- Tension member
Tension members are structural elements that are subjected to axial
tensile force s. They are usually used in different types of structures. Examples of tension members are: bracing for buildings and bridges, truss members, and cables in suspended roof systems. In an axially loaded tension member, the stress is given by:F = P/A
Where P is the magnitude of the load and A is the cross-sectional areaThe stress given by this equation is exact, knowing that the cross section is not adjacent to the point of application of the load nor having holes for bolts or other discontinuity. For example,
if you are given a 8 x ½ plate is used as a tension member (section a-a) and it is connected to a gusset plate with two 7/8-inch-diameter bolts (section b-b). So,
The area at section a - a (gross area of the member) is 8 x ½ = 4 in2
However, the area at section b - b (net area) is (8 – 2 x 7/8) x ½ = 3.12 in2 Knowing that the higher stress is located at section b - b due to its smaller area
To understand more about tension member, it would be useful to look at the stress-strain behavior of steel:
Where, Fy is the yield stress, Fsu is the ultimate stress, εy is the yield strain and εsu is the ultimate strain
Designing a tension member
In order to design tension members, it is important to analyze how the member would fail under both yielding (excessive deformation) and fracture which considered the limit states. The limit state that produces the smallest design strength is considered the controlling limit state; which also prevent the steel structure from failure.
Using AISC (American Institute of Steel Construction), we could obtain the recommended load and resistance factor design approaches.
The ultimate load on a structure can be calculated from one of the following combination:
1.4 D
1.2 D + 1.6 L + 0.5 (Lr or S)
1.2 D + 1.6 (Lr or S) + (0.5 L or 0.8 W)
1.2 D + 1.6 W + 0.5 L + 0.5 (Lr or S)
0.9 D + 1.6 W
Where:
D… is the dead load or the weight of the structure itself
L… is the live load which vary for different structures
S… is the snow load
W… is the wind load
the central problem of designing a member is to find a cross section for which the required strength doesn't exceed the available strength :
Pu < ¢ Pn where Pu is the sum of the factored loads.
to prevent yielding
0.90 Fy Ag > Pu
to avoid fracture,
0.75 Fu Ae > Pu
herefore, to design a tension member, we have to consider the loads applied to this member, the design forces acting on this member (Mu, Pu, and Vu), and the point where this member would fail.
ee also
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Compression member
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