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CE1201- MECHANICS OF SOLIDS/ UNIT IV/DEFLECTION OF BEAMS AND SHEAR STRESS
MECHANICS OF SOLIDS
(FOR III – SEMESTER)
UNIT - III
PREPARED BY
M.Chandrasekar,M.E
Assistant Professor/Mechanical
DEPARTMENT OF MECHANICAL ENGINEERING
CHETTINAD COLLEGE OF ENGINEERING AND TECHNOLOGY,
UNIT – IV
DEFLECTION OF BEAMS AND SHEAR STRESS
Deflection of beams – double integration method – Macaulay’s method – slope deflection using moment area method, Conjugate beam method – variation of shear stress – shear distribution in rectangular, I sections, solid circular, hollow circular sections, angle sections – shear flow.
S.NO / 2 MARKS / Page no1 / Define slope / 5
2 / Define – Deflection / 5
3 / Define flexural / 5
4 / What are the methods used to find the slope and deflection at a section of a beam? / 5
5 / Define Deflection curve / 5
6 / Write the differential equation for deflection (or) equation of elastic curve / 5
7 / What are the units of slope and deflection? / 6
8 / What are the assumptions made by double integration method? / 6
9 / Define Macaulay’s method. / 6
10 / What are the rules observed using Macaulay’s method? / 6
11 / When the Macaulay’s method is found to be very useful? / 6
12 / Name the method which employ BMD for the caluculation of slopoe and deflection / 7
13 / List out the two theorems of moment area method / 7
14 / State the Mohr’s theorem I / 7
15 / State the Mohr’s theorem II / 7
16 / What is the formula for slope and deflection at any section of a given beam by conjugate beam method? / 7
17 / Define – Spring. / 7
18 / List out some important purpose of springs / 7
19 / What is leaf spring? / 8
20 / Define stiffness of spring / 8
21 / Give the formula for deflection of a leaf spring / 8
22 / Give the formula for average shear stress of a rectangular section / 8
23 / Define Shear flow / 8
24 / What is the formula for maximum shear stress in a circular section? / 8
25 / What is the formula for total shear force acting on element area? / 9
26 / What is the strain energy stored density of a thin walled tube? / 9
S.NO / 16 MARKS / PAGENO
1 / Find the slope deflection of a given simply supported beam with UDL over the entire span of the beam using moment area method / 10
2 / Find the slope and deflection of a given beam by conjugate beam method / 11
3 / A beam of length 4.8 m and of uniform rectangular section is simply supported at its ends. It carries a uniformly distributed load of 9.375 kN/m run over the entire length. Calculate the width and depth the beam if permissible bending stress 7 N/mm2 and maximum deflection is not exceed 0.95 cm. take E = 1.05 X104 N/mm2 / 12
4 / Find the Maximum slope and deflection of simply supported beam with point load at the centre / 13
5 / A beam is made up of a chanel section 100X50 mm which has a uniform thickness of 15mm. draw the distribution of shear stress across a vertical section if the shearing force is 100 kN. Also find the ratio between the maximum shear and average shear / 14
6 / A cantilever of length 4m carries a u.d.l of 12kN/m for a length of 2.5m from fixed end and a point load of 10kN at free end. Determine the maximum slope and deflection using moment area method. Take EI = 6.3X104 kN/m2 / 15
7 / A cantilever beam of length of 3m carries three point loads of 1kN each at distances 1m, 2m and 3m from fixed end. Calculate the maximum slope and deflection in terms of flexural rigidity / 17
8 / A simply supported beam of length 3m carries a point load of 10kN at its centre. Determine the ,maximum slope and deflection of the beam. The cross-section of the beam is given below / 18
9 / A timber beam of rectangular cross section is simply supported at ends and carries a concentrated load at mid span. The maximum longitudinal stress = 12N/mm2 and maximum shear stress = 1N/mm2. find the ratio of the span to depth ratio / 19
2 MARKS QUESTIONS AND ANSWERS
1.Define slope.
