NG1S228 Engineering Mechanics 1

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).

• 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.

• 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).

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.

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)

ΣM_A = 0:   The addition of all moments at point A is equal to zero :
 – P₁*7.8 – P₂*47.5 – P₃*75 + R_B*89.5 = 0
ΣM_B = 0:  The addition of all moments at point B is equal to zero:
 – R_A*89.5 + P₁*81.7 + P₂*42 + P₃*14.5 = 0

Calculate the reaction at roller support at point B :
R_B = ( 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 :
R_A = ( 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 :

ΣF_y = 0:    R_A – P₁ – P₂ – P₃ + R_B =  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) =  + R_A
Q₁(0) =  + 7.60 = 7.60 (N)
Calculate the bending moment at a point (M):
M(x₁) =  + R_A*(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) =  + R_A – 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) =  + R_A*(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) =  + R_A – 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) =  + R_A*(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) =  + R_A – 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) =  + R_A*(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)

ΣM_A = 0:   The addition of all moments at point A is equal to zero:
 – P₁*7.8 – P₂*47.5 – P₃*75 + R_B*89.5 = 0
ΣM_B = 0:   The addition of all moments at point B is equal to zero:
 – R_A*89.5 + P₁*81.7 + P₂*42 + P₃*14.5 = 0
2. Solve this equations:

Calculate the reaction at roller support at point B:
R_B = ( 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:
R_A = ( 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: ΣF_y = 0:    R_A – P₁ – P₂ – P₃ + R_B =  14.39 – 9.84 – 9.82 – 4.91 + 10.18 = 0

First span of the beam: 0 ≤ x₁ < 7.8

Q(x₁) =  + R_A
Q₁(0) =  + 14.39 = 14.39 (N)
Calculate the bending moment at a point (M):
M(x₁) =  + R_A*(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) =  + R_A – 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) =  + R_A*(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) =  + R_A – 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) =  + R_A*(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) =  + R_A – 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) =  + R_A*(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)

ΣM_A = 0:   The addition of all moments at point A is equal to zero:
 – P₁*7.8 – P₂*52.3 – P₃*79.8 + R_B*89.5 = 0
ΣM_B = 0:   The addition of all moments at point B is equal to zero:
 – R_A*89.5 + P₁*81.7 + P₂*37.2 + P₃*9.7 = 0
2. Solve this equations:

R_B = ( 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:
R_A = ( 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: ΣF_y = 0:    R_A – P₁ – P₂ – P₃ + R_B =  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) =  + R_A
Q₁(0) =  + 12.07 = 12.07 (N)
Calculate the bending moment at a point (M):
M(x₁) =  + R_A*(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) =  + R_A – 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) =  + R_A*(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) =  + R_A – 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) =  + R_A*(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) =  + R_A – 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) =  + R_A*(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

ΣM_A = 0:   The addition of all moments at point A is equal to zero:
 – P₁*7.8 – P₂*37.9 – P₃*70.2 + R_B*89.5 = 0
ΣM_B = 0:   The addition of all moments at point B is equal to zero:
 – R_A*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:
R_B = ( 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:
R_A = ( 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: ΣF_y = 0:    R_A – P₁ – P₂ – P₃ + R_B =  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) =  + R_A
Q₁(0) =  + 15.69 = 15.69 (N)
Calculate the bending moment at a point (M):
M(x₁) =  + R_A*(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) =  + R_A – 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) =  + R_A*(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) =  + R_A – 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) =  + R_A*(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) =  + R_A – 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) =  + R_A*(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.

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.

• To obtain more accurate result from experiment , the surface /table where the experiment will be performing should be flat .If it is not flat then use a frame having legs adjusting knobs.
• The distance between left support and right support must be correct to get accurate readings. Use steel ruler for measuring distances.
• The measurement of cut section and location of hangers for carrying loads on beam should be accurate.
• Initially use spring gauge to obtain the datum value of apparatus (Hibbeler ,1985).

 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.