Example for Trusses,Beams,Frames and Arches

Module 2 : Analysis of Statically Determinate StructuresLecture 9 : Example for Trusses,Beams,Frames and Arches

Objectives In this course you will learn the following

Some examples of trusses.

Example 2.1 (a) Find the forces in AB , AD and AC in the following Figure E2.1.

(b) Find the forces in EG , FG and FH in the following Figure E2.1.

Figure E2.1

Solution:

FBD of the whole system:

(a) FBD of the joint B :

FBD of the joint A :

Force in AB = 22.5 kN (Compression).

Force in AD = 37.5 kN (Tension).

Force in AC = 30.0 kN (Compression).

(b) Take a section through EG , FG , FH and consider the equilibrium of part at the right side

Force in EG = 36.0 kN (Compression).

Force in FG = 22.5 kN (Tension).Force in FH = 18.0 kN (Compression).

Example 2.2 Find the forces in all members in the Figure E2.2.

Figure E2.2Solution:

From equilibrium of the whole body

Looking at joint A :

AB is a zero-force member and

Looking at joint B :

Both BC and BD are zero-force member.

Looking at joint D :

Both CD and DF are zero-force member.

Looking at joint C :

CF is a zero-force member and

Equilibrium of joint E :

and

Equilibrium of joint F :

and

Equilibrium of joint H :

Sign convention: Tension +ve, compression –ve.

Note that we have not obtained the support reactions before finding the member forces. It was not necessary for thisspecific problem. Find out these reactions at supports G and H and check if joint equilibrium is satisfied at these twojoints with the member forces that we have found already.

Recap In this course you have learnt the following

Some examples of trusses.

Example of Beams

Objectives

In this course you will learn the following

Some examples of beams.

Example 2.3 Obtain internal force diagrams for the beam in Figure E2.3.

Figure E2.3

Solution:


FBD for portion BC :

Consider a section at a distance x from A :

( )

( )

The two sets of equation can be written in a combined form by using singular function

where for

for

for

Note that bending moment is zero at the internal hinge location B .

Example 2.4 Obtain internal force diagrams for the beam in Figure E2.4.

Figure E2.4Solution:

FBD of AE :

Taking a section at a distance x from A :

( )

( )

( )

The internal forces remain same for as well, because .

Combining the expressions using singular functions:


Some examples of beams.

Example of Frames


Some examples of frames.

Example 2.5 Obtain internal force diagrams for the frame in Figure E2.5.

Figure E2.5

Solution:


Take a section at a distance from A along AB :

These expressions for internal force are valid from point A to B , using which we can obtain the internal forcediagrams for AB . Using these we can find the internal forces at point B ( ) from which we also get the

forces applied on member BD .

At

Internal forces on BD can be obtained from its free body diagram. Let us take a section at a distance from Balong BD .For

For

Example 2.6 Obtain internal force diagrams for the frame in Figure E2.6.

Solution:


From 6 static equilibrium equations:

For member AB :

For member BC :

For member CD :

For Member DE :

If we draw bending moments on tension side for the whole system


Some examples of frames.

Example of Arches


Some examples of arches.

Example for Arches

In this section we consider only three-hinged arches . A three-hinged arch has two hinges at the ends and oneinternal hinge along its curved span as shown in Figure 2.15. If we draw the free body diagram of the whole system,we have four unknown support reactions ( , , & ). Using the three static equilibrium conditions and the

internal hinge condition (total moment about that hinge is equal to zero), in addition, we can solve for these fourreactions. Therefore, three-hinged arches are statically determinate structures.

Figure 2.15 A three-hinged arch and its free-body diagram

Unlike other types of structural member we have dealt with so far, the centroidal axis of an arch is curvilinear.Therefore, the direction of axial force, which is aligned along the centroidal axis, does not remain same along thespan with respect to global coordinates. This is true for the direction of shear force as well. However, bendingmoment does not get affected due to the curved nature of the centroidal axis. In case of circular arches, theproblem with axial and shear force directions can be easily handled with circular coordinates as shown in

Example 2.7. The alternative is to consider internal forces in global x and y directions ( and ) in stead of

forces axial and transverse directions ( P and V ), as shown in Example 2.8.

Example 2.7 Obtain the internal force diagram ( AFD , SFD and BMD ) for the arch in Figure E2.7.

Figure E2.7

Solution:FBD of the whole system:

(i) (ii)

(iii)

FBD of part BC :

(iv)

From Equation (iii) and (iv):

Internal forces at an angle in AB :

At point B :

Internal forces in BC :

For curved structures, for example an arch, internal forces are often expressed as x and y components instead ofaxial and shear forces. Note that bending moment remains same.

Following same procedure as before we can obtain the variation of these internal forces as well, Try to find outdiagrams for , for the same arch. The next example illustrates such internal force diagrams.

Example 2.8 Obtain the internal force diagram ( , and bending moment diagram ) for the symmetric arch in

Figure E2.8.

Figure E2.8

Solution:FBD of the ABC :

It is a symmetric structure, so we need to consider only half of the system, either AB or BC .


Some examples of arches.

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Example for Trusses,Beams,Frames and Arches