C.K.Wang Intermediate Structural Analysis

[Tres]

Anyone having a copy of C.K.Wang Intermediate Structural Analysis.I was brushing up my analysis fundas for an important project coming up but have a very important doubt to clear from this book.This is making me lose sleep!!!!!Please help.If anyone has the book plz respond for further discussion.

 

[jheidt2543]

The two books by C.K. Wang that I have are:

"Elementary Therory of Structures", by Wang & Eckel, McGraw-Hill, 1957

"Statically Indeterminate Structures", by C.K. Wang, McGraw-Hill, 1953

So, I'm not sure if you are refering to one of these or not.
I'll help if I can.

jheidt2543@aol.com

 

[Tres]

Thanks very very much for the reply.May be the doubt is a fundamental one.In the analysis by slope deflection method, say we are analysing an unsymmetrical rectangular rigid frame(single bay), while writing the equilbrium equation for the sidesway to the right, considering the joints at the level of storey level we make the use of the equilbrium equation stating that:
"The sum of the horizontal forces to the left(left indicating the positive direction) acting on the combined free body of the joints is zero."(In my book page 212)
Wang has an example of two storey rigid unsymmetrical frame.Now suppose we take first storey level:
Equilbrium equation :
According to Wang's problem, fconsidering the free body of the joints at the first storey level Wang states:
-H9 - H11 -24 + H6 + H8=0;

H6=M5+M6/6.4 (acc. to Wang's problem)
H8=M7+M8/6.4 (acc. to Wang's problem)
H9=M9+M10/6.4 (acc. to Wang's problem)
H11=M11+M12/3.2 (acc. to Wang's problem)

H9, H11 are the horizontal shears corresponding to lower columns to left and right respectively and H6 , H8 are the horizontal shears corresponding to upper columns .Wang assumes that left indicates positive direction.He assumes H9 and H11 acting to the right and H9 and H11 to the left.Now, if we assume all H6,H8,H9,H11 acting to the right and say that:
-H9 - H11 -24 -H6 -H8=0;

H6=M5+M6/6.4
H8=M7+M8/6.4
H9=M9+M10/6.4
H11=M11+M12/3.2
the final results are varied?How?

Trevor.

 

[austim]

Trevor,

I don't think we need Wang's textbook to sort out your problem.

You cannot have positive H6,H8,H9,H11 all to the right, if you use a consistent sign convention for shear force. If (as Wang appears to assume) you consider positive shear to be clockwise (ie ----|___ ) in all columns, then the force on your free body due to positive shear in an upper column must be in the opposite direction to the force from positive shear in a lower column.

The four equations expressing H in terms of M are only correct as stated if you use Wang's sign conventions for positive shear force and bending moment consistently.

If you wish to change the sign convention for positive shear in the upper columns as you have suggested, then you also have to change the equations for H6 and H8 to

-H6=(M5+M6)/6.4 and
-H8=(M7+M8)/6.4.

(Note that I have replaced the brackets which you must surely have omitted from the formulae. An equation saying shear=moment+moment/distance is dimensionally inconsistent).

 

[Tres]

Hello :
Thanks very much for the reply. What i am trying to follow is:
I assume horizontal force acting to the right as positive(because we have assumed sidesway to the right as positive).Then based on clockwise moments at column ends(as we have for moment equations) i get horizontal forces .Now , if i get horizontal force to the left i take as negative and if i get it to the right i take it as positive.Then i form the shear equation.Doing so i also get symmetricity of coefficients in the system of linear equations.Is this what Wang is trying to expalin?This method works for all problems(i get symmetricity of coefficients too).On the other hand if i assume horizontal force acting to the left as positive the final answer comes to be same ofcourse but i do not get symmetricity of coefficients in the system of linear equations.Please send in your opinion!!Thanks again.

 

[austim]

trevor, you have clearly missed the point.

OK, by all means take "sidesway to the right as positive". But that doesn't mean that both the upper and lower columns will impose forces to the right.

Consider just the forces on a column. A horizontal force to the right at the top will generate sway to the right. How will this force be reacted? By an equal force to the left at the bottom.

