5 kip/ft 15 kip ft 15 kip · ft B- 6 ft 10 ft 6 ft

Q: 45° SG = 1.0 50 cm A Hinge 10 cm SG = 0.9 80 cm B

A:

Q: 15 ft – 15 ft 15 ft 5 kips 5 kips 9 kips |A J 8 ft K F I Bo E H.

A: Vertical force applied at pont D on the truss = 5kips Vertical force applied at pont G on the truss…

Q: 3 m C 3m E Cable 3 m 2000 N -3 m 4 m O 4 m F

A:

Q: F = 375 Ib F = 500 Ib B. 8 ft 6 ft 5 t- 30 F = 160 lb

A: given cantilever beam as shown below

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

A:

Q: = 4 x 10 *-6 m4 and A = 4 x 10"-3 m2 5 m 30

A:

Q: 5 kips 30 kip • ft I = 90 in.4. 6 in. B NA 2 in. 4 ft 4 ft

A: For solution refer below images.

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

A:

Q: 4. Ø.009MAB@ Answer:

A: refer generalized representation of geometric tolerance.

Q: 0.6 m 0.6 m acD =4 rad/s² 0.3 m wCD = 2 rad/s

A:

Q: Sm 8 m/s 15m 0 AI 6 m/s

A:

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

A:

Q: M т 12 ksi + 6 ksi

A: σx=Normal stress in x-direction=-12 ksi→Normal stress in x-direction is compressive stress.σy=Normal…

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

A:

Q: -20KN/m 10kN/m 2 m -5kN 3 m A B 2 m 2 m

A: Data given -  Two udl of 20 kN/m & 10 kN/m

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

A:

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

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: F-1200 +o 3t N A В 1.4 m

A: First of all take A+B as a system, Now apply force equation along and traverse to the inclined…

Q: h- 25 W/m2.C T-5'C 5 T- 100°C 7. 8. 6. L00 Insulated

A:

Q: W=4 kN/m 5 kN A 2.5 m + 1.5 m

A:

Q: 7. Where do the maximum BM óččurs for th 10 KN B A 2M 3M a) -54 kNm b) -92 kNm c) -105 kNm d) – 65…

A: Consider a cantilever beam that has fixed support at A. A transverse uniformly distributed load of…

Q: 6 ft y 9 + B E 4 ft- D H -4 ft 4 ft K +ft- G 3000 lb 1000 lb X

A: Given: Force applied at point H: →FH=3000 lb    (in downward direction) Force applied at point G:…

Q: 7kN m 5 kN/m |C B D = 3m to D/2 D/2 12KN

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: mN)(6 GN). 0 mm)(0.6 MN)²

A:

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: 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: 2Nm 15KN L. 12m 0.6 12m 18m 12m

A:

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

A: To determine the support reaction at J.

Q: 8 kN 6 kN/m А С 1m 4 m 3 m m E.

A: Calculating the reaction at the support,

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

A:

Q: 100 KH 20 KN/m B 2m C am RI

A:

Q: 15ON 50N 100N /m EI 8 midspan F2 6 12 El 8 over hang =

A:

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

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: 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: Determine 1. Thermal Efficiency 2. Mean Effective pressure.

A: To Determine  Thermal Efficiency Mean Effective pressure.

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: E= 103 GPa E = 75 GPa E= 190 GPa 0.8 mm 150 mm 100 mm 300 mm Bronze Al Steel

A: Stress produces in the material due to changes in temperature is known as thermal stress. Condition…

Q: A y FAD FAB FAC - 60 m- 60 m B 40 m C 40 m- 40 m

A: Writing the co-ordinates of the points as follows, Writing the position vector of the line AB, AC…

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: 5 applications

A: Write: Examples of application of shear stress

Q: 30N/m 0.4kg -Fa WwW

A: For solution refer below images.

Q: 10 F= 150 N !! F =100 N 90-15-10=6s

A: A parallelogram is formed by drawing a line from the head of F1 and F2 that will be parallel to each…

Q: 24 mm 5 mm 30 mm 8 mm 5 mm 30 mm Section AA 12,000 N

A: The correction factor is rc=3015=2 The correction factor for 2 is k=1.41. The moment of inertia is…

Q: 100 kN I 120 kN/m 60 kN/m Ic D 315 kN-m 3 m 3 m 3 m 3 m

A: Draw the free-body diagram of the beam.

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

A:

Free undamped vibrations

Whenever an object/system displaces in either side of its equilibrium position with the help of an external/internal means, then the object/system moves regularly or randomly on both sides of its equilibrium position. This motion is called vibration. Generally, there are three types of vibration: longitudinal vibration, transverse vibration, and torsional vibration.