Analysis of Building Frame: Introduction to Portal Method (Lecture 01)

The Portal Method is suitable for building which has a low elevation and uniform framing. The frames are considered to be a cantilever as they are fixed in the bottom. Moreover, to analyze the building frames using the portal method require following assumptions. 

              01. A hinge is placed at the center of each girder, since this is assumed to be a point of zero   moment.

              02. A hinge is placed at the center of each column, since this is assumed to be a point of zero   moment. 

                03. At a given floor level the shear of the interior column hinges is twice that at the exterior column hinges, since the frame is considered to be a super position of portals.   

Figure 01: A simple frame

Figure 02: Application of assumption 03:the shear of the interior column hinges is twice that at the exterior column hinges

Analysis of Truss: Lecture 02

Example 02: Calculate the force of member AB, AF, BF, BC, FC and FG of the truss shown in figure 01. Indicate whether they are in tension or compression?

Figure 01

Solution: Calculation of Reactions (Figure 02)

∑M_A = 0

= > 3X 4 + 6X8+ 9X12-R_EX16 =0

= > R_E= 10.5KN

∑F_Y = 0

= > R_A+ R_E -3- 6- 9 =0

= > R_A= 7.5 kN

Figure 02

Considering Joint A:

Here, ɵ = tan^-1 (¾) = 36.87⁰

∑F_Y = 0

= > 7.5 + AF Sin 36.87 =0

= > AF = -12.5 KN

            = 12.5 KN (C)

∑F_X = 0

= > AB – AF Cos 36.87 =0

= > AB – 12.5 Cos 36.87 =0

= > AB = 10KN (T)

Figure 03

Consider Joint B:

∑FY = 0

= > 3 - BF =0

= > BF = 3 KN (T)

∑FX = 0

= > BC- AB =0

= > AB = BC = 10KN (T)

Figure 04

Consider Section (1) ~ (1)

Figure 05

Here, β = tan-1 (2/4) = 26.560

∑MC = 0

= > -3X4 + 7.5 X8 + FG Cos 26.56 X 3 + FG Sin 26.56 X4 =0

= > -12 + 60 + 2.68 FG + 1.786 FG =0

= > FG = -10.76 KN

            = 10.76 KN (T)

∑MF = 0

= > 7.5 X 4 – BC X3 = 0

= > BC = 10KN (T)

∑FX = 0

= > 10 + FC Sin 53.13 – FG Cos 26.56 = 0

= > FC = -0.469

            = 0.50 KN (C)

Analysis of Truss: Lecture 01

Example 01: Calculate the force of member AB, AH, BH, BC, BG, and HG of the truss shown in figure 01. Indicate whether they are in tension or compression? 

Figure 01

Solution:  Calculation of reactions (figure 02)

           ∑M_A = 0

            = > 2x20+ 2x40+ 2X 60+ 6X40+5X60-R_E =0

            = > R_E = 9.75 kips

           ∑F_y = 2+ 2+ 2+6+5-9.75-R_A=0

           = > R_A= 7.25 kips

 

Figure 02 

Taking the joints A (Figure 03):

Figure 03

∑F_y = 0

= > 7.25 + AB_Y= 0

= > 7.25 + AB Sin30⁰=0

= > AB= -14.5kips

           = 14.5kips (C )

∑F_x = 0

= > AH – AB_X= 0

= > AH= 14.5 Cos 30⁰

            = 12.6 kips (T)

Taking Joint H (Figure 04):

Figure 04

∑F_x = 0

= > AH – GH= 0

= > AH= GH = 12.6 kips

∑F_y = 0

= > BH=0; Such type of member with zero force is considered as zero force member. 

Taking the section (1) ~ (1) of figure 5:

 

Figure 05

[We are in need of calculating the member force of BC, BG and HG. If we can take moment with respect to B, then the unknown load value of BC and BG will be zero and there will only be one load (HG) which will be easily calculated]

 ∑M_B= 0

= > HGx 11.54 – 7.25x 20= 0

= > HG = 12.56 kips (T)

[Now, If we can take moment with respect to G, then the unknown load value of BG and known load of HG will be zero and there will only be one unknown load (BC) which will be easily calculated. Moreover, BC will have two components and needed to be considered separately]

∑M_G= 0

= > 7.25 X 40 – 2X 20 + BC_X X 11.54 + BC_Y X 20 = 0

= > 250 + BC Cos 30⁰ X 11.54 + BC Sin 30⁰ X 20 = 0

= > 250 + 9.99 BC + 10 BC = 0

= > BC= -12.5 kips

           = 12.5 kips (C)

∑F_Y= 0

= > 7.25 – 2 – 12.5 Sin 30⁰ – BG Cos 60⁰ = 0

= > BG = 2 kips (T)

How to draw a shear force diagram - 02

In the previous lecture, A simple example of SFD for a point load was explained. Today we will learn how to draw shear force diagram for uniformly distributed load (UDL) for a simple beam. Figure 1 shows a simple beam with a P lb/ft uniformly distributed load. The reactions are calculated and place at A and C point. 

Figure 01

It is mentioned in the previous lecture that drawing of SFD should be started from the left side and it is seen that at A point, a upward load (reaction) is present. Hence the figure 2 is drawn at A point. Note that the upward load is considered as positive shear.

Figure 02

Now, it is seen that between A and B, there is a uniform load available. The point is how to draw SFD for Uniformly distributed load? Uniformly distributed load is nothing but a load acting per unit length. Consider figure 03, starting form the left, at the beginning point there is a 1 unit shear. When, 2nd line is considered, the total shear is 2 unit. The point is the more distance the more shear (UDL*distance). So the shear would be downward and UDL will be multiplied by distance.

Figure 03

Using the above mentioned techniques, the SFD between A and C can be drawn. We have a shear load of PL/2 at A point. So the shear at C point would be PL/2-P*L=-PL/2 [See figure 4]. If any point load is placed at B point or other, than the shear should have been drawn between the starting point and that particular point . 

Figure 04

Finally at point C, it is seen a upward load (PL/2) is available. Hence the calculation of shear of that point will be,  -PL/2 + PL/2 = 0 (figure5). 

Figure 05

 

For getting the lecture of drawing SFD for a point load please follow the link.

How to draw a shear force diagram - 01

Drawing a Shear Force diagram is a very important task is solid mechanics. A simple example of SFD for a point load is explained below.

Figure 01

Consider Figure 01: A point load is applied in the middle of the beam and the reactions in the two supports are given in the figure 01. 

Figure 02

01. Figure 02: Starting from the left side of the beam; the reaction at A is upward (see figure 02). As the shear force remains constant between A and B so the shear force diagram is constant between these points (figure 3).

Figure 03

• Note: actually when a structure / beam is in equilibrium then the shear force would be equal to the external load applied to a specific point. If a load in applied (fig 4.a) in a beam. In this point the shear would be the same (fig 4.). If the structure is in equilibrium then there would be an equal and opposite force to resist it (fig 4.c). Similarly, likewise of figure 4 (d) there will be an opposite force similar to the applied external load on the beam. This is the reason why the shear force are same/horizontal when there is no change of applied load between two points.

02.In the point B, there is a P load downward and we have a positive load of P/2 at this point. So the shear at B point would be - +P/2 – P = -P/2. Actually, the total amount of shear at B would be equal to the load applied on that point (P).

Figure 05

03.  Figure 6 is as Same as Number 01

Figure 06

04. A reaction at C, hence upward trend of the load meets the zero point (figure 07).

Figure 07