Since action and reaction are equal and opposite, the column will apply a force to right on the storey below, and a force to the left to the storey above.

 

[Tres]

Thanks again for the reply.Much obliged.I did not mean that both the upper and lower columns will impose forces to the right.What i was confused was the sign convention Wang was following.I assume that when we write the equation of equilbrium any force (irrespective whether in upper or lower column) if it acts to the right will be taken as positive(corresponding to positive sidesway to the right we assumed) if it acts to the left take it as negative, equating the sum of these forces to zero we get the equation of equilbrium corresponding to sidesway.Also in the system of the resultant linear equations all the coefficients are symmetric with respect to principal diagonal.On the contrary if i take the convention force to the right as negative and left as positive (for substituting in equilbrium equation) the coefficients are not symmetric though i get same answer for displacements.

 

[jheidt2543]

Trevor,

The slope-deflection method uses the sumation of H=0, the sumation of M=0 and consistant deformation to solve for the unknown forces in an indeterminate structure. The key is that you MUST use a consistant sign convention through out for both the external forces and the internal member and joint forces. I think if you work through the example again starting with the external known and unknown forces, then go through each member's free body diagram using the SAME sign convention through out, you will find the correct method.

 

[austim]

Trevor,

Now I would really prefer to see what Wang says to be sure of what I am about to say. Never mind, I have been wrong before.

It doesn't help that in your original post you wrote "He assumes H9 and H11 acting to the right and H9 and H11 acting to the left". What follows assumes that H9 and H11 are the lower columns.

I believe that I was correct in initially suggesting that Wang uses a convention for shear force that gives clockwise shear
(-----|_____) as positive. (Sketching that vertically is too tricky for this forum).

If that is so, then a sway "to the right" (meaning that the top of a column moves to the right of the bottom) will involve a positive shear.

In a lower column (Col 9, as an example), to generate a positive shear H, you must apply a force to the right at the top. That is, the shear force = +H9. But the column will exert a force to the left on the free body of the storey nodes. Thus Wang's equation for horizontal force equilibrium has the term -H9.

To generate a positive shear in an uper column (eg col 6), you must also apply a positive force at the top. But in this case, the column will exert a force to the right on the free body below it. So Wang's equation has the term +H6

This all takes much longer to describe than it would take for you to sketch it out. (As I have said elsewhere - when in doubt, draw it out).

I strongly advise you to sketch out the free body, plus the columns above and below, plus the floors above and below, with each floor swayed to the right of the one below. For simplicity, treat the floors as rigid, so the the columns remain vertical at each floor level. At midheight of each column, leave a gap so that you can draw in a positive shear force. That will be to the right at the top of the gap, and to the left at the bottom. Label the shears H6, H8,H9,H11.

When you look at your finished sketch, it should be absolutely clear why Wang has -H9-H11+H6+H8 in his equation of equilibrium, and why your suggestion of -H9-H11-H6-H8 cannot be right.

 

[Tres]

OK,fine.But still there is a problem. If there is a horizontal force of 20KN.acting to the right at that joint Wang puts it in the eqilbrium equation as -20??????Should'nt it be +20??

 

[austim]

Trevor,

No, the error is yours and mine (I did say I've been wrong before ), not Wang's.

When I re-read your original post you quoted Wang as saying "the sum of the forces to the left (left indicating the positive direction)...". The statement that positive is to the right was your own, and I did not pick up the inconsistency until just now.

I have also just picked up an error in my previous post, where I said draw the shear force "to the right at the top of the column gap, and to the left at the bottom". That would be the wrong way if clockwise shear were positive.

IF Wang's example is for sway to the right, and IF he is summing forces to the left as positive, and IF the terms for applied load to the right and shears in the lower columns have the same sign, then I have to conclude that he really treats anti-clockwise shear (____|----) as positive.

IF that is correct, then the argument goes like this - For anticlock positive, you should draw +shear to the right at the top, and to the left at the bottom of each gap (just draw a column sheared anticlockwise with the bottom to the right of the top to see that). So (entirely due to an inspired error on my own part) your/our sketch is correct as drawn, with H9 etc to the right at the top of each gap, and to the left at the bottom. (Note that the equation of horizontal equilibrium would not change even if you had sway to the left – the direction of positive H9 etc is solely dependent on the sign convention for shear, and you draw the variable H9 as I have indicated, even though its value could be positive or negative).

