being pushed one way on the top, and the other at the bottom, causing the cards to slide. Another example is when wind blows at the side of a peaked roof of a home - the side walls experience a force at their top pushing in the direction of the wind, and their bottom in the opposite direction, from the ground or foundation. William A. Nash defines shear force in terms of planes: "If a plane is passed through a body, a force acting along this plane is called a shear force or shearing force. OBJECTIVE  This experiment examines how shear forces varies with an increasing point load.
  • 2. 2 APPARATUS 1. Load 2. Digital force display 3. Beam tester
  • 3. 3 PROCEDURES 1. The Digital Force Display meter was checked read zero with no load. 2. A hanger with a 100g mass was placed to the left of the ‘cut’. 3. The digital force display were recorded. 4. Using masses of 200g, 300g, 400g, and 500g were repeated. 5. The mass was converted into a load ( in N). 6. The theoretical shear force were calculated at the cut.
  • 4. 4 RESULT Mass (g) Load (N) Experimental shear Force (N) Theoretical shear force (N) 0 0 0.0 0 100 0.98 0.6 0.67 200 1.96 1.1 1.34 300 2.94 1.7 2.0 400 3.92 2.2 2.68 500 4.90 2.7 3.34
  • 5. 5 CALCULATION Theoretical shear force. Formula = 𝑤𝑎 𝐿 For mass 100g = 0.98 𝑥 0.26 0.2 = 0.62N For mass 200g = 1.96 𝑥 0.26 0.2 = 1.34N For mass 300g = 2.94 𝑥 0.26 0.2 = 2.0N For mass 400g = 3.92 𝑥 0.26 0.2 = 2.68N For mass 500g = 4.90 𝑥 0.26 0.2 = 4.90N
  • 6. 6 Shear force For mass 100g ∑mb=0 FA(0.44) – 0.98(0.14) = 0 0.44FA=0.1372 FA=0.312N ∑Fy ↑ = ∑Fy↓ FB + 0.312 = 0.98 FB = 0.668N FA= 0.312N FA-w = 0.312N Fw = 0.312 – 0.98 = -0.668N FW-B = - 0.668N FB = -0.668 + 0.668 = 0 For mass 200g ∑mb=0 FA(0.44) – 1.96(0.14) = 0 0.44FA=0.274
  • 7. 7 FA=0.624N ∑Fy ↑ = ∑Fy↓ FB + 0.624 = 1.96 FB = 1.336N FA= 0.624N FA-w = 0.624N Fw = 0.624 – 1.96 = -1.336N FW-B = - 1.336N FB = -1.336 + 1.336 = 0 For mass 300g ∑mb=0 FA(0.44) – 2.94(0.14) = 0 0.44FA=0.412 FA=0.935N ∑Fy ↑ = ∑Fy↓ FB + 0.935 = 2.94 FB = 2.005N FA= 0.935N
  • 8. 8 FA-w = 0.935N Fw = 0.935 – 2.94 = -2.005N FW-B = - 2.005N FB = -2.005 + 2.005 = 0 For mass 400g ∑mb=0 FA(0.44) – 3.92(0.14) = 0 0.44FA=0.549 FA=1.247N ∑Fy ↑ = ∑Fy↓ FB + 1.247 = 3.92 FB = 2.673N FA= 1.247N FA-w = 1.247N Fw = 1.247 – 3.92 = -2.673N FW-B = - 2.673N FB = -2.673 + 2.673 = 0
  • 9. 9 For mass 500g ∑mb=0 FA(0.44) – 4.90(0.14) = 0 0.44FA=0.686 FA=1.560N ∑Fy ↑ = ∑Fy↓ FB + 1.560 = 4.90 FB = 3.34N FA= 1.560N FA-w = 1.560N Fw = 1.560 – 4.90 = -3.340N FW-B = - 3.340N FB = -3.340 + 3.340 = 0
  • 10. 10 OBSERVATION AND DISCUSSION From the experiment; a. Graph with compares your experimental results to theoretical. 0 0.3 0 0.28 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 mass 100g Experimental theoretical 0 1.1 0 1.34 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 mass 200g Experimental theoretical
  • 11. 11 b. Comment on the shape of the graph. What does it tell us about how shear force varies due to increased load? From the graph, the value of shear force increase when the load that has been place increase. The gradient of the shear force also increase due than increased load. 0 2.2 0 2.68 0 0.5 1 1.5 2 2.5 3 mass 100g Experimental theoretical 0 2.7 0 3.34 0 0.5 1 1.5 2 2.5 3 3.5 4 mass 100g Experimental theoretical
  • 12. 12 c. Does the equation we used accurately predict the behaviour of the beam? The value that we got from the equation determined the size of the beam so it accurately predict the behaviour of the beam. CONCLUSION In conclusion, aim of this task was to study the effect of different forces on the shear force in the beam and the result show that there is a linear relationship between shear force and applied load. Experimental and theoretical shear force shows perfect linear relationship with applied load with very little difference in the values of shear force.