The slope of point in loaded beam is defined as the angle of tangent drawn from that deflected point that makes with the beam axis.
2.Define – Deflection.
The deflection of a point in a loaded beam is defined as the vertical movement of that point, measured from beam axis before elastic loading, to elastic line after loading.
3.Define flexural
The product of Young’s modulus and area moment of inertia is called as flexural rigidity.
Flexural Rigidity = EI
Where, E = young’s modulus and I = Area Moment of inertia.
4.What are the methods used to find the slope and deflection at a section of a beam?
The methods that are used to find slope and deflection of a beam are:
- Double integration method
- Moment area method
- Macaulay’s method
- Strain energy method
- Conjugate beam method.
5.Define Deflection curve.
When the beam is loaded, the initially straight longitudinal axis is deformed into a curve called the deflection curve of the beam.
6.Write the differential equation for deflection (or) equation of elastic curve.
d2y/ dx2 = M / EI
7.What are the units of slope and deflection?
The unit of slope is radians.
The unit of deflection is mm or m.
8.What are the assumptions made by double integration method?
The assumptions made are;
- The whole deflection is due to BM only and that the deflection caused by SF is negligible.
- Deflection is small compared to cross sectional dimension of the beam.
- The beam is uniform cross section and straight before application of load.
- Modulus of elasticity in tension are equal.
9.Define Macaulay’s method.
Macaulay’s method is defined as a convenient method for determining deflections of a beam, subjected to point loads or in general discontinuous loads.
10.What are the rules observed using Macaulay’s method?
The rules observed using Macaulay’s method are:
- Always take origin on the extreme of the beam.
- Take left clockwise moment as negative and left counter clockwise moment as positive.
- Take a section in the least segment of the beam and take moment from the left.
- If the beam carries a UDL, extend it upto the extreme right and superimpose a UDL equal and opposite to that, which has been added while extending the given UDL.
- When the Macaulay’s method is found to be very useful?
1.When the problem of deflection in beams are a bit tedious and laborious.
2.When the beam is carrying several point loads.
3.It is used to find deflection where BM is discontinuous.
12.Name the method which employ BMD for the calculation of slope and deflection.
Moment area method employs BMD for the calculation of slope and deflection.
13.List out the two theorems of moment area method.
The two theorems are:
Mohr’s theorem I (Moment area theorem)
Mohr’s theorem II ( Moment area theorem)
14.State the Mohr’s theorem I
Mohr’s theorem I states that “ the change of slope between any two points on an elastic curve on an elastic curve is equal to the net area of BM diagram these points divided by EI”.
15.State the Mohr’s Theorem II.
Mohr’s theorem II states that “the interrupt vertical reference line of tangents at any points on an elastic curve is equal to the moment of BM diagram between these points about the reference line divided by EI”.
16.What is the formula for slope and deflection at any section of a given beam by conjugate beam method?
Slope = Wl2/ 16EI
Deflection = Wl3 / 48EI
17.Define – Spring.
A spring defined as an elastic body, whose function is to distort and absorb energy when loaded and recover its original shape release the stored energy, when the load is removed.
18.List out some important purpose of springs.
Springs are used to;
- Apply forces
- Measure Forces
- Store energy
- Absorb shock and vibration
- Control motion.
19.What is leaf spring?
Leaf spring consists of a number of parallel strips of a metal, having different length but same width and placed one over the other in laminations. The leaves are held together with clamp and bolt.
20.Define stiffness of spring
Stiffness of spring is defined as the load required to produce unit deflection.
S = W/δ; where, W = applied load, δ = deflection.
21.Give the formula for deflection of a leaf spring.
δ = 3Wl3/ 8Enbt3
where, W = load acting on the spring
l = span of the spring
E = young’s modulus
b = width of leaf spring
t = thickness of leaf spring
22.Give the formula for average shear stress of a rectangular section.
qave = shearing force / Area = F/ bd
23.Define Shear flow.
