Lab Report: Experimental Verification of Shear Force on a Beam

This report details an experiment conducted to determine the shear force at the cut section of a beam under various loading conditions. The experiment involved applying different sets of weights at varying locations on the beam and measuring the resulting shear force using a spring balance. Four test cases were performed, and the experimental shear force was calculated by subtracting the datum force from the actual force measured. Theoretical calculations were also performed for each test case to compare with the experimental results. The report includes an introduction to the concept of shear force and bending moment, a description of the equipment used, a detailed explanation of the experimental procedure, and a presentation of the results obtained. The theoretical calculations for each test case are presented with step-by-step explanations. The results from the experiment and theoretical calculations were slightly different by around 8%. Desklib offers a variety of similar reports and solved assignments for students.

SHEAR FORCE LAB REPORT
1 | P a g e
REPORT
ON
SHEAR FORCE ON A BEAM

SHEAR FORCE LAB REPORT
2 | P a g e
ABSTRACT
In this experiment, a complete set up is done to calculate the shear force at the
cut section in beam. For that purpose different sets of weight are tested on
the beams at different location. Four test were conducted and from spring
balance , the shear force in all those cases were obtained by subtracting the
datum force on balance to the actual force obtained using weights. Then the
theoretical calculations are done for same condition as in four cases. The result
obtained in actual calculation and experimental calculation were slightly
different by around 8 %.

SHEAR FORCE LAB REPORT
3 | P a g e
Table of Contents ................................................................................................ 3
1. Introduction .............................................................................................. 4
2. Background theory ................................................................................... 4
3. Equipment used ........................................................................................ 4
4. Experimental procedure ........................................................................... 5
5. Results ...................................................................................................... 6
6. Theoritical calculation .............................................................................. 7
7 Discussion ............................................................................................... 18
8. Conclusion .............................................................................................. 18
9. Recommendations .................................................................................. 19
10. Reference ................................................................................................ 19

SHEAR FORCE LAB REPORT
4 | P a g e
INTRODUCTION
AIM /OBJECTIVE
The main aim for this experiment is to calculate the shear force at the cut section of beam
using spring balance and multiple weights and compare the obtained experimental value
with the theoretical calculation .
BACKGROUND THEORY
The main theory behind this experiment is when a beam is supported at its ends and force
or load applied on it from the top then a resultant force act on each part of beam along its
length . This resultant force called shear force . Beam also observe bending moment .
Bending moment is product of shear force and displacement along the length of beam.
Units of Bending moment is Nm (Newton meter) and shear force is Newton( Nash , 1998).
EQUIPMENTS USED
 Beam :- On which all the loads and supports are applied and the shear force and
bending moment is obtain.
 Weights :- Multiple weights of different values are required to create different
experimental conditions.
 Spring balance :-Spring balance is attached at the cut section , initially the at no load
condition the datum value is obtained .
 Measuring scale :- To check the reading of length of beam where load is applied.
 Weight Hanger :- Weight hanger used to carry the weight , it hangs on beam to
transfer the weight to beam.

SHEAR FORCE LAB REPORT
5 | P a g e
EXPERIMENTAL PROCEDURE
 Set up all the equipments used as per provided picture and keep the beam straight
enough to get the accurate result.
 Initially measure the length of beam from where it is simply supported and length of
cut out section from left side from steel ruler and note it down.
 Place the three hanger (carry weight) at different position on the beam and measure
its position from left end of beam .
 Due to weight of hangers the beam slightly bend, so use the gauge to straight the
beam and find the datum value of force on beam . This datum value will be
subtracted at the end from the actual experimental value of shear force to get the
exact value.
 Now one by one place the weight on the hangers ,then the beam tends to bend.
 Both the spring need to be adjusted so that two parts of beam again bring back to
straight condition . Use a straight edge to check the straightness .
 Record a force that shown on spring balance after straightening the beam .
 Repeat the above steps by changing the location of hangers and weight on hangers .
 The result obtain should be checked again to remove human error (Ramsay,2017) .

SHEAR FORCE LAB REPORT
6 | P a g e
RESULT
According to the experimental procedure shown above , there the four cases need to be
tested and all the experimental values obtain are added in table.
Test 1 Test 2 Test 3 Test 4
Distance between both supports 89.5 89.5 89.5 89.5
Spring balance reading with no added
loads (datum)
6.87 6.87 6.87 6.87
Position of load 1 (cm from left support) 7.8 7.8 7.8 7.8
Magnitude of Load 1 (N) 4.92 9.84 9.82 9.82
Position of load 2 (cm from left support) 47.5 47.5 52.3 37.9
Magnitude of Load 2 (N) 4.92 9.82 4.91 9.83
Position of load 3 (cm from left/right side) 75.0 75.0 79.8 70.2
Magnitude of Load 3 (N) 4.91 4.91 9.84 4.92
Spring balance reading with load (N) 9.56 11.77 9.32 13.24
Shear force at cut(sb1-sb2) (N) 2.69 4.9 2.45 6.37
Distance to cut in the beam(cm from left
support)
7.8 7.8 7.8 7.8

