15 kN 25 kN 20 kN C В D 1 m 0.8 m 0.3 m 30 kN °0.4 m

Q: Q/ The force in the member (CA) is 12m 50 ку 490.50 N 500.44 N 200.12 N 2550 N 5m

A: None of the options are right. but Solution is 100% right.

Q: E 1 m 3 kN/m B G I m F -1 m→ -1 m- 1 2 m

A: Consider the Beam BC as shown below, VB+VG=3×1+1+1+1VB+VG=12    ....... (1)Taking moment about…

Q: 6 kN/m 20 kN 90 mm В C h 2.4 m -1.2 m –

A: Solution: To calculate support reactions Under static equilibrium RA+RB=6kN×2.4+20=34.4kN Taking…

Q: 10 kN 5 kN 3 kN/m F C D E 3 m 1 m 4 m 3 m 1 m

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Q: Q/ The force in the member (CA) is 12m 50 ку 5m BX

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Q: 2 m C 2.5 m 125 mm 1.5 m 25 kN/m 2 m 4 m

A: To determine the reaction at A and B, the beam can be dealt in two sections that are AD and DB. In…

Q: Lever Problems - effort Resistance and Mechanical Advantage  E = 35 Lbs a = 1.25 feet  A = 4.5…

A: Here, Effort = E = 35 lbs Distance (a) = 1.25 ft Distance (A) = 4.5 ft   Let us denote resistance by…

Q: 50 KN 100 KN/m 60 KN/m 40 KN/in 10 KN.m 2.0 m 6.0 m 2.0 m 4.0 m

A:

Q: 40 kN E 2 m B F 3 m 50

A: to find reaction component at B and C

Q: 120 mm 60 mm 90 mn D 80 mm C 45 mim 480 N

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Q: 38.3 kN-m F) 23.3 kN-m G 8.3 kN-m H. 13.8 kN-m E.

A:

Q: G 1 m H 2 m |C -2 m 2 m -2 m- -2 m 10 kN 15 kN 30 kN

A: Calculate the reaction at the support, RA+RE=10+30+15RA+RE=55   ....... (1)Taking moment about…

Q: 12 kN.m 3 m 12 kN/m 1 m 12 kN.m 1 m 4 m 3 m 12 kN/m

A: The shear force and bending moment diagram has been drawn below.

Q: 30 kN | 50 kN D A BA 1m - 3 m - 2 m

A:

Q: KIN 14 kN / m 10 kN / m A B ID 4 m 2.5 m 2.5 m 3 m

A: Finding support reactions:RA+RB=12*4*14+14*5+20+30RA+RB=148…

Q: F1 = 5 kN F2 13 kN -30 C. F3 = 4 kN 45 A 2m 7 m 5n F4 = 22 kN a30° F5 = 4 kN

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Q: BM at B is 3 kN 1 m D 1m 1 m B 1 m A

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Q: W= 75 Ib W= 150 lb W= 75 Ib R.

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Q: mN)(6 GN). 0 mm)(0.6 MN)²

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Q: 8' 14' 8' 6' 400 lbs' W, = 200 plf max W = 120 plf A

A: It is required to determine support reaction of given beam

Q: 15 kN 10 kN/m 20 kN · m A 2 m 1m 1m 2 m What is the reaction force at A?

A: Given: Point load=15kN Moment=20kNm Uniform load=10kN/m

Q: 15 KN 10 KN 10 KN 20 KN E A 5.0m

A: To determine the support reaction at J.

Q: 2 tór/m 8 ton F 4 ton-m E 3 ton/m 2 m 4 m Im1m 2 m 2 m

A: According to the principal of engineering mechanics  Reaction and moment at support is equal to zero

Q: 700 N 450 N 300 N 30° 60° B algo 1500 N•m 3 m -2 m- -4 m

A:

Q: In each case, draw the shear and moment diagrams for the beam.

