-

-

.

-*.

÷!I

or EMA = o

=

-

7

METHOD OF SECTIONS

Learning Objectives 1). To employ the method of sections to evaluate the axial force

carried by selected members in a truss. 2). To do an engineering estimate of the load in select members

of a truss. Procedure 1). Draw a FBD of the entire truss showing the reaction forces

at the supports and the external loads. Write the equilibrium equations and solve for as many unknowns as possible.

2). Locate the force members to be evaluated. Identify whether any of these forces can be determined by observation (e.g., zero-force members).

3). Identify section to be used and draw a FBD of the section including any support reactions, external loads and internal forces of sectioned members. Remember, the cutting plane must cut through the members of interest. Also the cutting plane need not be straight, it may be curved.

4). Write the equilibrium equations for one of the two sections. The equations for either half of the section will yield the same member forces.

5). Three equilibrium equations are available, so up to three unknowns can be solved with a single section.

6). At times more than one section may be necessary.

-

-

=

--_-

→ choose a suitable

point for moment

equation , leave

only one unknownforce in equation .

i

÷Cut

→

Ax T TAy My

ch) Emp =o : - PC b) - pl 2b) t My 1437=0

⇒ My' & p = 11.25 kips -

Efx =o : Ax--o

Zfy -

-o : Ay - p- P t My =o

⇒ Ay -- 18.75 kips

=

Au( b) at kips

b-- f O o I --myO O

(C) From FDD # 2 :

I Mk -0 : - fuel 6) t 15197T Ax (6) -Ay ( 181=0

⇒ IIE = - 33.75 kips ( compression )IMD " : Fu (b ) thx 112 ) -Ay 191=0

⇒ Fu = 28,13 kips (Tension )

ZFy=o :Ay - 15 - Tok Siro'⇒ Ex -

- 6.67 Kil's (Tension )-

--- n

---

-

or From f-BD # 3 :

EMk=o : TDE (f) 1-My (187=0

⇒ IIE = -33, 75 kips ( compression )

ZMp=o :- Fu (6) - 1519 ) + My (277=0⇒ He = 28,13 kik ( Tension )

Efy --o : Tpp Sino - 15 - My-

- o

⇒ fine -- 6.67 kips ( Tension )

471-0÷÷÷÷÷

UT'T TUy Vy

(A) ER, -0 : Ux -0[ Muro : - f- ( 2b) - 2FC3b) - 41--146)tVy( 64=0

⇒ Uy=4kips[ Ey -0 : Uy - F -25--41=+4--0

⇒ Uy -

-

3 kips

(b) Cut① :

E.Me --o : - Fadl o)+Vy 1.161⇒

⇒ Fac -- 6.4 kips

[ Tension)

I My ⇒ : foe ( lo) -Vy (I b)⇒

⇒ IIE = -6.4 kips

(compression)

Cut ② :s' s

's'

KITTEhly '-on. 5

' II

". .. . . ... .. *f¥¥¥÷÷÷÷÷.ZFt F ( 8) - Uy 1241=0 Ux

q⇒ Far -- o Uy

E Mk --o '.

- Foe ( Io ) - FDR CoD ( too ) t Ff 8) - Uy (243=0

⇒ Fpr -- o

Ep wtf I.

Ay Gy

( as -2M£ --o : - Pl b) - Q1 2b) - R ( 5-b) f- Gy ( 6 b) =o

⇒ Gg = 750. lbs

Z F -o : Ax --o

Zfy --o : Ay +Gy - p- Q- R-⇒ Ay -- 9500 lbs

•

Be

(b) IME? - FBC ( 12) -1400019 ) - Hoo ( 187=0⇒ Fisc = -11250 lbs (compression )

Efx -- o " Fy. 1- Fix =- ⇒ Fgy= 11250 ( Tension)

Ifj- o ! Ay - you 1- Fac --o

⇒ Fck = - 5500 lbs (compression )