Shear flow is defined as the product of shear stress and thickness t of the tube is the same at every point of the cross section. It is denoted by f.
f = t = constant
24.What is the formula for maximum shear stress in a circular section?
The formula is given as,
q max = (4/3) q ave.
25.What is the formula for total shear force acting on element area?
The formula is given is
dT = r f d s.
where, r – perpendicular distance from centre point 0 to line of action of force.
F – shear flow.
26.What is the strain energy stored density of a thin walled tube?
Strain Energy density= 2 / 2G.
16 MARKS QUESTION AND ANSWERS
1.Find the slope deflection of a given simply supported beam with UDL over the entire span of the beam using moment area method.
Consider a simply supported beam AB of length L carrying UDL of w/unit length over the entire span
(i). Now using the Mohr’s theorem for slope we get,
Slope at A = ( Area of BM diagram between A and C)/ EI
= Area of parabolic ACD / EI
= [(2/3)XACXAD]/ EI
= [(2/3)X(L/2)X(wL8)] / EI
= wL3/24EI
slope at A iA= wL3/24EI
Now using the mohr’s theorem for deflection, we get
y = A x / EI
where A = Area of BM diagram between A and C
= wL3/24
x = Distance of C.G of Area A from point A
= 5/8 X AC = (5/8) X (L/2)
yc = [(wL3/2) X 5L/16)] / EI
yc = 5 wL4/384EI.
- Find the slope and deflection of a given beam by conjugate beam method.
Consider a simply supported beam AB of length l carrying point load middle load W at C as shown in fig. now draw the conjugate beam ABC simply supported at A and B, and loaded with the bending moment diagram, as shown in fig C.
Total load on conjugate beam = Area of the load diagram.
= 1/2 X l X Wl /4EI = Wl2/8EI
Reaction at each support for the conjugate beam
RA1= RB1= (½ ) X (Wl2/8EI) = Wl2/16EI
But shear force at any section of the conjugate beam equal to the slope of the real beam.
Hence iA = - Wl2/16EI ( Negative sign due to left upward)
Similarly iB = Wl2/16EI
Deflection at C for the given beam = BM at C for the conjugate beam
= RA1 X l/2 – ( ½ X l/2 X Wl/4EI) X ( 1/3 X l/2)
= Wl3/48 EI.
But the BM at any section of the conjugate beam is equal to the deflection of the real beam.
Hence yc = Wl3/ 48EI.
- A beam of length 4.8 m and of uniform rectangular section is simply supported at its ends. It carries a uniformly distributed load of 9.375 kN/m run over the entire length. Calculate the width and depth the beam if permissible bending stress 7 N/mm2 and maximum deflection is not exceed 0.95 cm. take E = 1.05 X104 N/mm2.
Solution:
Length L= 4.8m = 4800mm
u.d.l = 9.375 kN/m
Bending stress f = 7 N/mm2
Central deflection yc = 0.95 cm = 9.5mm
Value of E = 1.05 X 104N/mm2
Let b = width of beam in mm
d = depth of the beam in mm
Moment of inertia (I) = bd3/12
Using equation
Yc = (5/384) X (wl4/ EI)
9.5 = (5/384) X [(9.375X48004) / 1X104 X (bd3/12) ]
bd3 = (5/384) X [(9.375X48004 X12) / 1X104 X 9.5 ]
818.52X107mm4 ------1.
The maximum BM for a simply supported beam carrying a uniformly distributed load is given by
M = wl2/8
= 9.375X48002/ 8
= 27,0000,000 Nmm.
Now using the bending equation as
M/I = f/y
27000000 / (bd3/12) = 7 / (d/2)
27000000 X12 / bd3 = 14/d
bd2 = 23142857.14 mm2 ------2
dividing Eq.(i) by Eq.(ii)
d = 818.52X107/ 2142857.14
= 335 mm
substituting the value of ‘d’in Eq (ii) we get,
b x(353)2 = 23142857.14
b = 185.72mm
4.Find the Maximum slope and deflection of simply supported beam with point load at the centre.
