1. In calculating the shear flow associated with the nail shown, which areas should be included in the calculation of Q? (3 points) Areas (1) and (5)

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In calculating the shear flow associated with the two nails shown, which areas should be included in the calculation of Q?

1 IDE 0 S08 Test 5 Name:. In calculating the shear flow associated with the nail shown, which areas should be included in the calculation of Q? ( points) Areas () and (5) Areas () through (5) Areas (), () and (4) Areas (), (), (4) and (5). In calculating the shear flow associated with the two nails shown, which areas should be included in the calculation of Q? ( points) Areas () through (6) Areas (), (), (5) and (6) Area (4) Areas (), (4) and (5). What is the alue of Q needed to determine the shear force acting on nail B? (6 points) 90,65 mm,40,65 mm 875,000 mm 750,000 mm

4. What is the alue of Q needed to determine the shear force acting on the two

If boards () and () were glued to board () instead of nailed, what would be

psi 98.5 psi 6. The allowable load for each weld is.

2 4. What is the alue of Q needed to determine the shear force acting on the two nails shown? (6 points) 480,000 mm,00 mm 96,000 mm 44,000 mm 5. If boards () and () were glued to board () instead of nailed, what would be the shear stress in the glue if V =,5 lb? (6 points) 6.5 psi 49.6 psi 99. psi 98.5 psi 6. The allowable load for each weld is.4 kip/inch in the longitudinal direction. What is the imum allowable shear force V? (8 points) 0.5 kips.8 kips 0.9 kips 5.6 kips

3 7. A.00-inch-diameter solid steel shaft supports loads P A = 00 lb and P D = 500 lb. Assume L = 5 in., L = 6 in., and L = 8 in. Determine the alues for V, Q, I, and b that would be used in the imum horizontal shear stress anywhere in the shaft. (0 points) V = lb Q = in. Ι = in. 4 b = in.

4 8. For the beam and loading shown, use the integration method to determine the equation of the elastic cure for segment AB of the beam. Do NOT sole for the integration constants. (8 points) 9. List the boundary, continuity, and/or symmetry conditions that could be used to sole the integration constants in the preious problem. (6 points)

5 0. Determine the beam deflection at point H. Assume that EI = 40,000 kn-m is constant for the beam. (6 points) H = mm. Determine the beam deflection at point H. Assume that EI = 40,000 kn-m is constant for the beam. (0 points) H = mm

6 . Determine the beam deflection at point H. Assume that EI = 40,000 kn-m is constant for the beam. (8 points) H = mm. Determine the beam deflection at point C. Assume that EI = 40,000 kn-m is constant for the beam. (0 points) C = mm

7 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 4, m 4 J N V M J solid circular shaft polar = πd 4 / moment J of inertia hollow circular shaft = π(d 4 o - d 4 i )/ normal force shear force bending moment equilibrium V = -w(x) dx M = V(x) dx ΣF = 0 ΣM (any point) = 0 in 4, m 4 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

8 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=6600 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 x +σ y )/ + (σ x )/ cos() + τ xy sin() σ = (σ x +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 ± 45 o σ, = (σ x +σ y )/ ± sqrt { [ (σ x )/ ] + τ xy } τ = sqrt { [ (σ x )/ ] + τ xy } = (σ -σ )/ σ ag = (σ x +σ y )/ = (σ +σ )/ strain transformation planar rotations ε u = (ε x +ε y )/ + (ε x )/ cos() + γ xy / sin() ε = (ε x +ε y )/ - (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 ± 45 o ε, = (ε x +ε y )/ ± sqrt { [ (ε x )/ ] + (γ xy /) } γ / = sqrt { [ (ε x )/ ] + (γ xy /) } ε ag = (ε x +ε y )/ 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 = (σ x -νσ y ) / E ε y = (σ y -νσ x ) / E ε z = -ν(ε x +ε y ) / (-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 = 0 σ radial, inside = -p failure theories imum shear stress theory τ < 0.5 σ yp

9 CANTILEVER BEAMS Beam Slope Deflection Elastic Cure PL EI PL EI Px = ( L x) 6EI ML EI ML EI Mx EI 6EI 4 8EI wx L Lx+ x 4EI (6 4 ) 0 4EI 4 0 0EI wx L L x+ Lx x 0LEI 0 (0 0 5 )

SIMPLY SUPPORTED BEAMS Beam Slope Deflection Elastic Cure PL 6EI PL 48EI Px (L 4 x ) 48EI for 0 x L Pb L 6LEI Pa L =+ 6LEI ( b ) ( a ) ML EI ML =+ 6EI Pba ( ) 6LEI x= a L b a 6LEI ML 9 EI @ x= L Pbx

10 SIMPLY SUPPORTED BEAMS Beam Slope Deflection Elastic Cure PL 6EI PL 48EI Px (L 4 x ) 48EI for 0 x L Pb L 6LEI Pa L =+ 6LEI ( b ) ( a ) ML EI ML =+ 6EI Pba ( ) 6LEI x= a L b a 6LEI ML 9 x= L Pbx ( L b x ) for 0 x a Mx L Lx+ x 6LEI ( ) 4EI EI wx L Lx + x 4EI ( ) wa ( L a) 4LEI wa =+ L a 4LEI EI 0 =+ 45EI ( ) wa a al L x= a 4LEI + ( 7 4 ) x= 0.59L 4 wx 4 ( a 4a L+ 4a L + a x 4LEI 4 alx + Lx ) for 0 x a wa a L+ L x+ a x Lx + x 4LEI for a x L ( 4 6 ) wx L L x + x 60LEI (7 0 )

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1. Sketch the ground reactions on the diagram and write the following equations (in units of kips and feet). (8 points) ΣF x = 0 = ΣF y = 0 =

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