A: The free body diagram of the first beam is as follows: Calculate the force in the vertical…

Q: Which of the following nearly gives the W = 200 kN W = 400 kN 1m DI 2 m B. 5.6 m

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Q: 80 kN 40 kN 60 kN -10 kN/m A B 2 mim. 3 m 3 m 2m 1 m (b) 10 k I k/ft B 12 8' (b)

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Q: Pc= 400 N 400 N/m 200 N/m A C E D - 2 m --Im 4 m - 2 m

A: Solution is attached below:

Q: 10 m 10 m 10 m 10 m 180 KN 8 150 KN 11 4 7 10 5 9. 12 12 m 1 3 6 14 20 m 13 200 KN 2 A

A:

Q: A 1.20m 18 KN 0.75m 12 kN/m 1.20m B

A: The given is as shown below,

Q: 100 kN/m 150 KN E (m) 5 7 10

A: "Since you have posted a question with multiple sub-parts, we will solve the first three sub-parts…

Q: 120 kN 7.5 kN/m 417.5 kNm 282.5 kNm A C +5m→5 m 10 m - RA = 115.5 kN Rg = 79.5 kN

A:

Q: 150 mm d 75 200 mm d 0.9 m Pump Power = y.Q. Ep P1 = 0 Ví = 0 P3 = 0 %3D 0.6 2.4 m 175 Hg mm H2…

A:

Q: 8 KN 3 AN/m 12 AN 6 KN/m BC D 2.0 m 1.0m 3.0 m 3.0 m

A: Apply the equilibrium condition of force in the vertical direction. Apply the equilibrium condition…

Q: 120 kN 7.5 kN/m 417.5 kNm 282.5 kNm A D 5 m 5 m 10 m- RA = 115.5 kN Rg = 79.5 kN

A: Shear force diagram represents shear force at every section of the beam. Shear force at any…

Q: س / استبدل نظام القوة الذي يعمل على الحزمة بقوة Bs محصلة مكافئة ولحظة عند 650 N 500 N 30° 300 N 60°…

A:

Q: Fi= 3 KN A <作る- 2KNT de 8 c=18 とるcm a= 22 m Fu= 20 KN Eト)=30° FB= 2 KN

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Q: 100 kN Half circle 6.0 m Pinned Pinned 4.0 m 40m

A: Let RA and RB are the reactions at pin support  By law of equilibrium of forces in vertical…

Q: 10 EN.M 12 kN 410(LN/M 3.

A:

Q: 5 kN 1m 2m E 2 m 3 kN D B 2 m

A: GIVEN DATA- A PLANE TRUSS GIVEN  WE HAVE TO FIND FORCE IN MEMBER BC

Q: إذا كان هناك حاجة لعزم دوران )لحظة( .P تستخدم الكماشة لإمساك نهايات أنبوب الحفر \ \ Q4: ((A و التي…

A: Note: As per the guidelines we will answer the first question for you, for the rest of the question…

Q: 12 kN/m 40 kN C A B 2 m 1m-

A: Let us draw the reaction force at A and C

Q: members DE, DJ, EJ, and IJ. 2 KN XNK 8m 8m 8m 8m L K J H 1 kN 8m 3 KN 1.5m 2m 8m

A: To find force in member DE,DJ,EJIJ

Q: 20 kN D 20 kN 20 kN 3 m 10 kN 3 m 3 m A 4 m 4 m E 4 m 4 m B,

A:

Q: C 30 kN 50 kN 70 kN 50 kN d=20 mm d=40 mm 1m 2 m 1m

A: draw FBD of each section and find stresses.for solution refer below image.

Q: F3 = 52 lb 13 12 F1 = 80 lb 25°

A: Given that, The component of forces in the x-direction,

Q: t A Cable 2m 75 N B 2m C 30

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Q: 120 1b 60 lb/R

A: given:

Q: In fig. shown. Pind the reaction force on paít(Bc) . OZ07 kpd L = 107 HRir %3D oiL D=75mm Y.d =0.88…

A:

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.