UKSSSC- Structure Mechanics

 INTRODUCTION

Structure mechanics is the field of applied mechanics in which we compute stress, strain, share force, bending moment, deflection etc. The purpose of determine above all quantities for finding the strength of various demands like beam, column, slab, bridge etc. to prevent the structure from failure. The field of structure mechanics is too large but here I will discussed only upto the purpose of UKSSSC JE.

Light of focus point 👇

♀ Properties

♀ Shear force & Bending Moment

♀ Moment of inertia 

♀ Slop & Deflection in Beam

♀ Column

♀ Truss

📝CHAPTER-1: PROPERTIES

◆ Class of materials

●- Elastic- stress is proportional to strain.

●- Plastic- Having a definite yield and stress is not directly proportional to strain.

●- Malleable- That can easily fold & rolled.

●- Ductile- can be converted into wire form without loosing strength.

●- Brittle- Direct failure occurs under the load.

◆ Stress and strain Theory-

● stress- stress is defined as the ratio of load to cross sectional area. If force act in perpendicularly, stress is called normal stress. Denoted by σ (sigma) and if force act in some angle, stress is called shear stress. Denoted by  τ (tou). 

σ or τ = P/A

Unit- KN/m^2 or N/mm^2

Stress also defined as various nature by act. Like tensile, compressive etc.

● strain- strain is defined as the ratio of change in length to original length. Denoted by ε.

ε = ∆L/L (Dimension less quantity)

Strain is further divide into various category in the same reference.. as like volumetric, tensile, compressive,shear etc.

◆ Hook's law- Hook's state that the stress is directly proportional to strain in elastic limits.

 It means- σ ∝ ε

 Therefore  σ = Eε , E= σ/ε 

 E = young modules of elasticity

 Value of E for mild steel is 2×10^5 N/mm2

 and E for concrete is 5000√Fck N/mm2

From the above relation, we can find the change in length ∆L, ∆L = PL/AE

◆ Shear Modulus or Modulus of rigidity (G)

G = shear stress(τ)/ shear strain(£Φ)

◆ Bulk Modulus (K) = normal stress/Vol.strain

K = σ/£v

◆ Poisson's ration(μ or 1/m) - lateral strain/longitudinal strain

👉 Relation b/w E,G,K and μ or 1/m

E= 2G(1+μ)

E= 3K(1-2μ)

E= 9KG/3K+G

◆ Ultimate stress = max. load/ cross-sectional area

◆ Working stress = ultimate stress/ factor of safety

◆ Yield stress = load at yield point/ cross sectional area

◆ Breaking stress = load at breaking point/ Cross sectional area

σ or τ

📝 CHAPTER-2 : SHEAR FORCE & BENDING MOMENTS

◆ Shear force- The algebraic sum of all the forces in vertical direction called shear force.

There are two methods for taking sign convention for shear force are- 

◆ Bending Moment- Algebraic sum of all the moments called bending moment.

There are also two methods for taking sign convention for bending moment are-

◆ Types of Beam- total six types of beam lightly described here are- 

A- Cantilever Beam

B- Simply supported Beam

C- Overhang Beam

D- Continuous Beam

E- Fixed End Beam

F- Propped Cantilever Beam 

◆ Type of Loading-

A- Point or Concentrated load

B- Distribution load

I)- udl - uniformally distributed load- W= w×l

II)- uvl - uniformally varying load- W=w. + ky

C- couple loading

◆ Cantilever Beam shear force & Bending moment- 

◆ Simply supported Beam shear force & Bending moment- 

● Where shear force  value is zero, there occurs maximum Bending moment.

● where Bending moment is zero of converting sign +ve to –ve or vice-versa, called point of contrafluxure.

● point of contrafluxure never found in cantilever beam.

📝CHAPTER-3: MOMENT OF INERTIA

◆M.O.I = moment of inertia is the product of mass of a body and square of perpendicular distance from C.G to reference line.

I = mass×y^2

Unit -mm^4, m^4, cm^4

◆Radius of gyration k = √I/A

Unit- m,cm,mm

◆Section modulus Z = I/y

Y = c.g of section from reference line

Unit- m^3, cm^3,mm^3

●Parallel axis theory Iab =Ig + Ah^2

●Perpendicular theorem = Izz =Ixx + Iyy

◆ Bending Equation- 

M/I = σ/y =E/R

Here M= Moment, I=moment of resistance

σ =stress, y= extreme line E= youn modulus

R= radius

◆Moment of Resistance or flexural strength 

M/I = σ/y

M = I/y*σ , because Z =I/y

Therefore M=Z.σ

📝 CHAPTER-4 SHEAR STRESS IN BEAM

◆ Shear stress at a section in beam varies along the depth

◆ Shear stress of rectangular section occurs maximum in the neutral axis

◆In Rectangular beam, maximum shear stress is equal to 1.5 time of average shear stress.

τmax =1.5τavg

◆In circular beam, maximum shear stress is equal to 4/3 time of average shear stress

τmax =4/3 τavg

◆shear stress at top of rectangular beam is equal to zero.

