Text:A beam is made using the HE240A profile. The beam is 7 meters long and secured against displacement at node A and restrained against displacement in the vertical direction at node B, which indicated on the sketch.The beam is supported by a uniformly distributed load q= 6,40 kN/m between junction point A and B. and concentrated downward load P. Modulus of elasticity for steel is 2,0x105N/mm2. Cross-sectional data is obtained from steel table.A)Calculate how large the force can be when the transverse displacement (see image)apply superposition principleCalculate how large the transverse displacement (see image), , under the concentrated load will be for this the load effect. When the transverse displacement u(x) for a freely supported beam with uniform distributed load is: SEE IMAGE OPPGAVE NR. 1:En bjelke er utført ved bruk av HE240A profil. Bjelken er 7 meter lang og fastholdt motforskyvning i knutepunkt A og fastholdt mot forskyvning i vertikal retning i knutepunkt B, somangitt på skissen.A#P =3000fra ståltabell.3500*A₁4₂*Bjelken er påkjent av en jevnt fordelt last99 = 6, 40KNmPB40003500**Nkonsentrert nedadrettet last P. Elastisitetsmodul til stål er 2, 0 × 10²BkN=6, 40 mellom knutepunkt A og B, og enmmm[mm]. Tverrsnittsdata hentes a) Beregn hvor stor kraften P kan være når tverrforskyvningen A₂ ≤2(anvendsuperposisjonsprinsippet).b) Beregn hvor stor tverrforskyvningen, A₁, under den konsentrerte lasten blir for dennelastvirkningen. Når tverrforskyvningen u(x) for en fritt opplagret bjelke med jevntfordelt last er:u(x)=4qL424EI LX+c) Beregn rotasjonene 4 og 3 for denne lastvirkningen (anvend superposisjons-prinsippet).

NOTE : As the translation of part C is not provided, solution of only parts A & B is provided.…

Answered: A beam is made using the HE240A…

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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Text:A beam is made using the HE240A profile. The beam is 7 meters long and secured against displacement at node A and restrained against displacement in the vertical direction at node B, which indicated on the sketch.

The beam is supported by a uniformly distributed load q= 6,40 kN/m between junction point A and B. and concentrated downward load P. Modulus of elasticity for steel is 2,0x10⁵N/mm². Cross-sectional data is obtained from steel table.

A)Calculate how large the force can be when the transverse displacement (see image)apply superposition principle

Calculate how large the transverse displacement (see image), , under the concentrated load will be for this the load effect. When the transverse displacement u(x) for a freely supported beam with uniform distributed load is: SEE IMAGE

OPPGAVE NR. 1:
En bjelke er utført ved bruk av HE240A profil. Bjelken er 7 meter lang og fastholdt mot
forskyvning i knutepunkt A og fastholdt mot forskyvning i vertikal retning i knutepunkt B, som
angitt på skissen.
A
#
P =
3000
fra ståltabell.
3500
*
A₁
4₂
*
Bjelken er påkjent av en jevnt fordelt last
9
9 = 6, 40KN
m
PB
4000
3500
*
*
N
konsentrert nedadrettet last P. Elastisitetsmodul til stål er 2, 0 × 10²
B
kN
=
6, 40 mellom knutepunkt A og B, og en
m
mm
[mm]
. Tverrsnittsdata hentes

a) Beregn hvor stor kraften P kan være når tverrforskyvningen A₂ ≤2(anvend
superposisjonsprinsippet).
b) Beregn hvor stor tverrforskyvningen, A₁, under den konsentrerte lasten blir for denne
lastvirkningen. Når tverrforskyvningen u(x) for en fritt opplagret bjelke med jevnt
fordelt last er:
u(x)
=
4
qL4
24EI L
X
+
c) Beregn rotasjonene 4 og 3 for denne lastvirkningen (anvend superposisjons-
prinsippet).

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