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 .
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).
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 |
All the four different cases are calculated with theoretical formulas and concepts are shown below (Timoshenko,1953)
TEST 1 CALCULATION
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
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)
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)
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)
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)
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)
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
First span of the beam: 0 ≤ x1 < 7.8
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)
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)
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)
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)
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:
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
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) = + 12.07 = 12.07 (N)
Calculate the bending moment at a point (M):
M(x1) = + RA*(x1)
M1(0) = + 12.07*(0) = 0 (N*cm)
M1(7.80) = + 12.07*(7.80) = 94.16 (N*cm)
Second span of the beam: 7.8 ≤ x2 < 52.3
Calculate the shear force at a point (Q):
Q(x2) = + RA – P1
Q2(7.80) = + 12.07 – 9.82 = 2.25 (N)
Q2(52.30) = + 12.07 – 9.82 = 2.25 (N)
Calculate the bending moment at a point (M):
M(x2) = + RA*(x2) – P1*(x2 – 7.8)
M2(7.80) = + 12.07*(7.80) – 9.82*(7.80 – 7.8) = 94.16 (N*cm)
M2(52.30) = + 12.07*(52.30) – 9.82*(52.30 – 7.8) = 194.35 (N*cm)
Calculate the shear force at a point (Q):
Q(x3) = + RA – P1 – P2
Q3(52.30) = + 12.07 – 9.82 – 4.91 = -2.66 (N)
Q3(79.80) = + 12.07 – 9.82 – 4.91 = -2.66 (N)
Calculate the bending moment at a point (M):
M(x3) = + RA*(x3) – P1*(x3 – 7.8) – P2*(x3 – 52.3)
M3(52.30) = + 12.07*(52.30) – 9.82*(52.30 – 7.8) – 4.91*(52.30 – 52.3) = 194.35 (N*cm)
M3(79.80) = + 12.07*(79.80) – 9.82*(79.80 – 7.8) – 4.91*(79.80 – 52.3) = 121.24 (N*cm)
Calculate the shear force at a point (Q):
Q(x4) = + RA – P1 – P2 – P3
Q4(79.80) = + 12.07 – 9.82 – 4.91 – 9.84 = -12.50 (N)
Q4(89.50) = + 12.07 – 9.82 – 4.91 – 9.84 = -12.50 (N)
Calculate the bending moment at a point (M):
M(x4) = + RA*(x4) – P1*(x4 – 7.8) – P2*(x4 – 52.3) – P3*(x4 – 79.8)
M4(79.80) = + 12.07*(79.80) – 9.82*(79.80 – 7.8) – 4.91*(79.80 – 52.3) – 9.84*(79.80 – 79.8) = 121.24 (N*cm)
M4(89.50) = + 12.07*(89.50) – 9.82*(89.50 – 7.8) – 4.91*(89.50 – 52.3) – 9.84*(89.50 – 79.8) = 0 (N*cm)
REACTIONS AT A & B
ΣMA = 0: The addition of all moments at point A is equal to zero:
– P1*7.8 – P2*37.9 – P3*70.2 + 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*51.6 + P3*19.3 = 0
2. Solve this equations:
Calculate the reaction at roller support at point B:
RB = ( P1*7.8 + P2*37.9 + P3*70.2) / 89.5 = ( 9.82*7.8 + 9.83*37.9 + 4.92*70.2) / 89.5 = 8.88 (N)
Calculate the reaction at roller support at point A:
RA = ( P1*81.7 + P2*51.6 + P3*19.3) / 89.5 = ( 9.82*81.7 + 9.83*51.6 + 4.92*19.3) / 89.5 = 15.69 (N)
