A box girder has the singly symmetrical trapezoidal cross section shown in Fig. P.10.23. It supports a vertical shear load of 500 kN applied through its shear centre and in a direction perpendicular to its parallel sides. Calculate the shear flow distribution and the maximum shear stress in the section. Ans. qon=0.25sa 9AB = 0.21sg–2.14×10-4+250 9вс — 0.17sс + 246 Tmax=30.2 N/mm²,

A box girder has the singly symmetrical trapezoidal cross section shown in Fig. P.10.23. It supports a vertical shear load of 500 kN applied through its shear centre and in a direction perpendicular to its parallel sides. Calculate the shear flow distribution and the maximum shear stress in the section. Ans. qon=0.25sa 9AB = 0.21sg–2.14×10-4+250 9вс — 0.17sс + 246 Tmax=30.2 N/mm²,

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

Braced cuts

The braced cut is a method of providing lateral support to the soil in foundation trenches, to prevent the trenches from deformation by the caving in of soil and resisting it from moving laterally. During shallow excavation in firm soil, generally, slopes are provided to resist the caving in of soil. During shallow excavation in soft soil and deep excavation, the chances of caving in of soil increase. Hence, it becomes necessary to make a provision to resist the deformation of trenches due to caving in of soil. The braced cut is the method used for such conditions. The excavation using braced cuts is a bit uneconomical as compared to the traditional open-cut excavation method but provides better safety than the cut excavation method.

Question

A box girder has the singly symmetrical trapezoidal cross section shown in Fig. P.10.23.
It supports a vertical shear load of 500 kN applied through its shear centre and in a direction
perpendicular to its parallel sides. Calculate the shear flow distribution and the maximum shear
stress in the section.
Ans. qOA=0.255A
9AB = 0.21sg-2.14×10-4+250
9BC=-0.17sc +246
Tmax=30.2 N/mm?.
!!

500 kN
SC
8 mm
10 mm
10 mm
120°
120°
12 mm
sB
1m
SA
А
E
F
2m

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