  • 13. 13 Experiment 1.2: shear force variation with for Various loading conditions. THEORY The shearing force (SF) at any section of a beam represents the tendency for the portion of the beam on one side of the section to slide or shear laterally relative to the other portion. The diagram shows a beam carrying loads . It is simply supported at two points where the reactions are . Assume that the beam is divided into two parts by a section XX. The resultant of the loads and reaction acting on the left of AA is F vertically upwards, and since the whole beam is in equilibrium, the resultant force to the right of AA must be F downwards. F is called the Shearing Force at the section AA. It may be defined as follows:- The shearing force at any section of a beam is the algebraic sum of the lateral components of the forces acting on either side of the section. Where forces are neither in the lateral or axial direction they must be resolved in the usual way and only the lateral components are used to calculate the shear force. 1. Vertical equilibrium (total force up = total force down) 2. Horizontal equilibrium (total force right = total force left) 3. Moment equilibrium (total clockwise moment = total anticlockwise moment)
  • 14. 14 OBJECTIVES  This experiment examines how shear forces varies at the cut position of the beam for various.
  • 16. 16 8. Digital force display 9. Beam tester PROCEDURES Figure 1 force diagram W1 RA 140mm cut RB
  • 17. 17 W1=3.92 N (400g) Figure 2 force diagram W1= 1.96 N (200g) W2= 3.92 N (400g) W2 RA cut RB W1 220mm 260mm
  • 18. 18 Figure 3 force diagram W1 = 4.91N (500g) W2 = 3.92N (400g) 1. The digital force display meter was checked read zero with no load. 2. Carefully load the beam with the hangers in the position shown in figure 1. 3. The digital force display reading were recorded. 4. The support reactions RA and RB were calculated and the theoretical shear force were calculated at the cut. 5. The procedures with the beam loaded as in figure 2 and 3 were repeated. RA cut RB 240mm 400mm W1 W2
  • 19. 19 RESULT Figure W1 (N) W2 (N) Force (N) Experimental bending moment (Nm) RA (N) RB (N) Theoretical bending moment (Nm) 1 3.92 1.3 0.16 5.167 -1.247 0.37 2 1.96 3.92 3.2 0.40 2.584 3.296 0.64 3 4.91 3.92 3.5 0.44 2.588 6.242 0.68
  • 20. 20 OBSERVATION AND CALCULATION Experimental bending moment For mass 400g =1.3 x 0.125 =0.16Nm For mass 200g and 400g =3.2 x 0.125 =0.40Nm For mass 500g and 400g =3.5 x 0.125 =0.44Nm Bending moment For mass 400g ∑mb = 0 RA(0.44) – 3.92(0.58) = 0 0.44RA = 2.274 RA = 5.167N ∑Fy ↑ = ∑Fy↓
  • 21. 21 5.167 + RB = 3.92 RB = -1.247N RA = 5.167N RA-W = 5.167N RW = 5.167 – 3.92 = 1.247N RW-B = 1.247N RB = 1.247 – 1.247 = 0 For mass 200g and 400g ∑mb = 0 RA(0.44) – 1.96(0.22) – 3.92(0.18) = 0 0.44RA= 1.137 RA= 2.584N ∑Fy ↑ = ∑Fy↓ 1.584 + RB = 1.96 + 3.92 RB = 3.296 RA = 2.584N RA-W1 = 2.584N
  • 22. 22 RW1 = 2.584 – 1.96 = 0.624N RW1-W2 = 0.624N RW2 = 0.624 – 3.92 = -3.296 N RW2-B = -3.296N RB = -3.296 + 3.296 = O For mass 500g and 400g ∑mb = 0 RA(0.44) – 4.91(0.2) – 3.92(0.04) = 0 0.44RA = 0.681 RA = 2.558N ∑Fy ↑ = ∑Fy↓ 2.588 + RB = 4.91 + 3.92 RB = 6.242N RA = 2.558N RA-W1 = 2.558N RW1 = 2.322 – 4.91 = -2.322N RW1-W2 = -2.322N RW2 = -2.322 – 3.92
  • 23. 23 = -6.242 N RW2-B = -6.242N RB = -6.242 + 6.242 = O 0 0.3 0 0.28 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 mass 100g Experimental theoretical
  • 26. 26 DISCUSSION 1. Comment on how the result of the experiment compare with those calculated using the theory. The result of the experiment we get is the value from the experiment is less than the value from theoretical. 2. Factors that effected in both experiment and theory. a. The way we handle the apparatus when doing the experiment. b. The way we calculated the theoretical equation. 3. The importance of shear force in civil engineering. a. We can use a shear force to analyse the beam. b. The shear force indicates the shear force resisted by the beam section along the length of the beam.
  • 27. 27 CONCLUSION From experiment, we can use the theoretical value to determine the beam, we can learn more details about the beam and load base on the graph and we also can analyse the type of beam and how the beam response when the load apply by look at shear force diagram.
  • 28. 28 REFERENCES 1. Braja M.Das (2011). Principles of structure engineering (9nd ed). British 2. Robert D.Holtz (2010). A Introduction to structure Engineering (2nd ed). British. 3. Robert W.Day (2009). structure engineers handbook (2nd ed). Us