Why is the force Fd pointing downwards?By force equilibrium, shouldnt it be3 + Fd = 9Fd = 6kN, thus pointing upwards? 8.44 The steel pipe AB has a 72-mm outer diameter and a 5-mm wall thickness.Knowing that the arm CDE is rigidly attached to the pipe, determine the principalstresses, principal planes, and the maximum shearing stress at point H.Use stress transformation eqns.Fo9- 3 = 6 kN b%3D%3DTo= (12 x10)(120 »10)%3D3 kN%3D1440N.mD| 9 kN120 mm150 mm120 mm1440 N-m '6 kN 900 N.At the section contain ing point H,P=0, V= 6 kN,(6x103)(150x10')ET = 1440 N.m%3D900 Nomdo = 72 mm C, d. - 36 mn C; = Co-t = 31 mmT7 (C*- C;* ) I - (c.*-c:") J = 2IFor hadf-pipe,At point HQ = (c.³ - c.' )%3DVQItTC= O55.0 MPa55.0 MPaエ

yes, you are right by equalibrium condition 3+Fd=9 we calculate reactions using the equailibrium…

Answered: Why is the force Fd pointing downwards?…

Why is the force Fd pointing downwards? By force equilibrium, shouldnt it be 3 + Fd = 9 Fd = 6kN, thus pointing upwards?

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Concept explainers

Laws of Equilibrium

A body is said to be in equilibrium if the system of forces acting on the body tends to keep the body at rest, in other words, there is no translation motion or net torque on the body. Under such conditions, the equilibrium conditions are to be applied to analyze the condition. The branch of mechanics that deals with this condition is known as 'statics'. Under equilibrium, there are no net forces on the body.

Question

Why is the force Fd pointing downwards?

By force equilibrium, shouldnt it be

3 + Fd = 9

Fd = 6kN, thus pointing upwards?

8.44 The steel pipe AB has a 72-mm outer diameter and a 5-mm wall thickness.
Knowing that the arm CDE is rigidly attached to the pipe, determine the principal
stresses, principal planes, and the maximum shearing stress at point H.
Use stress transformation eqns.
Fo
9- 3 = 6 kN b
%3D
%3D
To= (12 x10)(120 »10)
%3D
3 kN
%3D
1440
N.m
D
| 9 kN
120 mm
150 mm
120 mm
1440 N-m '6 kN 900 N.
At the section contain ing point H,
P=0, V= 6 kN,
(6x103)(150x10')
E
T = 1440 N.m
%3D
900 Nom
do = 72 mm C, d. - 36 mn C; = Co-t = 31 mm
T7 (C*- C;* ) I - (c.*-c:") J = 2I
For hadf-pipe,
At point H
Q = (c.³ - c.' )
%3D
VQ
It
TC
= O
55.0 MPa
55.0 MPa
エ

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