Methods of Structural Design

Basically two methods can be used for designing a structure, i.e., Permissible Stress and Limit State Method. In this article I will explain the process of using the two design methods in detail. I will also show you how forces in a structure can be determined using a graphical method. Explanation will be provided in such a simple way that the people with limited knowledge can also get benefit from this article. An example will be used to explain the two methods. Engineering students and those who want to learn structural engineering will get benefit from this article.

Permissible Stress Method

In the permissible stress method the maximum stress that a material can bear elastically (yield stress, fy) is multiplied by a coefficient, say 0.6. This reduced stress is called a permissible stress or allowable stress. The design stress that is calculated using the actual applied load is compared with the permissible stress. In this method we make sure that the design stress does not exceed the permissible stress.

Limit State Method

In the method of strength design the design force (force or moment), developed in the member/structure, is multiplied by a factor, say 1.5 foe example. The force multiplied by a factor is called a factored force. The applied load can also be multiplied prior to analysis. In this case the multiplying factor is called a load factor. The same term is used even if the internal forces are multiplied. Capacity of the member that is calculated using the yield stress is reduced by a capacity reduction factor, say 0.9. Ratio of Capacity to force is calculated. In this method we make sure that this ratio is not less than 1. Because of the load factor used, this method is also called Load factor method.

Different load factors are used for different limit states and different load combinations. There are basically three limit states, namely, Strength, Serviceability and Stability limit states. In the Strength design we use a highest load factors. When we combine live load (LL) with dead load (DL), also called self load, we use a lesser load factor for self load than that for the live load. For example, 1.5*LL+1.2*DL.

Reason being, live load is not constant and there is a significant possibility for live load exceeding the estimated amount. Whereas, the same possibility for the dead load  is low.

In the Serviceability design we basically, make sure that the structure is usable. For example, deflection of a beam should not be more than a certain value, say span/250. In the process of calculating the deflection a load factor equal to 1.0 is used in general.

In the Stability design we make sure that the structure does not move due to applied load. For example, when we analyse a dam we basically check its stability against overturning, sliding, floatation and bearing capacity failure, etc. For stability analysis we use generally the same load factor as strength. In the similar way, a capacity reduction factor is used to reduce the capacity of a member, a stabilizing capacity is multiplied by a reduction factor, such as 0.9 for weight and 0.45 for bearing capacity of the foundation soil, etc. In the stability design we make sure that the ratio of reduced stabilising capacity to factored destabilising capacity is not less than one.

Example of Strength Design: Now let's see through an example how we design a member. Consider two rods holding 20 kN of weight as shown below. We need to design the rods A and B.

Arrangement of rods

In order to design the rods we need to analyse the structure first to determine the forces that develop in in the rods due to applied load 20 kN. For the purpose of determining forces we can use either a calculations or a graphical method. I will show you how the forces in the rods can be determined using a graphical method.

Cable car hanging in a cable

Determination of forces using graphical method:

Since weight is acting vertically, draw a vertical line for the 20 kN force to some reasonable scale, say 20 mm. Draw the lines for rods A and B starting from beginning and end of the vertical line for 20 kN force. The lines for rods A and B must be drawn precisely to 60 and 45 degrees with vertical respectively. Distance measured along the lines for rods from their intersection point to line for vertical force gives the force developed in the respective rods. The measured distance shall be converted to forces using the scale that was used to draw the vertical force. If you have used 20 mm =20 kN ,i.e., 1mm for 1 kN scale for drawing the vertical force, then you will measure about 14.5 mm and 18 mm for rods A and B respectively. Therefore, convert the measurements to forces approximately to 14.5 kN (14.5 mm*1 kN)  and 18.0 kN (18.0 mm*1 kN) for Rod-A and Rod-B respectively. Watch the arrows used in the lines for rods A and B. The arrows shall always be in the spinning direction.In this case it should be in anti clockwise direction. The arrow shows whether the forces are compression or tension. In the following example the forces in rod are tension. Following figure describes this process.

Graphical method for determination of forces

Determination of forces by calculation:
In order to find the forces more precisely, in this example, a Sin formula can be used which is described below.

