Shear stress, what is a flat tie bar ?

In summary, a flat tie bar is a flat strip of metal used to connect two lengths of material. In this problem, the maximum pull that can be applied to the tie-bars is being calculated, taking into account the shear stress in the rivets, which is not to exceed 60 MN/m^2. The lap riveted joint connecting the two bars has three rivets, each 10mm in diameter, and the cross-sectional area of these rivets will bear the shear stress. The stress is calculated by dividing the load by the area bearing that load.
  • #1
scotthands
25
0
shear stress, what is a flat tie bar...?

Homework Statement


two lengths of flat tie bar are connected together by a lap riveted joint with three rivets each 10mm in diameter. Calculate the maximum pull that can be applied to the tie-bars if the shear stress in the rivets is not to exceed 60MN/m^2


The Attempt at a Solution



i can't attempt this because i don't know that a flat tie bar is or what it would look like if they were connected together by a lap riveted joint with three rivets...
does anyone know what it is?
 
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  • #2
scotthands said:
i can't attempt this because i don't know that a flat tie bar is or what it would look like if they were connected together by a lap riveted joint with three rivets...
does anyone know what it is?

This is a tie bar. http://en.wikipedia.org/wiki/Image:Tieslide.jpg
 
  • #3
Well a flat tie bar is flat strip of metal, usually uniform thickness with a rectangular cross-section - basically looks like a ruler, but thicker. The two bars are overlapped and held together by 3 rivets. However that is irrelevant to this problem.

What is relevant is the fact that there are three rivets, each of 10 mm dia, which will bear the load P to be determined such that the shear force does not exceed 60 MN/m2.

State assumptions, e.g. not credit given to the friction in the lap joint of the lapped tie bars, i.e. one assumes the rivets bear the full load. What else?
 
  • #4
so have i just to work out P? as in: P=T(tau) * Area
?

if so then I'm not really sure about the area, do i take the 10mm diameter and work out the area that the rivets go through?(or is it the perimeter of the 10mm diameter the hole that the rivets go through?)

thanks for the help.
 
  • #5
The cross-sectional area of the three rivets will bear the shear stress.

The stress is just the load divided by the area bearing that load (force).

Remember stress, which has the units of pressure (lbf/in2, or MPa), is simply force divided by area, whether it's a tensile, compressive or shear force.
 

Related to Shear stress, what is a flat tie bar ?

1. What is shear stress?

Shear stress is a type of stress that occurs when two surfaces slide against each other in opposite directions. It is a result of forces acting parallel to the cross-sectional area of an object.

2. How is shear stress calculated?

Shear stress is calculated by dividing the force acting on a material by its cross-sectional area. The resulting value is typically measured in units of force per area, such as pounds per square inch (psi) or newtons per square meter (N/m^2).

3. What is a flat tie bar?

A flat tie bar, also known as a shear reinforcement bar, is a structural element used to resist shear forces in reinforced concrete structures. It is typically a steel bar or plate that is placed perpendicular to the direction of the shear force and is anchored to the concrete on both ends.

4. How does a flat tie bar work?

A flat tie bar works by transferring the shear force from one part of the structure to another, effectively reinforcing the concrete and preventing it from failing under the applied stress. It acts as a connector between two parts of the structure, helping to distribute the shear force more evenly.

5. Where are flat tie bars commonly used?

Flat tie bars are commonly used in reinforced concrete beams, slabs, and walls. They are also used in other structures where shear forces are present, such as bridges and retaining walls. In addition, they can be used in retrofitting existing structures to increase their resistance to shear stress.

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