◆shear stress distribution diagram for a rectangular beam is a parabolic curve.

◆ shear stress formula for Rectangular beam-

τ = 3F/2b×d

📝CHAPTER-5 : SLOPE & DEFLECTION

There are various method to find the Slope & Deflection in beam but here I will discuss only about Moment area method or Mohr's method. This is easiest way to find slope and deflection up to diploma level. Mohr was the scientist who discovered the formula for slope and deflection through bending moment diagram, that's why called Mohr's theorems.

I)- Mohr's first theorem:- Mohr first theorem is an about the slope. Theorem stand that-

θ = Area of bending moment diagram/EI

II)- Mohr's second theorem:- Mohr second theorem ia an about the deflection. Theorem stand that-

∆ = θ× x̄ , here  x̄ = Distance of slope from free or supported end.

◆ slope and deflection for cantilever beam-

● in case of point load in free end 

We know the the bending moment diagram of this case is-

We know that the C.G of right angle triangle act h/3 from base. So in this case h=L there fore the C.G= L/3

Therefore slope = Area of B.M diagram/EI

We know that area of triangle = 1/2×base×heigth

Area of B.M diagram = 1/2×wL×L = wL^2/2

Therefore θ = wL^2/2EI

Now we know that Mohr's 2nd theorem,

∆ = θ× x̄, here x̄ = 2L/3

Therefore ∆ = wL^2/2EI × 2L/3 = wL^3/3EI

Same above solution in handwriting

● in case of udl entire beam

We know the the bending moment diagram of this case is-

We know that the C.G of right angle triangle act h/4 from base. So in this case h=L there fore the C.G= L/4

Therefore slope = Area of B.M diagram/EI

We know that area of triangle = 1/3×base×heigth

Area of B.M diagram = 1/3×wL^2/2×L = wL^3/6

Therefore θ = wL^3/6EI

Now we know that Mohr's 2nd theorem,

∆ = θ× x̄, here x̄ = 3L/4

Therefore ∆ = wL^3/6EI × 3L/4 = wL^4/8EI

Handwriting solution as same as above

● Combined case of cantilever-

Simply add above both cases value

◆ slope and deflection for simply supported beam-

● in case of centre point loading beam

We know the the bending moment diagram of this case is-

We know that the C.G of right angle triangle act h/3 from base. So in this case h=L there fore the C.G= L/3

Therefore slope = Area of B.M diagram/EI

We know that area of triangle = 1/2×base×heigth

Area of B.M diagram = 1/2×wL/4×L/2 = wL^2/16

Therefore θ = wL^2/16EI

Now we know that Mohr's 2nd theorem,

∆ = θ× x̄, here x̄ = 2/3×L/2

Therefore ∆ = wL^2/16EI × L/3 = wL^3/48EI

Complete solution as same as above

● in case of uld loading entire beam

We know the the bending moment diagram of this case is-

We know that the C.G of right angle triangle act 3h/8 from base. So in this case h=L there fore the C.G= 3L/8

Therefore slope = Area of B.M diagram/EI

We know that area of triangle = 2/3×base×heigth

Area of B.M diagram = 2/3×wL^2/8×L/2 = wL^3/24

Therefore θ = wL^3/24EI

Now we know that Mohr's 2nd theorem,

∆ = θ× x̄, here x̄ = 5/8×L/2

Therefore ∆ = wL^3/24EI × 5L/16= wL^4/384EI

Complete solution as above mentioned

📝CHAPTER-6: COLUMN 

Short column- fail under the crushing

L/d < 12 or slenderness ration less than 32

Long column- fail under the buckling

L/d > 12 or slenderness ratio greater than 120

Intermediate column- slenderness ration b/w 32 to 120. fails under crushing and buckling both.

◆ End condition for effective length of concrete column is- 

◆ Euler's Buckling or Crippling load = π^2EI/L^2

● if slenderness ratio for mild steel column is less than 80, Euler's formula is not valid

◆ Rankine's Formula for axial compressive load/stress- Pc= σ×A ,

 here σ = ultimate compressive strength

● no limit of slenderness ratio for Rankine's formula

● Rankine formula use in all types of column.

📝CHAPTER-7: TRUSS ANALYSIS

"Number of members connected to each other in such a systematic manner, called truss."

◆Truss classifieds into following three frames are-

● Perfect Frame

● Deficient Frame

● Redundant Frame

● Perfect frame- if the frame satisfied the condition of N = (2j – 3) 

Here N= total number of mamber in frame

j = no of joint in frame

● Deficient frame- if the frame satisfied the condition of N < (2j – 3)

● Reduntant frame- if the frame satisfied the condition of N > (2j – 3)

Keep in mind 👇

♂ all the joints of the truss are hinged or pin-joint.

♂ all loads are supposed to act on the joints o the frame structure.

♂ members are subjected either tension or compression only. Never subjected to any bending moment.

♂ all the members lie in one plane is called perfect frame.

♂ cantilever truss start to solve from free end.

♂ truss having span about 30m, called Bel-Fast truss.

♂ a queen post truss suitable for the span length of 8m to 12m.