A 13KN force is applied to the 60 mm diameter post ABD. Determine the principle stress at point H. 13 300m館 100 m 195 min 150mm 2 Ox + Oy 01,2 士 + T2 xy %3D
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A: Normal stress acting at point A is σA=FA=4Fπd2=43 kipπ1.5748 in2=1.540 kip/in2 4 cm=1.5748 in
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A: Given:Average normal stress, σ=3.8 MPaLoad applied, P=1200 Nouter diameter, d=25 mm
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A: GIVEN :- b1=85mmb2=130mmb3=60mmd=28mmP=1850N
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A: Given: F1 = 7KN F2 = 6KN F3 = 2KN F4 = 9KN
Q: If F1 = 1070 Ib and F2 = 2095 Ib, determine the magnitude of the normal stress produced at point H.…
A: Given: The force, F1 = 1070 lb The force, F2 = 2095 lb
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A: Given ThatA=270mm2△Lmax=3mmL1=3.8m L2=2.2mL3=0.85m E=100GPa
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A: given data σx = σy = σ, ζxy = 0 to determine stress on the plane inclines at 45 to x plane
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Q: 1.4 kNm C H KI 10 kN 240
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Q: Determine the maximum normal stress in the bar. The diameter is 1.5cm. F1 = 7KN F2 = 6KN F3 =…
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Q: 5. Two forces act on the solid circular rod (r=.005 m) as shown. Using Mohr's circle, determine the…
A: Draw mohr circle and find principal stresses and maximum shear stress.
Q: 5/5find the normal stress at left of point C (1 A 100 C 40 kN 50 kN 40 mm 70 10 kN 1 20 mm 30 mm 100…
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A: Fifth option is correct combined shear stress=4.805Mpa
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Q: B. SOLVE FOR THE STRESS IN THE ROD IF ITS CROSS-SECTIONAL AREA 1S 2500mm^2. 1.5m P=25KN
A: Given A=2500mm2
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A: GivenA=410mm2w=9kN/mP=3kNQ=5kNx=0.5m
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Q: A body is subjected to multiple applied forces and moments. The force present on a 10 m area with a…
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Q: The rectangular bar has a width of w = 3.50 in. and a thickness of t= 1.50 in. The normal stress on…
A: Given Normal stress = 6 ksi Thickness, t = 1.5 in Width, W = 3.5 in Find Load, P…
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A: GivenA=145568956-50258925-230MPaprincipal stresses λ1=-250.81λ2=-64.93λ3=180.74
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Q: The two portions of member AB are glued along a plane forming an angle 0 = 65°with the horizontal.…
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Q: The specification of a cylindrical vessel are as follows: D = 350 mm %3D t = 12 mm P = 4.5 MPa…
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Q: The solid circular rod has a cross-sectional area of 460 mm². It is subjected to a uniform axial…
A: Solution.jpg
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Q: For the given state of stress, determine the normal and shearing stresses exerted on the oblique…
A: givenσx=6ksiσy=10ksiτxy=4ksioblique plane angle θ=75o
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- Problem 3. What is the moment of the 1.4 kN force along the direction of the 1.0 kN force? Express your answer as a magnitude and sketch the direction of rotation about the axis defined by the 1.0 kN force TIGBAO TK 1.0KN X 539 indi 21 150 mm 195 mm 2 4 160 mm 21% ARD 19TO 1.4 kN MEBIRT 0 DAW ITOM 00 1400Problem 5 Determine the magnitudes of the components of the force acting along the axis AB of the wrench handle and perpendicular to it. Son Can toa 300 mm x B 9 = 128° (0,0,300) 500 mm UF = COS 30%. Sin us i = -0.6124; + 0.6124j F = 80 N 30 5 45" + Cos 30° cos 45j+ sin 30 + 0.5k (0,500, 300) UAB = -j Cos 0 = UF • UAB = (-.61241 - 0.61245 +95 k) (-j) I How do we go from this = -0.6124 to this?Warm Up Activity • Resolve each force into x and y components in order to calculate Ax and Ay. Express your answer in kN with direction arrow. F₂ = 450 N 45° y F₁ = 300 N Ay 5 3 4 Ax F3 = 600 N ·X
- Answer for the 1st part of the problem. magnitude of force A = 74 N direction of force A = 25o north of west second force = P direction of force P = α south of west arrow_forward The resultant of two forces is horizontal , it means that the y component of resultant = 0 Ry = ASin(25o) - PSin(α) 0 = (74×0.42) - PSin(α) PSin(α) = 31.08 NFor, the smallest force P , the value of Sinα must be maximum which is 1 when the value of αis 90 degree Thus, P×Sin(90o) = 31.08 N P = 31.08 N The magnitude of smallest force P = 31.08 N the direction of force P = 90 degree south of west = along - y axis= arrow pointing vertically downwA force F of magnitude 680 lb is applied to point C of the bar AB as shown. Determine both the x-y and the n-t components of F. Answers: Fx= Fy= Fn = Ft= 46 68° i 630.48 254.73 i 489.15 i 472.36 B F = 680 lb lb lb lb lb'Two Forces Shown in Fig. (1) acts on the bracket, determine the effect of these two forces (using Parallelogram method). 100 F₂=150N F₁-200 N
- The force P of magnitude 50 kN is acting at 215° from the x-axis. Find the components of P in v 157° from x, and u negative 69° .from x Pv367.44 kN & Pu358.95 kN O Pv=58.95 kN & Pu=67.44 kN O3m Find the rectangular representation of the force F, given that its magnitude 240 N. * F = -135.76 i+ 169.71 j– 101.82 k F = 135.76 i + 169.71 j+ 101.82 k F = 135.76 i - 169.71 j+ 101.82 k F = -135.76i+ 169.71 j+ 101.82 k200 400 B 200 N m E D 200 F Dimensions in mm Determine the magnitude of the force (in N) of the x-component of the internal pin at D.
- MACARTHUR ESIA KELVIN SASIC MECHANICS Goli s8 7221 Answers Q1- Qュ Q 3 point v. determne bthe magnitude or the resultant force the angles Ore, Q and Oz the resultant Tuwo furces act 9. the companent bolt at the resultkat on makes X,4 and z axis FI=FON 30 55 F=300NFind the magnitude of the force F1 and its direction 0 given that the resultant force is 430 N and directed along the positive y axis. Set F2= 860 N F1 15° F2 1056 N, 0= 51.85 deg. 856.2 N, 0= 38.15 deg. 856.2 N, 0= 77.84 deg. 1056 N, 0= 38.15 deg. 1056 N, 0= 77.84 deg. 856.2 N, 0= 51.85 deg.In your response, please display the full computation and all FBDs that were used. Only round off at the final answer with 4 decimal places. P=2kN, F=3kN, and moment M=0.5kN-m are acting on the structure in the illustration. A pin at Z and a cable RQ connected by a connecting pin R support it. Im ● F ● P S TO M Im R lu W OZ a' 0.8m 3m If the pin support Z's factor of safety against shear stress is 3, determine whether a single or double shear should be provided for an economical and safe design. 4 m The mas allowable shear stress in the pin is 50MPa. Consider the following specifications. Single shear design: 25mm pin diameter Double shear design: 17mm pin diameter