MTE STATICS Example Problem P. Beer & Johnston, 004 by Mc Graw-Hill Companies, Inc. The structure shown consists of a beam of rectangular cross section (4in width, 8in height. (a Draw the shear and bending moment diagrams for the beam AB (b Determine the maximum flexural stress in sections just to the left and just to the right of point D Solution: (a Shear and bending moment diagrams FBD of the entire beam + = 0 : B ( 3in ( 480lb( 6in ( 400lb( in = 0 By M A =+ 365lb B = 365lb y y + = 0 : ( 480lb ( 6in + ( 400lb ( 0in A( 3in = 0 M B A =+ 55lb A = 55lb F x = 0 : B x = 0
MTE STATICS The 400-lb load is replaced by an equivalent force-couple system acting on the beam at point D. F D = 400 lb M D = ( 4in( 400lb = 600lb-in Shear Force and Bending Moment Diagrams: FBD from A to C: 0 < x < in + F y = 0 : 55 40x V = 0 V = 55 40x + M = 0 : ( 55x + 40x x + M = 0 M = 55x 0x FBD from C to D: in < x < 8in + F y = 0 : 55 480 V = 0 V = 35lb + M = 0 : 55x + 480( x 6 + M = 0 M = ( 880 + 35x lb in
MTE STATICS 3 FBD from D to B: 8in < x < 3in + F y = 0 : 55 480 400 V = 0 V = 365lb + M 3 = 0 : 55x + 480( x 6 600 + 400( x 8 + M = 0 M = ( 680 365x lb in now plotted: The shear and bending diagrams can be Between A and C: ( 0 < x < in Shear: V = 55lb at x = 0 V = 55 480 = 35lb at x = in Bending Moment: M = 0 at x = 0 M = 55( 0( = 3300lb in Between C and D (in < x < 8in Shear: V = 35lb Bending Moment: M = 880 + 35( = 3300lb in at x = M = 880 + 35( 8 = 350lb in at x = 8in at x = in
MTE STATICS 4 Between D and B: (8in < x < 3in Shear: V = 365lb Bending Moment decreases linearly: M = 680 365( 8 = 50lb in at x = 8 M = 0 at x = 3in (b Maximum flexural stress to the left and right of point D The section modulus for a rectangular cross section is given by: I S = c 3 bh h where I = and c = Hence: bh ( 4in ( 8in 3 S = S = = 4.67in 6 6 To the left of D: M = 350lb-in M 350 lb σ m = = = 8.3 S 4.67 in To the right of D: M = 50lb-in M 50 lb σ m = = =.8 S 4.67 in
MTE STATICS 5 Example Problem P. The figure shows a cross section through a tank with a rectangular sloping gate that is 0m long ( L = 0m and 4m wide ( b = 4m, width is measured into the page. The gate is hinged along its top edge and held closed by a force at its bottom edge at A. Friction in the hinge and the weight of the gate can be neglected. Goal: Find the total load on the gate and the magnitude of the force acting on the bottom edge of the gate. Solution: We want to find the magnitude of the force acting at A needed to keep the gate closed FBD of the sloped gate with the linear fluid force distribution At point B: The load per unit length is the hydrostatic pressure at B multiplied by the width b of the gate: ω = ρ gh b min liq B ω min = kg m 000.8 ( 7.5m( 4m 3 m s kn ω min = 4.3 m
MTE STATICS 6 At point A: The load per unit length is the hydrostatic pressure at A multiplied by the width b of the gate: ω = ρ gh b max liq A kg m ω max = 000.8 (.5m( 4m 3 m s ω = max kn 40.5 m Equivalent fluid force R: Rectangle A : F = ( 0m ( ω min kn F = 0m 4.3 = 43kN m Point of application: d 0 = = 5.0m Rectangle A : F ( 0m ( ωmax ωmin = kn 0m 6. m F = = 8kN Point of application: d 0 = = 3.33m 3 The magnitude of the total force is then: R = F + F
MTE STATICS 7 R = 43kN + 8kN = 34kN Use equilibrium equation to find the magnitude of the force at A: + = 0 : M B 43( 0 5 + 8( 0 3.33 F A ( 0 = 0 F A = 6kN Example Problem P.3 Beer & Johnston, 004 by Mc Graw-Hill Companies, Inc. A thin steel plate which is 4mm thick is cut and bent to form the machine part shown. Goal: Knowing that the density of steel is kg ρ = 7850 m 3, determine the moment of inertia of the machine respect to the coordinate axes. Solution: Step : Computation of masses Semicircular plate: V = πr t 6 3 V = π ( 0.08 ( 0.004 = 40. 0 m m = ρv 3 6 m = 7.85 0 40. 0 = 0.356kg
MTE STATICS 8 Rectangular plate: 6 3 V = ( 0.( 0.6( 0.004 = 8 0 m 3 6 m = 7.85 0 8 0 =.005kg Circular plate: V3 = πa t ( 6 3 V3 = π 0.05 ( 0.004 = 3.4 0 m 3 6 m3 = 7.85 0 3.4 0 = 0.466kg Step : Moments of Inertia Semicircular plate: For a circular plate of mass m and radius r : = mr, = Iz = mr 4 Because of symmetry: = ( mr, = Iz = ( mr 4 Since the mass of the semicircular plate is m = m, we have: ( 0.356 ( 0.08 3 = m r = =.00 0 kg m = Iz = mr = ( 0.356 ( 0.08 4 4 = Iz 3 = 0.505 0 kg m Rectangular plate: (.005 ( 0.6 3 = m c = =.44 0 kg m (.005 ( 0. 3 = m b = = 3.40 0 kg m 3 3 3 3 3 I = I + I =.44 0 + 3.40 0 I =.60 0 kg m z x y Circular plate: z 3 0.466 0.05 3 = m a = = 0.54 0 kg m 4 4 = m a + m d = 0.466 0.05 + 0.466 0. 3 3
MTE STATICS 3 =.774 0 kg m Iz = m a + m d = 0.466 0.05 + 0.466 0. 4 4 Iz 3 3 3 =.60 0 kg m Entire machine part: 3 = (.00 +.44 0.54 0 kg m Iz Iz 3 = 3.00 0 kg m 3 = ( 0.505 + 5.544.774 0 kg m 3 = 3.8 0 kg m 3 = ( 0.505 + 3.400.60 0 kg m 3 =. 0 kg m