Refer the first figure- Arrangement of rods

FA/Sin 135 = FB/Sin 120 = 20/Sin (60+45)
Using this equation, force in Rod-A, FA=Sin135*20/Sin 105=14.64 kN

Similarly, force in Rod-B, FB=Sin120*20/Sin 105=17.93 kN

Design of Rods: After determining forces in the rod we can design the rod. Let's design the rods using permissible stress method first then we will design it also using the limit state method. We will chose steel rod for this design. Most of the structural designs are prepared basing on Hook's law. Therefore, we use the yield stress of steel rod, fy = 250 N/mm2.

Permissible Stress method:

Let's say permissible stress for the steel, σp=0.6*250= 150 N/mm2. Here 0.6 is a coefficient for permissible stress.

Let's assume 10 mm diameter solid rod.

Then cross section area of rod, A=π /4*10^2=3.14/4*10^2=78.5 mm2

Rod-B

Stress in Rod-B, σB=FB/A=17.93*1000/78.5=228.41 N/mm2 > 150 N/mm2 No good, choose a larger rod

use 14 mm dia. rod, Area=π /4*14^2=153.86 mm2

Here the force is in kN and permissible stresses is in N/mm2. In order to keep the same unit system we need to convert the force to N by multiplying it by 1000. 

Now, stress in Rod-A, σB=17.93*1000/153.86=116.53 N/mm2<150 N/mm2 Good, use 14 mm dia. rod.

The optimal diameter would be 12.34 mm which would give exactly 150 N/mm2 stress in the rod. Due to availability reason 14 mm shall be chosen.

Rod-A

Similarly, force in Rod-A, FA=14.6 kN which is less than 17.93 kN the same diameter as in Rod-B will be safe to use in Rod-A. However, using the same method of calculation, a smaller rod may also be designed if available. 

Different coefficients for permissible stresses for different purposes are specified by the national standard.

Limit State method:

Internal load in Rod -B, NB=17.93 kN

Load factor=1.5, Yield stress, fy=250 N/mm2

Factored load for rod-B, NB*=1.5*17.93=26.89 kN.

Here, the asterisk (*) next to NB symbolises the force NB to be a factored load.

Let's assume 10 m dia. rod, cross section area= 78.5 mm2

Capacity reduction factor, ф =0.9

Tensile capacity of the rod, фN=0.9*250*78.5=17662.5 N or 17662.5/1000=17.662 kN

Capacity/Load=17.662/26.89<1 No good.

Let's try with 14 mm dia. rod, then cross section area= 153.86 mm2

Tensile capacity of the rod, фN=0.9*250*153.86=34618.5 N =34618/1000=34.62 kN

Capacity/Load=34.62/26.89=1.29>1 Good.

Use 14 mm dia. rod for the both rods.

As mentioned earlier, because of the availability, a larger rod has been used here. For the Rod-A a smaller rod can also be be designed in the similar way if available. 

Load factors and capacity reduction factors are to be used as specified by the national standard.

Design for connections at the supports and at intersection point need to be designed as well. I will explain in subsequent article about how to design a connection.

Design process of designing compression members such as columns are different to that of tension members. I will explain in subsequent article about how to design columns, beams,etc. In design of all these structures/members we compare the reduced capacity to factored applied load or in permissible stress method, we compare the developed force (moment or force) with the respective capacity that is calculated using permissible stresses. 

Serviceability Limit State:

In the serviceability check we generally apply a load factor=1 and then determine the actual deflection of the structure, such as deflection of a beam. Depending on type of structure we have different deflection limits that make sure the intended use of the structure. For example, for a beam carrying a wall, we can limit a lower deflection, say span/200. Because we don't want to see any cracks in the wall. Crack width of a concrete structures retaining water can be limited to max. 0.2 mm. This limitation makes sure that the corrosion of reinforcement is minimised. Theses deflections, serviceability limits are specified in the national standard.

With this, in general, I have explained the two methods of structural design through an example of design of a tension member. In the next article I will explain briefly the basics of designing a  compression member.

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