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Q: Determine the normal stress at corners A and B of the column.

A: Draw the figure.

Q: 200 mm 150 mm 100 mm 200 mm

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Q: 230 mm 60 mm 25y

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Q: 90mm 60mm 45mm R15mm

A: Let’s divide the given section into 3 sections. Section 1 is a rectangular OABC. Section 2 is the…

Q: please answer part g last part

A:

Q: 9/16 inch is approximately equivalent to . mm. 19 17 14 12

A: The dial indicator is a traditional mechanical kind with a dial display that resembles a clock face…

Q: 80 mm 50 mm 2 1 7 15 mm 15 mm 3.

A: The Center of mass is a point where the sum of the moment of the weight of each particle is zero. It…

Q: 4000 N 3000 N B C 300 mm 100 mm 100 mm 十+

A: For question 1 option B is correct.  For question 2 option A is correct.  Steps are given in next…

Q: y 200 mm 200 mm + 150 mm- 125 mm 60 mm 125 mm 1.

A:

Q: 200 mm C 600 mm 200 mm

A: Area moment of inertia is a geometrical property that defines how different points are arranged with…

Q: 150 200 mm D mm 150 150 v2 mm mm 0 = 45° E 12.5 rad/s

A: Given data,θ=45°ω=10 rad/sOB=1502mm=0.2155mAB=350mm=0.3556mO'A=150=0.1524m

Q: 1. 100 mm 225 mm 75 mm 125 mm 50 mm

A: According to the details provided in the question, we need to calculate centroid.

Q: 600 mm 600 mm

A: given:

Q: 50 mm 30 mm 20 mm 40 mm * -60 mm- 40 mm x

A: GIVEN DATA- A COMPOSITE AREA GIVEN  WE HAVE TO FIND CENTROID OF AREA

Q: Fx A 75 mm 50 mm 150 mm 350 mm В F2 250 mm y

A:

Q: 100 mm 100 mm

A: GivenP=9kNd=6mm

Q: A 150 mm B 240 mm 80 mm

A:

Q: 100 mm- 11.5 mm 7.5 mm N A 225 mm 11.5 mm (a)

A:

Q: Locate the centroid of the plane area shown.

A: The centroid of a 2d plane is the point at which the entire plane can be balanced on the tip of a…

Q: 25 mm R50mm 100 mm

A: Given

Q: how would this look like if written?

A: The feature control frame is required to describe the condition and tolerances of the geometric…

Q: - 100 mm 20 mm yo 140 mm -20 mm 20 mm - 100 mm

A: Rectangle Area (mm2) x¯ (in) y¯ (in) x¯y¯A (in4) I 2000 -50 70 -7000000 II 2400 0 0 0 III 2000…

Q: 150 mm 50 mm 200 mm 50 mm 30 mm 200 mm -50 mm

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Q: Take a = 330 mm. y a - 70 mm 120 mm a

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Q: 10 mm 20 mm 50 mm

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Q: 300 mm 150 mm 150 mm 75 mm 100 mm 100 mm

A:

Q: 50 mm 150 mm 150 mm 300 mm *

A: A1=300×50A1=15000 mm2y¯1=3002=150 mm   from bottom,A2=300×50A2=15000 mm2y¯2=300+502=325 mm,

Q: Y. 20 mm 20 mm 260 mm 20 mm 200 mm

A:

Q: 6 mm 55 mm 10 mm

A: Impact loads are those loads that are acting for a short span of time. Toughness is defined as the…

Q: a) 10mm IG mm IG mm 1050mm IG mm 16mm 250 mm 300 mm

A:

Q: 9 cm cm S cm 7 cm

A: Answer- 5.9

Q: 300 mm 300 mm Figure Q2

A: Given data:- Area =1950 mm^2 Maximum deflection of point C = 0.6mm

Q: Identify the drawings as isometric, trimetric, oblque or perspective. A. B

A: Isometric projection is a technique used in technical and engineering drawings to visually…

Q: 50 N 100 mm 200 mm

A: As per our guidelines, experts can provide solution to only one problem in case of multiple…

Q: F=20 lb a Q6 25 mm 70 mm 100 mm- 70 25 mm mm

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Q: mm 40 mm 1.5m

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Q: H, = 0.40 Hx = 0.30 D 150 mm 150 mm B E 300 mm 300 mm \F 150 mm 150 mm 250 mm

A:

Q: b. Classify each of the following materials as to whether it is a metal, ceramic, or polymer.…

A: Brass – Metal (Alloy) Magnesium oxide (MgO) – Ceramic Plexiglas – Polymer Polychloroprene – Polymer…

Q: 200 mm 300 mm Steel 75 mm Brass

A: Givenρb=8.75 Mg/m3=8750 kg/m3ρs=7.87 Mg/m3=7870 kg/m3

Q: 200 mm B 300 mm 400 mm 100 D G 600 mm

A:

Q: C 110 mm F В 150 mm E 300 mm A D 200 mm 150 mm

A: There is a circumferential contact force between gears B and EF=TABrB=TEFrETEF=TABrErBTAB=170…

Q: A thermoformed disc ending at a diameter of 2 cm and a thickness of 1 cm started from a disc with a…

A: Thermoforming is a general word for a procedure that starts with an extruded plastic sheet. The…

Q: F125 mm - 17 mm Yo +. -Xo 175 mm 17 mm 17 mm -125 mm- Answers: (10) mm4 i (106) mm4 i Iyo

A:

Q: F1.5 in-| 1.5 in. 1.5 in. 1,5 in. 1.5 in.

A:

Q: 100 mm 25 mm 75 mm

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Q: 120 mm 600 mm 10-N 4N 600 mm 4N

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Q: D C 2 mm 5 mm

A: GivenAD=15mmAB=15mm

Q: 90 mm, 90 mm 170 mm 50 mm 37.5 mm Figure Q1 (a)

A: Find the maximum tensile stress and maximum compressive stress Moment=100KN-m

Q: 2n X1 = 3.75 sin (100rt + mm 9. 2n X2 = 4.42 sin (100nt 5 mm

A: Find the attachment for the solution.

Free Body Diagram

The system of forces acting on a body tends to either rotate it or make the body undergo motion. In general, when an external agent exerts a force on a body, the body undergoes translational motion and rotational motion. The body has six degrees of freedom under the influence of that force. Here, three degrees of freedom are from the translational motion while the other three are from the rotation motion. However, sometimes the bodies are not allowed to undergo any motion and are fixed, under such conditions the body is said to be constrained.