SHEAR FORCE LAB REPORT
7 | P a g e
THEORITICAL CALCULATION
All the four different cases are calculated with theoretical formulas and concepts are shown
below (Timoshenko,1953)
TEST 1 CALCULATION

SHEAR FORCE LAB REPORT
8 | P a g e
REACTIONS AT A & B (Gere, 1997)
ΣMA = 0: The addition of all moments at point A is equal to zero :
- P1*7.8 - P2*47.5 - P3*75 + RB*89.5 = 0
ΣMB = 0: The addition of all moments at point B is equal to zero:
- RA*89.5 + P1*81.7 + P2*42 + P3*14.5 = 0
2. Now Solve the equations :
Calculate the reaction at roller support at point B :
RB = ( P1*7.8 + P2*47.5 + P3*75) / 89.5 = ( 4.92*7.8 + 4.92*47.5 + 4.91*75) / 89.5 = 7.15 (N)
Calculate the reaction at roller support at point A :
RA = ( P1*81.7 + P2*42 + P3*14.5) / 89.5 = ( 4.92*81.7 + 4.92*42 + 4.91*14.5) / 89.5 = 7.60 (N)
3. The addition of all forces is zero :
ΣFy = 0: RA - P1 - P2 - P3 + RB = 7.60 - 4.92 - 4.92 - 4.91 + 7.15 = 0
SHEAR FORCE & BENDING MOMENT CALCULATION
First span of the beam: 0 ≤ x1 < 7.8
Calculate the shear force at a point (Q):
Q(x1) = + RA
Q1(0) = + 7.60 = 7.60 (N)
Calculate the bending moment at a point (M):
M(x1) = + RA*(x1)
M1(0) = + 7.60*(0) = 0 (N*cm)
M1(7.80) = + 7.60*(7.80) = 59.25 (N*cm)
Second span of the beam: 7.8 ≤ x2 < 47.5
Calculate the shear force at a point (Q):
Q(x2) = + RA - P1
Q2(7.80) = + 7.60 - 4.92 = 2.68 (N)
Q2(47.50) = + 7.60 - 4.92 = 2.68 (N)
Calculate the bending moment at a point (M):
M(x2) = + RA*(x2) - P1*(x2 - 7.8)
M2(7.80) = + 7.60*(7.80) - 4.92*(7.80 - 7.8) = 59.25 (N*cm)
M2(47.50) = + 7.60*(47.50) - 4.92*(47.50 - 7.8) = 165.46 (N*cm)
Third span of the beam: 47.5 ≤ x3 < 75
Calculate the shear force at a point (Q):
Q(x3) = + RA - P1 - P2
Q3(47.50) = + 7.60 - 4.92 - 4.92 = -2.24 (N)
Q3(75) = + 7.60 - 4.92 - 4.92 = -2.24 (N)
Calculate the bending moment at a point (M):
M(x3) = + RA*(x3) - P1*(x3 - 7.8) - P2*(x3 - 47.5)

SHEAR FORCE LAB REPORT
9 | P a g e
M3(47.50) = + 7.60*(47.50) - 4.92*(47.50 - 7.8) - 4.92*(47.50 - 47.5) = 165.46 (N*cm)
M3(75) = + 7.60*(75) - 4.92*(75 - 7.8) - 4.92*(75 - 47.5) = 103.74 (N*cm)
Fourth span of the beam: 75 ≤ x4 < 89.5
Calculate the shear force at a point (Q):
Q(x4) = + RA - P1 - P2 - P3
Q4(75) = + 7.60 - 4.92 - 4.92 - 4.91 = -7.15 (N)
Q4(89.50) = + 7.60 - 4.92 - 4.92 - 4.91 = -7.15 (N)
Calculate the bending moment at a point (M):
M(x4) = + RA*(x4) - P1*(x4 - 7.8) - P2*(x4 - 47.5) - P3*(x4 - 75)
M4(75) = + 7.60*(75) - 4.92*(75 - 7.8) - 4.92*(75 - 47.5) - 4.91*(75 - 75) = 103.74 (N*cm)
M4(89.50) = + 7.60*(89.50) - 4.92*(89.50 - 7.8) - 4.92*(89.50 - 47.5) - 4.91*(89.50 - 75) = 0 (N*cm)
TEST 2 CALCULATION

