Perfect Truss (Stable Truss) Condition: n = 2j – R j = Number of joints n = Number of members R = No. of reaction components (external) Imperfect Truss (unstable) n ≠ 2j – R Deffecient Truss n < 2j - R Redundant Truss n > 2j – R Table to remember: $\table Sr.No. ,Member ,Magnitude,Nature$
Rectangular Components of a Force in Space $F_x$ = F cos$θ_x$ , $F_y$ = cos$θ_y$ , $F_z$ = cos$θ_z$ $\ov{F}$ = $F_x$i + $F_y$j + $F_z$k l = cos$θ_x$ = ${F_x}/{F}$ Similarly for others. $l^2 + m^2 + n^2 = 1$ Unit Vector $\ov{F}$ = F $\ov{e}$ where, $\ov{e}$ = unit vector in the direction of force F = (cos$θ_x$)i + (cos$θ_y$)j + (cos$θ_z$)k Unit Vector when Force is Specified by Two Points $\ov{F}$ = F $[{(x_2 – x_1)i+(y_2 – y_1)j+(z_2 – z_1)k}/{√{(x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2}}]$ Components of Force when Orientation of Planes ...
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Types of Beams (1)Simply supported beam, (2)Overhanging beam (3)Cantilever beam, (4)A beam with hinge and roller supports (5)A fixed beam, (6)A continuous beam (7)Prop Cantilever Types of loads on the beam (1)Point load or Concentrated load Always represented in Newton or kN (2)Uniformly Distributed Load (UDL) Always represented in N/m or kN/m Converted in point load by taking product of intensity of udl and spreading distance. This acts on centre of udl (3)Uniformly Varying Load (UVL) $\table Shape,\text"Point load",\text"Point of action (From left & Right)";\text"Right triangle", 1/2 wl,{2l}/3 \text" &" l/3;\text"Trapezoidal",w_1l \text" & " (w_2 – w_1)l/2,l/2 \text"&" 2/3 l;\text"Pyramidal ...
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Conditions of Equilibrium (a) For co-planar concurrent forces ∑$F_x$ = 0 and ∑$F_y$ = 0 Thereforce, R = 0 (b) For co-planar non-concurrent forces ∑$F_x$ = 0 and ∑$F_y$ = 0 and ∑M = 0 Types of supports and Corresponding Reactions: $\table Support\/Connection,Specification,No. of unknowns;Rollers,\text"Known reaction which is perp. to plane of roller",One;\text"Smooth Surface",\text"Reaction is perp. to the surface",One;\text"Rough Surface",\text"Two reactions components with unknown directions",Two;\text"Smoth pin or Hinge",\text"Two reaction components with unknown directions",Two;\text"Flexible cord,rope or cable of negligible weight",\text"One axial force acting away from body",One;\text"Fixed support",\text"Two reaction components and one moment with all components unknown in directions",Three;\text"A smooth pin in ...
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Section I – Impulse – Momentum Theorem Linear Momentum: Linear momentum = mass * velocity = m * v Momentum is vector quantity and its SI unit is kg.m/sec. Angular Momentum Angular momentum = I . w Angular momentum is vector quantity and its SI unit is kg . $m^2$ rad/s Linear Impulse If force is constant, Impulse = F Δt Impulse is a vector quantity and its SI unit is N.sec Angular Impulse If torque is constant, Angular Impulse = T Δt It is a vector quantity and its SI unit is N.m.sec Impulsive – Momentum Theorem for Particles ...
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The S.I. unit of work done is N.m or Joule W.D. is a scalar quantity has only magnitude but no direction. Work-Energy Principle for a Particle $U_{1→2} = KE_2 – KE_1$ where, $U_{1→2}$ = Algebric sum of work done by all forces from position 1 → 2 $KE_1$ = KE of system at position 1 $KE_2$ = KE of system at position 2 Work Done Calculations for Different Forces: W.D. by External Force: (constant) W.D. = F * S (Positive if work is done in the direction of force. W.D. by Variable Force: (When F v/s S curve is given) ...
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Steps for application: 1. Draw F.B.D of particle 2. Apply D’Alembert’s force m x $a_N$ = ${m v^2 }/{ρ}$ in opposite direction of $a_N$ (i.e. away from centre of curvature) 3. Apply Σ$F_N$ = 0 Some important concepts: (a) When particle is moving along a horizontal circle, mark y axis and normal axis in F.B.D. (b) When particle is moving along a vertical circle, mark normal and tangential axis in F.B.D. (c) When a particle attached by a string and whirled in verticle circle $T_{top} = {mv^2}/{r} – mg$ $T_{bottom} = {mv^2}/{r} + mg$ (d) When an aeroplane flying along ...
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For projectile on H.P. ⋆ Time of flight $t = {2u sinα}/{g}$ ⋆ Horizontal Range $R = {u^2 sin2α}/{g}$ when α = 45°, Range is maximum: $R_{max} = {u^2}/{g}$ ⋆ Max Height $H_{max} = {u^2 sin^2 α }/{2g}$ ⋆ Equation of path (Trajectory) y = x tanα – ${g x^2}/{2 u^2 cos^2 α}$ ⋆ When projectile is projected with a horizontal velocity from top of a tower of heigh h $y = – {g x^2}/{2 u^2}$ [as α = 0] -h = $-{g x^2}/{2 u^2}$ Horizontal distance, $x = v_o √{{2h}/{g}}$ Time of flight, t = $√{{2h}/{g}}$ ⋆ Same range, ...
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$\table Direction,Velocity-Component,Acceleration-Component;x,v_x={dx}/{dt},a_x={dv_x}/{dt};y,v_y = {dy}/{dt},a_y = {dv_y}/{dt};Tangential,v_T = V, a_T = {dv}/{dt};Normal, v_N = 0, a_N = {v^2}/{ρ};Radial, v_r = r’,a_r = r” – r θ’^2;Transverse, v_θ = r θ’, a_θ = 2r’θ’ + r θ”$ Note: ⋆ $a_T$ represents change in speed while $a_N$ represents change in direction. ⋆ At point of inflection ρ -> ∞ so $a_N$ = 0 ⋆ Always use kinematic equations with $a_T$ in curvilinear motion Methods to find Radius of curvature (1) When speed v and $a_N$ are known $|ρ| = {v^2}/{a_N}$ (2) When path equation y = f(x) is given, $|ρ| = {[a + ...
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No important or extra ordinary formula needed in this chapter