Consider a simply supported beam AB of length l carrying point load W at the centre of the beam C as shown in fig.
Consider a section X – X at a distance x from A,
The BM at this section is given by
Mx = RA x = (W/2)x
EI (d2y/dx2) = Wx/2
Integrating the above equation
EI (dy/dx) = (W/2) X (x2/2) + C1------2
Where C1 is constant of integration
W.K.T, x = l /2 ; dy/dx = 0. (slope)
Sub the above equation (2).
0 = (W/4) x (l/2)2 + C1
C1 = - Wl2/16
Again sub. The C1 value in equation 2
EI (dy/dx) = Wx2/4 – Wl2/16------3
The above equation is known as slope equation.
Sub at x = 0 ( max. slope)
(dy/dx) = - Wl2/ 16EI.
iA = iB = - Wl2/ 16EI ( negative sign indicates anti clock wise direction)
Deflection at any point is obtained by integrating the slope equation.
EI y = (W/4) X (x3/3) – (Wl2 /16) x + C2 ------4.
We know At x = 0 ; y =0, sub. These values in equation 4 we get.
EI y = (Wx3/12) – (Wl2/16)x
At x = l/2 max deflection will occur
EI yc = (W/12) (l/2)3 - (Wl2/16)(l/2)
= - Wl3/ 48.
yc = - Wl3/ 48EI.
5.A beam is made up of a channel section 100X50 mm which has a uniform thickness of 15mm. draw the distribution of shear stress across a vertical section if the shearing force is 100 kN. Also find the ratio between the maximum shear and average shear.
Shear Force S.F= 100kN = 100X103N
Moment of inertia of the channel section with reference to neutral axis.
I = (50X1003/12) – (35X703/12)
I = 3.17X106mm4
To determine the shear stress at the flange section:
Area of flange = 50X15 = 750mm2.
Distance of centre of gravity of the flange section from neutral axis.
= (100/2) – (15/2) = 42.5mm
shear stress τ = Fay / Ib
= (100X103X750X42.5) / (3.17X106X50) = 20.11 N/mm2
to determine the shear stress at the junction of flange and web
τ = Fay / Ib
= (100X103X750X42.5) / (3.17X106X15) = 67.03 N/mm2
(Note: the value of b is taken as 15 times the width the web. All other values are same as in the previous calculation for τ at flange.
This Value 67.03 N/mm2 can be cross checked by multiplying the shear stress at a flange with the width of flange (B) disided byt eh width of web (b).
τ = 20.11X(50/15) = 67.03 N/mm2
to determine the shear stress at web:
Area of the section till neutral axis
Area of flange = 50X15 = 750 mm2
Area of web till neutral axis = (50 – 51)X15 = 525mm2
Distance of centre of gravity from neutral axis
y = ( 50 – (15/2) ) = 42.5mm (flange)
y = [(50 -15)/2] = 17.5mm (web)
Due to symmetry the shear stress will be maximum at neutral axis
τ max = QAy / Ib
= 100X103 [ (750X42.5) + (525X17.5)] / ( 3.17X106X15)
Total area of the section = (50X100) – (35x70) = 2550mm2
τ max = SF/Area of section = 100X103/ 2550 = 39.22 N/mm2
Ratio between τ max and τ ave
τ max / τ ave = 86.36 / 39.22 = 2.2
The ration between τ max and τ ave is 2.2
6.A cantilever of length 4m carries a u.d.l of 12kN/m for a length of 2.5m from fixed end and a point load of 10kN at free end. Determine the maximum slope and deflection using moment area method.Take EI = 6.3X104 kN/m2.
Given:
Length of cantileverL = 4m.
EI = 6.3X104 kN/m2.
Now B.M at fixed end due to u.d.l = 12X2.52 / 2 = 37.5 kNm.
Centroidal distance of the above B.M diagram from free end
= (3/4)X2.5 + 1.5 = 3.375 m
B.M due at fixed end due to point load = 10X4 = 40kNm.
