Question

Problem 7.21 The load D has a mass of 250 kg and is being hoisted by the motor M with constant velocity (Figure 1) Figure < 1 of 1 -2m- 2m- 2m- 1-2 0.1 m -1m -1.5 mm

Part B

Determine the resultant shear force acting on the cross section at point CC in the beam.

Express your answer to three significant figures and include appropriate units.

Part C

Determine the resultant bending moment acting on the cross section at point CC in the beam.

Express your answer to three significant figures and include appropriate units.

Show all steps and solution clearly, with correct units as asked in problem:

Answer 1

Solution :

Given Data :-

Mass of load D (mp) = 250 kg

Free body diagram of the whole beam, with section at C and E:

. | Axc 1 ۱۳: دا اسم ۱۳ ،2 a malay .ا250x9 . 8 أ . 25 : ا250x9 . 8 - 2152.5N 252.5M 5 عدد EINE 3 . 5Nواد 0 . 1M ہ مه ) ہو، دیہا2 fہ sN.یہ2 (ن) "0 . 6 (ن)

For the FBD (i) applying (SMA = 0);

:: RB X 4 - 2452.5 x 6.1 – 2452.5 x 1.9 = 0

:: RB = 4905 N = 4.905 kN

Now , ΣΕ, = 0 + Α, + RB – 2 x 2452.5 = 0

Ay=0

Now , ΣF = 0 = A, = 0

.

.

For the FBD (ii)

ΣF = 0 + - Να – 2432.5 = 0

... c = -2452.5 V = -2.4525 kV

The normal force acting at point C is -2.4525 kN

.:. ΣF = 0 + V - 2432.5 = 0

.:Vc = 2452.5 = 2.4525 kN

The shear force acting at point C is 2.4525 kN.

:: Mc=0=-Mc - 2452.5 x 0.6 + 2452.5 x 0.1 = 0

:: Mc = -1226.25 N/m= -1.2263 kN/m

The resultant bending moment acting at point C is -1.2263 kN/m .

.

For the FBD (iii) :

ΣΕ = 0 + NE + 2432.5 = 0

Ng= -2452.5 N = -2.4525 kN

The normal force acting at point E is -2.4525 kN.

... ΣΕ = 0 + -VE - 2432.5 = 0

Ve= -2452.5 = -2.4525 kn

The shear force acting at point Eis -2.4525 kN

:: SME=0 Mg + 2452.5 x 1 = 0

: ME= -2452.5 N/m = -2.4525 kN/m

The resultant bending moment acting at point C is -2.4525 kN/m