Now sum the forces to the LEFT as positive, and I think you/I have it. At least I do hope so - even I am beginning to lose sleep now.

 

[Tres]

H9, H11 are the horizontal shears corresponding to lower columns .In Wang's problem he has directed H9 and H11 to the right.H6 , H8 are the horizontal shears corresponding to upper columns .In Wang's problem he has directed H6 and H8 to the left.
This means clockwise shears positive????
Then he says that a force directed to the left is to be taken as positive and to the right as negative.He then finally writes the eqilbrium equation where obviously a horizontal force of 20KN. to the right is negative.
H6+H8-H9+H11-24=0 (equation 1)
Fine!!!
What i say if we find the horizontal forces in these upper and lower columns with clockwise moments at the column ends we get in the lower column a force directed to the left(for rotational eqilbrium to be satisfied)[H9 and H11] and in the upper column a force directed to the right (for rotational eqilbrium to be satisfied)[H6 and H8]
Now the equilbrium equation becomes:
Using left as positive:
H9+H11-24-H6-H8=0 (equation 2)
Equation 1 and Equation 2 are not similar????
This is my problem
I know Wang is right cause i have verified answers with STAAD analysis too which are same as that of Wang's

 

[austim]

this is my very last gasp.

Try to understand what I wrote in my penultimate paragraph, about the difference between the variables H6, etc and their values.

When Wang shows an equation -H9-H11-24+H6+H8 = 0 he is using the shear forces in the columns as variables, with anti-clockwise shear having a positive value. For sway to the right, the values of these variables will all be negative.

Suppose that both lower columns have shear forces of -10, and the upper columns have -15.

Using Wang's formula, you would write
-(-10)-(-10)-24+(-15)+(-15)=0.


That is exactly the same as your own conclusion, except you want to write it as 10+10-24-15-15=0, and then say the equation is H9+H11-24-H6-H8=0. But that is just not so, when H9=-10, H11=-10, H6=-15 and H8=-15.

The basic point is that having a negative value for any variable in an equation does not change the equation.

 

[Tres]

Austim,
My profound gratitude to you!!Thanks a lot---Much grateful
Suppose i take the left hand side upper column as a free body with colckwise moments at the ends.(Lets us say for a while that both the upper and lower colun are fixed at the ends)
Let us say for the left hand side upper column,denoting the fixed end as 5 and other end as 6.The horizontal reaction at the fixed end(M5+M6)/L( directed towards the left).This is shear in the upper left column.

Similarly for the right hand side upper column ,denoting fixed end as 7 and other end as 8 The horizontal reaction at the fixed end(M7+M8)/L( directed towards the left) .This is shear in the upper right column.


Now lower columns :
Let us say the left hand side fixed end as 10and other end as 9.The horizontal reaction at the fixed end(M9+M10)/L( directed towards the right) . This is shear in the lower left column.

Let us say the right hand side fixed end as 12 and other end as 11.The horizontal reaction at the fixed end(M11+M12)/L( directed towards the right) .This is shear in the lower right column.


To get sum of horizontal forces equal to zero (take left as positive) we add the above shears in the upper and lower columns (with load if any) and equate to zero?Right?

 

[austim]

TRevor.

I think you have now got it. Well done.

 

[jheidt2543]

Austim,

Well done too!

 

[Tres]

What i did earlier was earlier that i was taking the equilbriating force at the beam column junction , that is in the upper columns is (M5+M6)/L and (M7+M8)/L(directed towars right(negative)
and lowe columns the equilbriating force as (M9+M10)/L and M11+M12)/L( directed towards the left(positive))
There is also a external horizontal force of 24 KN. at the beam column junction directed towards right which i took as negative.
When i added the above to obtain the horizontal equibrium equation to equate to zero , i did not get displacements(slopes and sidesway) as above.
Just last one, what was the error(just to make myself doubly sure of what i have in mind)
Please do'nt get impatient.Austim, i run short of words to express my gratitude.