IDE S8 Test 6 Name:. Sketch the ground reactions on the diagram and write the following equations (in units of kips and feet). (8 points) ΣF x = = ΣF y = = ΣM A = = (counter-clockwise as positie). Sketch the ground reactions on the diagram and write the following equations (in units of kips and feet). (8 points) ΣF x = = ΣF y = = ΣM A = = (counter-clockwise as positie)
. For the beam and loading shown, derie an expression for the reaction at support A. Assume that EI is constant for the beam. ( points) A y =. A timber [E =,8 ksi] beam is loaded and supported as shown. The cross section of the timber beam is in. wide and 8 in. deep. The beam is supported at B by a.5-inch-diameter steel [E =, ksi] rod, which has no load before the distributed load is applied to the beam. After a distributed load of 9 lb/ft is applied to the beam, determine the force carried by the steel rod. ( points) B = kips
5. What is the correct alue of the normal stress shown on the stress element? ( points) - - + + 6. What is the correct alue of the shear stress shown on the stress element in the preious problem? ( points) - - + + 7. The stresses shown act at a point in a stressed body. Determine the normal and shear stresses at this point on the inclined plane shown. Show these on an appropriate sketch. ( points)
8. Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and ertical planes at the point are shown. Determine the principal stresses acting at the point. Show these stresses on an appropriate sketch. ( points) 9. Mohr s circle is shown for a point in a physical object that is subjected to plane stress. Determine the following alues. (6 points) σ x = ksi σ y = ksi τ xy = ksi p = deg σ = ksi σ = ksi τ in-plane = ksi τ absolute = ksi
. Sketch Mohr s circle for the following state of stress. ( points) σ x = MPa σ y = -5 MPa τ xy = 5 MPa. Determine the following alues from the Mohr s circle in the preious problem. ( points) σ center = MPa radius = MPa σ = MPa σ = MPa τ in-plane = MPa
TRIGONOMETRY sin = opp / hyp cos = adj / hyp tan = opp / adj STATICS Symbol Meaning Equation Units x, y, z centroid position y = Σy i A i / ΣA i in, m I moment of inertia I = Σ(I i + d i A i ) in, m J N V M J solid circular shaft polar = πd / moment J of inertia hollow circular shaft = π(d o - d i )/ normal force shear force bending moment equilibrium V = -w(x) dx M = V(x) dx ΣF = ΣM (any point) = in, m lb, N lb, N in-lb, Nm lb, N in-lb, Nm MECHANICS OF MATERIALS Topic Symbol Meaning Equation Units σ axial = N/A σ, sigma normal stress τ cutting = V/A σ bearing = F b /A b axial ε, epsilon normal strain ε axial = ΔL/L o = δ/l o ε transerse = Δd/d in/in, m/m γ, gamma shear strain γ = change in angle, γ = c rad E Young's modulus, modulus of elasticity σ = Eε (one-dimensional only) G shear modulus, modulus of rigidity G = τ/γ = E / (+ν) ν, nu Poisson's ratio ν = -ε'/ε δ, delta deformation, elongation, deflection in, m δ = NL o /EA + αδtl o α, alpha coefficient of thermal expansion (CTE) in/inf, m/mc F.S. factor of safety F.S. = actual strength / design strength
Topic Symbol Meaning Equation Units τ, tau shear stress τ torsion = Tc/J torsion φ, phi angle of twist φ = TL/GJ rad, degrees, theta angle of twist per unit length, rate of twist = φ / L rad/in, rad/m P power P = Tω r T = r T watts = Nm/s hp=66 in-lb/s ω, r ω angular speed, speed of rotation = r ω rad/s omega f frequency ω = πf Hz = re/s K stress concentration factor τ = KTc/J flexure σ, sigma σ, sigma τ, tau normal stress σ beam = -My/I composite beams, n = E B /E A σ A = -My / I T σ B = -nmy / I T shear stress τ beam = VQ/Ib where Q = Σ(y bar i A i ) q shear flow q = V beam Q/I = nv fastener /s or y beam deflection = M(x) dx / EI in, m Topic Equations Units stress transformation planar rotations σ u +σ y )/ + (σ x )/ cos() + τ xy sin() σ +σ y )/ - (σ x )/ cos() - τ xy sin() τ u = -(σ x )/ sin() + τ xy cos() principals and in-plane shear tan( p ) = τ xy / (σ x ), s = p ± 5 o σ, +σ y )/ ± sqrt { [ (σ x )/ ] + τ xy } τ = sqrt { [ (σ x )/ ] + τ xy } = (σ -σ )/ σ ag +σ y )/ = (σ +σ )/ strain transformation planar rotations ε u = (ε x )/ + (ε x )/ cos() + γ xy / sin() ε = (ε x )/ - (ε x )/ cos() - γ xy / sin() γ u / = -(ε x )/ sin() + γ xy / cos() ε z = -ν (ε x + ε y ) / (-ν) principals and in-plane shear tan( p ) = γ xy / (ε x ), s = p ± 5 o ε, = (ε x )/ ± sqrt { [ (ε x )/ ] + (γ xy /) } γ / = sqrt { [ (ε x )/ ] + (γ xy /) } ε ag = (ε x )/ in/in, m/m Hooke's law D strain to stress σ = Eε D strain to stress σ x = E(ε x +νε y ) / (-ν ) σ y = E(ε y +νε x ) / (-ν ) τ xy = Gγ xy = Eγ xy / (+ν) D stress to strain ε x -νσ y ) / E ε y = (σ y -νσ x ) / E ε z = -ν(ε x ) / (-ν) γ xy = τ xy /G = (+ν)τ xy / E in/in, m/m pressure σ spherical = pr/t σ cylindrical, hoop = pr/t σ cylindrical, axial = pr/t imum principal stress theory σ, < σ yp σ radial, outside = σ radial, inside = -p failure theories imum shear stress theory τ <.5 σ yp
CANTILEVER BEAMS Beam Slope Deflection Elastic Cure PL EI PL EI Px = ( L x) 6EI EI EI Mx EI 6EI 8EI wx L Lx+ x EI (6 ) EI EI wx L L x+ Lx x LEI ( 5 )
SIMPLY SUPPORTED BEAMS Beam Slope Deflection Elastic Cure PL 6EI PL 8EI Px (L x ) 8EI for x L Pb L 6LEI Pa L =+ 6LEI ( b ) ( a ) EI =+ 6EI Pba ( ) 6LEI x= a L b a 6LEI 9 EI @ x= L Pbx ( L b x ) for x a Mx L Lx+ x 6LEI ( ) EI 5 8EI wx L Lx + x EI ( ) wa ( L a) LEI wa =+ L a LEI 7 6EI =+ 5EI ( ) wa a al L x= a LEI + ( 7 ).65 EI @ x=.59l wx ( a a L+ a L + a x LEI alx + Lx ) for x a wa a L+ L x+ a x Lx + x LEI for a x L ( 6 ) wx L L x + x 6LEI (7 )