3 m 2 m B 4 m F A 30 kN 18 kN

Q: 40 kN 30 kN 20 kN 16 kN-m 24 kN/m 10 kN/m A 15 kN 3 m 4 m 2 m 4 m 2m 1.5 m

A: Loads acting on a beam is given. Dimensions at which loads acting and the magnitudes of the load are…

Q: 2m A * 4m 4m E 4M 4200 kN

A:

Q: A B 60° 3 kN 40° 2 kN

A:

Q: 1 kN 2 kN 2 kN 2 kN 1 kN H 0.46 m F D 2.62 B I A C| E G| 2.4 m 2.4 m 2.4 m 2.4 m Figure 2

A:

Q: 4 kN/m 2 m

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

Q: 20 kN 50 kN 50 kN/m 25 kN/m B. 2 m 3 m

A: Consider the free body diagram. The cantiliver support will be replace by two forces FH and FV and a…

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

A:

Q: 2 kN/m 4 kN

A: Given data:

Q: -5 ft 3 ft- 2 ft 4 ft B В 5 3 12 13 4 5 500 lb 200 lb 260 lb

A:

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

A: to find reaction component at B and C

Q: A 'B 2 ft I 4 ft H. ft E G F

A:

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

A:

Q: 2 ft 2.5 ft 1.5 t

A: Centroid, also called the center area of anybody is the point at which the whole area of the body…

Q: F B A 6 m 4m 4m

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

Q: H G 60 E D 7 kN 7 kN 9 kN 6m 6m 6m 6m

A:

Q: 200 kN 100 kN 100 kN в F 30° 60° 60° 60 60° 30° A G 2 m 2 m 2 m 50 kN 50 kN

A:

Q: 2000 lb M = 4800 lb-ft В C A D 3 ft RA 6 ft 3 ft Ro 2.

A: Given DataP=2000lbMx=4800lb-ftTo find SFD and BMD **As question demanding for solving Question no 2…

Q: 1. 500 N-m/sec to HP 2. 85 joules/sec to KWatt

A: Hi and thanks for the questions. Since you have asked questions with multiple sub-parts, we will…

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: 10 kN 4 Smart 3 soi non 12 kN 8 kN 6 kN Aqu.21 Azn Iml F

A: Given

Q: 100 kN 64 kN 82 kN 82 kN 55 kN 2m A 3 m 3m 3 m 3 m 3m 3 m 91 kN 73 kN 46 kN

A:

Q: 1.5 m 4 kN A B 2 m 8 kN C D 2 m F E

A: Let  RFx and RFy are the horizontal and vertical reaction at F RE is vertical reaction at E By using…

Q: H G 0° 60 6 kN 6 kN 8 kN Sm Sm Sm Sm

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

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: 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: 2 KN/m 2 KN/m URN A 30 3.0m 3m 2.0m 2.0 m 4,0 m

A: There are three different types of loads are there on this beam which are Uniformly varying,…

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

A: The given is as shown below,

Q: F 105° 60° 45° 300 lb 60° 400 lb 200 lb

A:

Q: 2 KN 20 20 5 KN X 45° D F 2 KN

A:

Q: 15 kN 30 kN-m Figure 2 ID в 1.5 m 3 m 1.5 m

A: The fixed support resists all the moments that is translational motion and the rotational motion of…

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

A:

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

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

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: 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: F2-13 KN C A4 kN A 2m 4- 22 KN 30 FS-4 KN

A: Since on the system, some of applied forces are acting at some angle, so it is better to resolve…

Q: 5 kN 4 m B 5 kN 4 m C D 2 m

A: The free-body diagram of the strut is given as,

Q: 2 kN/m 10 kN t. 5 m 3 m 2 m Figure 2

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: 20 KN 40 kN/m 150 kN m 3m

A: According to the company's guidelines when multiple questions are asked in a single request,  only…

Q: F3 = 52 lb 13 12 F1 = 80 lb 25°

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

Q: 2.6 kN 2.4 kN-m B A 1.0 m 1.0 m 1.0 m

A: To draw the shear force or bending moment diagram, one should first determine the reaction at the…

Q: 3.kN 30 kN/m B.

A: given; upper left beam; ⇒lets take reaction force at A=RA⇒lets take reaction force at B=RB⇒load…

Q: 2.5 kN 1.5 kN 3 kN 30 A B. 2m 4m 2m

A: Resolve the two forces along the beam and perpendicular to the beam as shown. Write the x component…

Combined Loading

Any functioning component, such as machine parts, structural members, pressure vessels, and so on, are all under the influence of more than one force. When anybody is under the influence of more than one force such as shear forces, bending moments, axial forces, torsion, and so on, then such body is said to be under combined loading. A very practical example of combined loading is the crankshaft of an internal combustion engine (IC engine). The crankshaft of the IC engine is under high rpm, which is caused due to the torsional forces. Due to the presence of piston, counterbalances, and internal imbalances, the crankshaft experiences lateral forces and due to the rise in temperature, the crankshaft undergoes thermal expansion, which pulls the crankshaft along the axial direction and lateral direction. The presence of unbalanced loads along the periphery of the crankshaft also induces a bending moment.