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

Q: 10 kN 2.5 kN•m 30° 1.5 m

A: The correct option is b.

Q: 6 m 3m 4 m 4 m 4 m 6 m 3 m m 4 m

A: Write coordinates of A,B,C and DA=0,0,3.5B=3,4,4C=-6,-3,6D=4,-3,4

Q: 5 kN P 60 40 20 Al diniensions in mm Fat

A: The given section is, For the given section calculate the moment the fixed joint is,…

Q: 4 kN/m 2 m

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

Q: 2 kN/m 4 kN

A: Given data:

Q: 60 Ib 90 lb 30 45 150 lb

A: GIVEN DATA- A 150lb BOX GIVEN ACTED BY TWO FORCES WE HAVE TO FIND ACCELERATION OF IT

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: 50 KN/m 120 KN.m B- 9m 4m

A: For solution refer below images.

Q: 8 &N 25 kN m 5 KN Fig (1)

A:

Q: A Mansard roof truss is loaded as shown. Determine the force in members GI, HI, and HJ.

A: Given data To determine the forces in the members GI,HI and HJ

Q: F1 = 5 kN F2 = 13 kN 0 = 45° B - 60° C F3 = 4 kN B 2 m 3. 7m F4 = 22 kN E lJa = 30° F5 = 4 kN

A: In case of multiple force on a system, resolve the forces in horizontal and vertical direction. This…

Q: 1.0 m 2.6 m 2.6 m 1600 N/m 900 N/m

A: Prismatic cantilever beam EI=1200 KN-m2 Taking section at a distance x from the free end. By similar…

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: 15 kN 20 kN 2 m m

A:

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

A:

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

A: Q7.35 Given: Load at A (P2)=15kNLoad at C (P1)=25kNLoad at D P3=30kNLoad at E (P4)=20kN

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

A:

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

A:

Q: 30 EN 100 kN.m M-1200 KN m 2 m 3m 4 m 5m

A: Given: To draw SFD and BMD to calculate shear force and bending moment.

Q: 2 kN 50 mm 5 kN• m D 80 mm B C 1.0 m 1.0 m 1.0 m

A:

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: 100 lb 30° 200 lb 2 ft 300 ft-lb 2 ft 2 ft 2 ft-

A: First we resolved 100 lb force, and calculated horizontal force, vertical force and moment due to…

Q: 20 kN 40 kN 10 kN/m 1 m 2 m 2 m 2 m 1 m w

A: Before going through the calculation for shear force and bending moment calculation first, calculate…

Q: Using unity conversion ratios, show that 1.00 lbm weighs 1.00 lbf on earth

A: Given Data: The given mass is M = 1 lbm. The expression for the weight (W) is, W=Mg Where g is the…

Q: 8 kN 6 kN/m А С 1m 4 m 3 m m E.

A: Calculating the reaction at the support,

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

A:

Q: 6 m 6 m 4 m A

A:

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

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: 1.0 m 2.0 m 05 m 05 m 20 kN 4 KN 10 kN 4.0 m

A: Given Data: Force (F) = 10 kN

Q: RB 50KN bm 100 KN 4m B 4m

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 15 kN/m A 2 m 4 m B.

A: The reactions on the supports of beam can be drawn as follows,

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: 10 KN/m 1a KN (m Ra 1.5 m 4m 3 m 1.5m

A: For solution refer below images.

Q: (4) 18 kN 4 kN/m B 6 m -2 m

A: Question 4) is a propped cantilever beam

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: 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: F126KN 180KN 26 -3m 1.5m 48 65° 145KN V165 KN

A: Drawing the system of force as shown below – Let us take point A as an origin of the coordinate…

Q: Identify the zero-force members in each truss

A: There are two rule to identify zero force in the truss members:- (a) Let's consider a joint which is…

Q: 15 KN 18 kN 40 kN/m 3 m 6 m 4 m Fig.

A: To draw the SFD and BMD and determine the maximum bending moment

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

A: For solution refer below images.

Q: 50 kN 500 40 kN 30 kN 50 mm. 200 mm 200 mm 50 mm 250 mm 30 kN

A:

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: 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…

Forming and Shaping

Metal forming is a category of metalworking process through which the material can be given the desired shape and size by subjecting it to plastic deformation, where the component undergoes bulk deformation process upon application of loads. Unlike the shearing and machining process, the material does not undergo up to a failure point as indicated in the stress-strain curve. Hence, the stresses induced during the forming process are greater than yield strength but less than the failure strength of the material. The process does not lead to the loss of material, so it can be said to be an economical process. During forming, the material is subjected to tensile forces, compressive forces, bending, and shearing loading conditions. The material should possess the property of ductility to undergo the forming process.