3.The addition of all forces is zero: ΣFy = 0: RA – P1 – P2 – P3 + RB = 15.69 – 9.82 – 9.83 – 4.92 + 8.88 = 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) = + 15.69 = 15.69 (N)
Calculate the bending moment at a point (M):
M(x1) = + RA*(x1)
M1(0) = + 15.69*(0) = 0 (N*cm)
M1(7.80) = + 15.69*(7.80) = 122.40 (N*cm)
Second span of the beam: 7.8 ≤ x2 < 37.9
Calculate the shear force at a point (Q):
Q(x2) = + RA – P1
Q2(7.80) = + 15.69 – 9.82 = 5.87 (N)
Q2(37.90) = + 15.69 – 9.82 = 5.87 (N)
Calculate the bending moment at a point (M):
M(x2) = + RA*(x2) – P1*(x2 – 7.8)
M2(7.80) = + 15.69*(7.80) – 9.82*(7.80 – 7.8) = 122.40 (N*cm)
M2(37.90) = + 15.69*(37.90) – 9.82*(37.90 – 7.8) = 299.16 (N*cm)
Calculate the shear force at a point (Q):
Q(x3) = + RA – P1 – P2
Q3(37.90) = + 15.69 – 9.82 – 9.83 = -3.96 (N)
Q3(70.20) = + 15.69 – 9.82 – 9.83 = -3.96 (N)
Calculate the bending moment at a point (M):
M(x3) = + RA*(x3) – P1*(x3 – 7.8) – P2*(x3 – 37.9)
M3(37.90) = + 15.69*(37.90) – 9.82*(37.90 – 7.8) – 9.83*(37.90 – 37.9) = 299.16 (N*cm)
M3(70.20) = + 15.69*(70.20) – 9.82*(70.20 – 7.8) – 9.83*(70.20 – 37.9) = 171.34 (N*cm)
Fourth span of the beam: 70.2 ≤ x4 < 89.5
Calculate the shear force at a point (Q):
Q(x4) = + RA – P1 – P2 – P3
Q4(70.20) = + 15.69 – 9.82 – 9.83 – 4.92 = -8.88 (N)
Q4(89.50) = + 15.69 – 9.82 – 9.83 – 4.92 = -8.88 (N)
Calculate the bending moment at a point (M):
M(x4) = + RA*(x4) – P1*(x4 – 7.8) – P2*(x4 – 37.9) – P3*(x4 – 70.2)
M4(70.20) = + 15.69*(70.20) – 9.82*(70.20 – 7.8) – 9.83*(70.20 – 37.9) – 4.92*(70.20 – 70.2) = 171.34 (N*cm)
M4(89.50) = + 15.69*(89.50) – 9.82*(89.50 – 7.8) – 9.83*(89.50 – 37.9) – 4.92*(89.50 – 70.2) = 0 (N*cm)
Initially the experimental value of shear force is noted for four different conditions and then the actual value is calculated and added in table.
|
TEST 1 |
TEST 2 |
TEST 3 |
TEST 4 |
Experimental shear force at 7.8 cm from left (N) |
2.69 |
4.9 |
2.45 |
6.37 |
Actual shear force at 7.8 cm from left (N) |
2.68 |
4.55 |
2.25 |
5.87 |
From the whole experiment done for 4 different set of values of shear force are compared to actual calculated values and marginal error between values are obtain as shown below.
|
TEST 1 |
TEST 2 |
TEST 3 |
TEST 4 |
Experimental shear force at 7.8 cm from left (N) |
2.69 |
4.9 |
2.45 |
6.37 |
Actual shear force at 7.8 cm from left (N) |
2.68 |
4.55 |
2.25 |
5.87 |
Error (deviation from actual value) |
0.37 % |
7.69 % |
8.88 % |
8.51 % |
William A. Nash (1 July 1998). Schaum’s Outline of Theory and Problems of Strength of Materials. McGraw-Hill Professional. p. 82. ISBN 978-0-07-046617-3. Retrieved 20 May 2012.
Ramsay, Angus. “The Influence and Modelling of Warping Restraint on Beams”. ramsay-maunder.co.uk. Retrieved 7 May 2017.
Timoshenko, S., (1953), History of strength of materials, McGraw-Hill New York
Gere, J. M. and Timoshenko, S. P., 1997, Mechanics of Materials, PWS Publishing Company.
Hibbeler, R.C (1985). Structural Analysis. Macmillan. pp. 146–148.
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