The steel plane truss shown in the figure is loaded by three forces P, each of which is 490 KN. The truss mem- bers each have a cross-sectional area of 3900 mm² and are connected by pins each with a diameter of dp = 18 mm. Members AC and BC each consist of one bar with thickness of AC = TBC = 19 mm. Member AB is composed of two bars [see figure part (b)] each having thickness LAB/2 = 10 mm and length L = 3 m. The roller support at B, is made up of two support plates, each having thickness tsp/2 = 12 mm. (a) Find support reactions at joints A and B and forces in members AB, BC, and AB. (b) Calculate the largest average shear stress T.max in the pin at joint B, disregarding friction between the members: see figures parts (b) and (c) for sectional views of the joint. (c) Calculate the largest average bearing stress 45° max acting against the pin at joint B. C IP = 490 KN P L=3 m by Support plate and pin 45 B 600 By A a a b P Member AB Pin By Fac at 45 Member BC -Support plate B (b) Section a-a at joint B (Elevation view) Member AB (2 bars, each) Pin Support plate (2 plates, cach) Load Pat joint & is applied to the two support plates (c) Section b-bat joint B (Plan view)

The steel plane truss shown in the figure is loaded by three forces P, each of which is 490 KN. The truss mem- bers each have a cross-sectional area of 3900 mm² and are connected by pins each with a diameter of dp = 18 mm. Members AC and BC each consist of one bar with thickness of AC = TBC = 19 mm. Member AB is composed of two bars [see figure part (b)] each having thickness LAB/2 = 10 mm and length L = 3 m. The roller support at B, is made up of two support plates, each having thickness tsp/2 = 12 mm. (a) Find support reactions at joints A and B and forces in members AB, BC, and AB. (b) Calculate the largest average shear stress T.max in the pin at joint B, disregarding friction between the members: see figures parts (b) and (c) for sectional views of the joint. (c) Calculate the largest average bearing stress 45° max acting against the pin at joint B. C IP = 490 KN P L=3 m by Support plate and pin 45 B 600 By A a a b P Member AB Pin By Fac at 45 Member BC -Support plate B (b) Section a-a at joint B (Elevation view) Member AB (2 bars, each) Pin Support plate (2 plates, cach) Load Pat joint & is applied to the two support plates (c) Section b-bat joint B (Plan view)

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Concept explainers

Axial Load

When a bar of the uniform cross-section is acted by a load, such that the load acts along the longitudinal axis of the bar, such kinds of loadings are known as axial loadings. In general terms, a load is an external force acting on a component. An axial load acts perpendicular to the geometric cross-section of the component. Based on the direction of the application of the load, an axial load can be tensile or compressive. The analysis of bodies under the action of axial loads is generally studied under the branch of mechanics, known as statics.

Question

The steel plane truss shown in the figure is loaded by three forces P, each of which is 490 kN. The truss mem-
bers each have a cross-sectional area of 3900 mm and are connected by pins each with a diameter of d, = 18 mm. Members AC
and BC each consist of one bar with thickness of tac = trc = 19 mm. Member AB is composed of two bars (see figure part (b)]
each having thickness tAg/2 = 10 mm and length L = 3 m. The roller support at B, is made up of two support plates, each having
thickness fp/2 = 12 mm.
(a) Find support reactions at joints A and B and forces in members AB, BC, and AB.
(b) Calculate the largest average shear stress T„max in the pin at joint B, disregarding friction between the members;
see figures parts (b) and (c) for sectional views of the joint.
(c) Calculate the largest average bearing stress o max acting against the pin at joint B.
|P = 490 kN
P.
Member AB
a
(2 bars, each
Fac at 45
Member AB L Member BC
by
L = 3 m 45
-Support plate
(2 plates, cach
Pin plate
Pin
Luoddng-
A
45
B
By
Support
Load Pat joint B is applied
to the two support plates
(b) Section aa at
joint B (Elevation view)
plate
By
and pin
(e) Section b-b at
joint 8 (Plan view)
A

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