Centroidal distance above B.M diagram form free end = (2/3)X4 = 2.67m
Maximum slope:
max = 1/EI ( Area of B.M diagram only for u.d.l +
Area of B.M diagram only for point load)
= (1/6.3X104) X[ 1/3X2.5X37.5 + 1/2X4X40 ] = 0.0018 radian
Maximum deflection:
Ymax= 1/EI ( Moment of the B.M diagram for u.d.l about B +
Moment of the B.M diagram for point load about B)
= ( 1/6.3X104)X [ 1/3X2.5X37.5X3.375 + 1/2X4X40X2.67]
= 5.06X10-3m = 5.06mm.
7.A cantilever beam of length of 3m carries three point loads of 1kN each at distances 1m, 2m and 3m from fixed end. Calculate the maximum slope and deflection in terms of flexural rigidity.
Given:
Span L= 3m
Point load applied W = 15kN
Length of AB, a1= 1m
Length of AC, a2= 2m
Let EI flexural rigidity of the cantilever.
For the given cantilever, maximum slope at free end,
max = Slope at due to ( point load at B+Point load at C + Point load at D)
= (Wa12/ 2EI)+(Wa22/ 2EI)+(WL2/2EI)
= (15X12/2EI)+(15X22/2EI)+(15X32/2EI)+(7/EI)
Again, maximum deflection at free end,
y max= { (Wa13/3EI) + (Wa12/2EI)(L-a1)} +
{ (Wa13/3EI) + (Wa12/2EI)(L-a2)} + WL3/3EI
= { (15X13/3EI) + (15X12/2EI)(3-1)} +
{ (15X23/3EI) + (15X22/2EI)(3-2)} + 15X33/3EI
= 225/EI.
8.A simply supported beam of length 3m carries a point load of 10kN at its centre. Determine the ,maximum slope and deflection of the beam. The cross-section of the beam is given below.
Given:
Central load,W= 10kN.
Span of the beam L= 3m.
Moment of inertia of the steel section about the neutral axis,
Is =50X1003/ 12 = 4.17X106 mm4.
Moment of inertia of the concrete section about the neutral axis,
Ic = 2X [] = 29.06X106mm4
Flexural rigidity of the entire beam section,
ΣEI = EsIs + EcIc = (2X105) X (4.17X106)+ (2X104) X(29.06X106)
= 1.415X1012Nmm2 = 1415kNm2.
Maximum slope of the beam= slope at its end supports.
= (WL2/ 16ΣEI) = (10X32)/ (16X1415) = 0.00397 radians
= 0.00397 X (180/п) = 0o14’ (approx)
Maximum deflection of the beam= deflection at its mid span.
= (WL3/ 48ΣEI) = (10X33) / (48X1415) = 3.97X10-3m = 3.97mm
9.A timber beam of rectangular cross section is simply supported at ends and carries a concentrated load at mid span. The maximum longitudinal stress = 12N/mm2 and maximum shear stress = 1N/mm2. find the ratio of the span to depth ratio.
Given:
Let Breadth of the beam= b
Depth of the beam= d
Span of the beam= l
Maximum bending moment due to concentrated load W at the centre,
M = Wl/4
Moment of inertia I = bd3/12.
Maximum bending stress f = 12N/mm2
Maximum shear stress q = 1N/mm2
Maximum S.F at support = W/2
Average shear stress, q ave = S.F/Area = W/2bd
Maximum shear stress q = 1.5X qave = 1.5 X (W/2bd) = 0.75W/bd
By problem,
q = 0.75W/bd = 1
Or, 0.75W = bd
W/bd = 1.33 ------1
Again from the bending equation,
M/I = f/y
[ (Wl/4) / (bd3/12)] = [12 / (d/2) ]
(Wl / bd3 ) x (12/4) = 12X2 / d
W/bd = 8d/l ------2
Substituting the value of W/bd from equation
8d / l = 1.33
l /d = 8 / 1.33
Required span to depth ratio= 6:1