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

Q: 2 ft F D E -1.5 ft 600 lb 300 lb -3 ft -3 ft-

A:

Q: N 20 mm 200 mm 20 mm 300 mm pmri 20 mm A 1.5 m C N N 站 B D TTTTTTT 1.5 m 5m

A: given; load at point c (Pc)=430Nload at point D(PD)=634NAC=1.5mBD=1.5mAB=5mCD=2mlets take reaction…

Q: E 600 mm B Rigid P. 80 mm 160 mm 80 mm

A:

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: 150 lb/ft A В -6 ft -8 ft

A: Let us draw the reaction at the supports  Weight of the rectangular load = 150 lb/ft(8ft)=1200 lb…

Q: 120 mm 600 mm 10-N 4N Fig. 5 600 mm 4N

A:

Q: What are the V and M diagrams? Thanks! 50 kN 125 kN 50 kN B C D A E -0.5 m 0.3 m 0.4 m 0.2 m

A: As per the guidelines of BARTLEBY we are subjected to solve only one question at a time. Please…

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

A:

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: 75 mm D. 1.5 m E F B 2 m 150 mm 5 deg 2 deg 4 deg 1.5 deg 1 deg 3.5 deg

A: Write the twist angle of point A respect to point B.   φA/B=φA-φBφA-φB=3°…

Q: 25 kN 500 mm 400 mm 700 mm

A: Basics of Mechanics.

Q: 1 GPa is O a. 1x 106 N/mm² O b. 1x10° N/cm² O c. 1x106 kN/mm² O d. 1x103 N/mm²

A:

Q: Vertically 300lb/pulg2 and horizontally 150 lb/pulg2 I have only this data please answer.

A: Giver:- My=300 lb/pulg2 Mx=150 lb/pulg2   To find:- Resultant

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

A:

Q: F 1000 lb y 500 lb 45° 30°

A:

Q: 1000 Ib -12 ft 12 ft- B 2000 lb 8 ft D. ton -12 ft- 6 ft 6 ft

A:

Q: 100 kN 50 mm 50 mm 75 mm 75 mm 40 kN 30° 400 mm

A:

Q: erage 1.5 kN 1.5 kN A B Uniform Thickness-Smm 20 mm 0.4 m 0.4 m 0.2 m 20 mm

A:

Q: The material has a specific weight y= 61 lb/ft (Eigure 1) Figure K 1 of 1> 0.25 t 0.25 ft 2 ft 0.5…

A: Given data: γ=61 lb/ft3 Need to determine the moment of inertia of the wheel about an axis which is…

Q: Determine the maximum shear stress acting at section a–a in the beam.

A: Given beam with reactions at 'A' and 'D' are as shown below: Applying equilibrium conditions: (i)…

Q: 120° F= 400 lb F = 600 lb

A: Here for force F1, Angle with x-Axis (alpha) = 60o Angle with Y-Axis (beta) = ? Angle with Z-Axis…

Q: y -9 cm- -3 cm 2 cm 19 cm 2 cm C 15 cm 9 cm- 3 cm→

A: Divide given figure in 3 section/shape,

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

A: To determine the support reaction at J.

Q: 500 sq mm в 300 sq mm 400 sq mm A D 10KN 40 KN 50 KN- 20 KN 200 mm 600 mm 300 mm

A: →For part AB:-Force, P=10kN=104NLength(L)=200mm, Area(A)=300mm2→For part BC:-Force,…

Q: In the Figure 6, the absolute force value in the bar AC is (in kN): Figure 6:

A: Introduction: Absolute force: Absolute force, often known as absolute strength, is the maximum…

Q: 6 kN 9 kNm 4 kN/m E 2 m 0.5 m 0.5 m 2 m

A:

Q: C 3 ft B 0.8 ft A

A:

Q: 4 m A) 12 kN.m B) 16 kN m C) 20 kN m D) 22 kN.m 4 KN co 2 m 10 kN 2 m 2 m 2 kN/m E

A:

Q: In the Figure 6, the absolute force value in the bar AC is (in kN): Figure 6: 5 m/ 5 m 5 m A 20 KN B

A:

Q: 24 kN/m 64 kN • m D A B - 2 m -2 m

A: Let us draw the reaction force

Q: - 2- 45 100lbs/ft 300lb 30 200lbs

A:

Q: A psychrometer (an instrument used to measure humidity) weighing 4 ounces (1/128 slugs) is whirled…

A:

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

A:

Q: 5 panels at 3 m E 3 m G 1.8 kN 15 1.8 kN 3 m 15° 3 m 5 m

A: given data: we consider only calculating side figure: TAB=force in member ABTAE=force in member…

Q: -30 mm- 5 4 3 6. 80 mm 50 mm 2

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

Q: mm 40 mm 1.5m

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: 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: 7 panels at 4 m B C D 4 m A H. 8 kN 8 kN

A:

Q: Determine the x and y components of each of the forces shown.

A: To find out the x component of the forces Resolve all the force in the x-axis…

Q: 2 kN 2 kN/m A B (a) 4 m 2 m RA Re

A: A beam is defined as a structural member subjected to transverse shear load during its…

Q: 400 lb F1 = 400 F2 = -600 k F3 = 500 sin 45 i-s00 cos 45k 7t Y600 lb 45 M- rxF,. M,- r2x , M,- rgxf,…

A: "Since you have asked multiple questions, we will solve the first question for you. If you want any…

Q: 120 kN 50 kN/m 4 m 1m 8 m 2 m

A: Calculate the support reaction as follows, RA+RB=120+50×5RA+RB=370 kN     .......... (1)Taking…

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: In the Figure 6, the absolute force value in the bar AC is (in kN): Figure 6: 5 m 5 m A 20 KN 5 m B

A: FAC = 23.0940 kN

Q: 8 kN/m | 20 kN 4 kN/m 2.0 m 1.5 m

A: For solution refer below images.

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

Axial Load

When a bar of the uniform cross-section is acted by a load, such that the load acts along the longitudinal axis of the bar, such kinds of loadings are known as axial loadings. In general terms, a load is an external force acting on a component. An axial load acts perpendicular to the geometric cross-section of the component. Based on the direction of the application of the load, an axial load can be tensile or compressive. The analysis of bodies under the action of axial loads is generally studied under the branch of mechanics, known as statics.