SHEAR FORCE LAB REPORT
10 | P a g e
REACTIONS AT A & B
ΣMA = 0: The addition of all moments at point A is equal to zero:
- P1*7.8 - P2*47.5 - P3*75 + RB*89.5 = 0
ΣMB = 0: The addition of all moments at point B is equal to zero:
- RA*89.5 + P1*81.7 + P2*42 + P3*14.5 = 0
2. Solve this equations:
Calculate the reaction at roller support at point B:
RB = ( P1*7.8 + P2*47.5 + P3*75) / 89.5 = ( 9.84*7.8 + 9.82*47.5 + 4.91*75) / 89.5 = 10.18 (N)
Calculate the reaction at roller support at point A:
RA = ( P1*81.7 + P2*42 + P3*14.5) / 89.5 = ( 9.84*81.7 + 9.82*42 + 4.91*14.5) / 89.5 = 14.39 (N)
3.The addition of all forces is zero: ΣFy = 0: RA - P1 - P2 - P3 + RB = 14.39 - 9.84 - 9.82 - 4.91 + 10.18 = 0
SHEAR FORCE & BENDING MOMENT CALCULATION
First span of the beam: 0 ≤ x1 < 7.8
Calculate the shear force at a point (Q):
Q(x1) = + RA
Q1(0) = + 14.39 = 14.39 (N)
Calculate the bending moment at a point (M):
M(x1) = + RA*(x1)
M1(0) = + 14.39*(0) = 0 (N*cm)
M1(7.80) = + 14.39*(7.80) = 112.21 (N*cm)
Second span of the beam: 7.8 ≤ x2 < 47.5
Calculate the shear force at a point (Q):
Q(x2) = + RA - P1
Q2(7.80) = + 14.39 - 9.84 = 4.55 (N)
Q2(47.50) = + 14.39 - 9.84 = 4.55 (N)
Calculate the bending moment at a point (M):
M(x2) = + RA*(x2) - P1*(x2 - 7.8)
M2(7.80) = + 14.39*(7.80) - 9.84*(7.80 - 7.8) = 112.21 (N*cm)
M2(47.50) = + 14.39*(47.50) - 9.84*(47.50 - 7.8) = 292.70 (N*cm)

SHEAR FORCE LAB REPORT
11 | P a g e
Third span of the beam: 47.5 ≤ x3 < 75
Calculate the shear force at a point (Q):
Q(x3) = + RA - P1 - P2
Q3(47.50) = + 14.39 - 9.84 - 9.82 = -5.27 (N)
Q3(75) = + 14.39 - 9.84 - 9.82 = -5.27 (N)
Calculate the bending moment at a point (M):
M(x3) = + RA*(x3) - P1*(x3 - 7.8) - P2*(x3 - 47.5)
M3(47.50) = + 14.39*(47.50) - 9.84*(47.50 - 7.8) - 9.82*(47.50 - 47.5) = 292.70 (N*cm)
M3(75) = + 14.39*(75) - 9.84*(75 - 7.8) - 9.82*(75 - 47.5) = 147.67 (N*cm)
Fourth span of the beam: 75 ≤ x4 < 89.5
Calculate the shear force at a point (Q):
Q(x4) = + RA - P1 - P2 - P3
Q4(75) = + 14.39 - 9.84 - 9.82 - 4.91 = -10.18 (N)
Q4(89.50) = + 14.39 - 9.84 - 9.82 - 4.91 = -10.18 (N)
Calculate the bending moment at a point (M):
M(x4) = + RA*(x4) - P1*(x4 - 7.8) - P2*(x4 - 47.5) - P3*(x4 - 75)
M4(75) = + 14.39*(75) - 9.84*(75 - 7.8) - 9.82*(75 - 47.5) - 4.91*(75 - 75) = 147.67 (N*cm)
M4(89.50) = + 14.39*(89.50) - 9.84*(89.50 - 7.8) - 9.82*(89.50 - 47.5) - 4.91*(89.50 - 75) = 0 (N*cm)
TEST 3 CALCULATION

SHEAR FORCE LAB REPORT
12 | P a g e
REACTIONS AT A & B
ΣMA = 0: The addition of all moments at point A is equal to zero:
- P1*7.8 - P2*52.3 - P3*79.8 + RB*89.5 = 0
ΣMB = 0: The addition of all moments at point B is equal to zero:
- RA*89.5 + P1*81.7 + P2*37.2 + P3*9.7 = 0
2. Solve this equations:
Calculate the reaction at roller support at point B:
RB = ( P1*7.8 + P2*52.3 + P3*79.8) / 89.5 = ( 9.82*7.8 + 4.91*52.3 + 9.84*79.8) / 89.5 = 12.50 (N)
Calculate the reaction at roller support at point A:
RA = ( P1*81.7 + P2*37.2 + P3*9.7) / 89.5 = ( 9.82*81.7 + 4.91*37.2 + 9.84*9.7) / 89.5 = 12.07 (N)
3.The addition of all forces is zero: ΣFy = 0: RA - P1 - P2 - P3 + RB = 12.07 - 9.82 - 4.91 - 9.84 + 12.50 = 0