treliça statica.pdf

Batch: Jan - May 2008 R. Ganesh Narayanan 1
Engineering Mechanics Statics
InstructorR. Ganesh Narayanan
Department of Mechanical EngineeringIIT Guwahati

Batch: Jan - May 2008 R. Ganesh Narayanan 2
-These lecture slides were prepared and used by me to conduct lectures for 1st year B. Tech.
students as part of ME 101 Engineering Mechanics course at IITG.
- Theories, Figures, Problems, Concepts used in the slides to fulfill the course requirements are
taken from the following textbooks
- Kindly assume that the referencing of the following books have been done in this slide
- I take responsibility for any mistakes in solving the problems. Readers are requested to rectify
when using the same
- I thank the following authors for making their books available for reference
R. Ganesh Narayanan
1. Vector Mechanics for Engineers Statics & Dynamics, Beer & Johnston; 7th edition
2. Engineering Mechanics Statics & Dynamics, Shames; 4th edition
3. Engineering Mechanics Statics Vol. 1, Engineering Mechanics Dynamics Vol. 2, Meriam &
Kraige; 5th edition
4. Schaums solved problems series Vol. 1: Statics; Vol. 2: Dynamics, Joseph F. Shelley

R. Ganesh Narayanan 3
Engineering mechanics- Deals with effect of forces on objects
Mechanics principles used in vibration, spacecraft design, fluid flow, electrical, mechanical m/c design etc.
Statics: deals with effect of force on bodies which are not moving
Dynamics: deals with force effect on moving bodies
We consider RIGID BODIES Non deformable

Scalar quantity: Only magnitude; time, volume, speed, density, mass
Vector quantity: Both direction and magnitude; Force, displacement, velocity, acceleration, moment
V = IvI n, where IvI = magnitude, n = unit vector
n = V / IvI
n - dimensionless and in direction of vector V
In our course:y
x
z
j
i
k
i, j, k unit vectors

Dot product of vectors: A.B = AB cos ; A.B = B.A (commutative)
A.(B+C) = A.B+A.C (distributive operation)
A.B = (Axi+Ayj+Azk).(Bxi+Byj+Bzk) = AxBx+AyBy+AzBz
Cross product of vectors: A x B = C; ICI = IAI IBI Sin ; AxB = -(BxA)
C x (A+B) = C x A + C x B
i j k
A
B
i . i = 1
i . j = 0
k i
j
k x j = -i; i x i = 0
AxB = (Axi+Ayj+Azk)x(Bxi+Byj+Bzk) = (AyBz- AzBy)i+( )j+( )k
i j k
Ax AY AZ
BX BY BZ

Force:- action of one body on another- required force can move a body in the direction of action, otherwise no effect
- some times plastic deformation, failure is possible- Magnitude, direction, point of application; VECTOR
Force < P kN
Force, P kN
Direction of motion
Body moves
Body does not move
P, kN
bulging

Force system:
PWIREBracket
Magnitude, direction and point of application is important
External effect: Forces applied (applied force); Forces exerted by bracket, bolts, foundation.. (reactive force)
Internal effect: Deformation, strain pattern permanent strain; depends on material properties of bracket, bolts

Transmissibility principle:A force may be applied at any point on a line of action without changing the resultant effects of the force applied external to rigid body on which it acts
Magnitude, direction and line of action is important; not point of application
PP
Line of action

Concurrent force:
Forces are said to be concurrent at a point if their lines of action intersect at that point
AF1
F2
R
F1, F2 are concurrent forces
R will be on same plane
R = F1+F2
Plane
Parallelogram law of forces
Polygon law of forces
AF1
F2
R
F2
F1A
F1
F2R
Use triangle law
A F1
RF2
R does not pass through A
R = F1+F2 R = F1+F2

Two dimensional force system
Rectangular components:
Fx
Fy
j
i
F
F = Fx + Fy; both are vector components in x, y direction
Fx = fx i ; Fy = fy j; fx, fy are scalar quantities
Therefore, F = fx i + fy j
Fx = F cos ; Fy = F sin
F = fx2 + fy2 ; = tan -1 (fy/fx)+ ve
+ ve
- ve
- ve

Two concurrent forces F1, F2
Rx = Fx; Ry = Fy
DERIVATION
F2F1
R
i
j

Moment: Tendency to rotate; torque
Moment about a point: M = Fd
Magnitude of moment is
proportional to the force F and
moment arm d i.e, perpendicular
distance from the axis of rotation
to the LOA of force
UNIT : N-m
Moment is perpendicular to plane about axis O-O
Counter CW = + ve; CW = -ve
B
A
F
d
r
O
O
M

Cross product:
M = r x F; where r is the position vector which runs from the moment reference point A to any point on the LOA of F
M = Fr sin ; M = Fd
M = r x F = -(F x r): sense is important
B
A
d
r
Sin = d / r

Varignons theorem:
The moment of a force about any point is equal to the sum of the moments of the components of the forces about the same point
oQ
P R
r
B Mo = r x R = r x (P+Q) = r x P + r x Q
Moment of PMoment of Q
Resultant R moment arm d
Force P moment arm p; Force Q moment arm q
Mo= Rd = -pP + qQ
Concurrent forces P, Q
Usefulness:

Pb:2/5 (Meriam / Kraige):
Calculate the magnitude of the moment
about O of the force 600 N
1) Mo = 600 cos 40 (4) + 600 sin 40 (2)
= 2610 Nm (app.)
2) Mo = r x F = (2i + 4j) x (600cos40i-600sin40j)
= -771.34-1839 = 2609.85 Nm (CW);
mag = 2610 Nm
o
600N4
2
A
in mm
40 deg
r
i
j

Couple: Moment produced by two equal, opposite and non-collinear forces
-F
+Fa
d
o=>-F and F produces rotation
=>Mo = F (a+d) Fa = Fd; Perpendicular to plane
Independent of distance from o, depends on d only
moment is same for all moment centers
M

Vector algebra method
-F
+F
orb
rar M = ra x F + rb x (-F) = (ra-rb) x F = r x F
CCW Couple
CW Couple
Equivalent couples
Changing the F and d values does not change a given couple as long as the product (Fd) remains same
Changing the plane will not alter couple as long as it is parallel

M
-F +Fd
M
-F+F
d
M
-F+F d
-2Fd/2+2F
M
EXAMPLE
All four are equivalent couples

Force-couple system
=>Effect of force is two fold 1) to push or pull, 2) rotate the body about any axis
Dual effect can be represented by a force-couple syatem
a force can be replaced by a force and couple
F
A
B
F
A
B F
-F
B F
M = Fd

o
80N
o
80N
80 N80 N o80 NMo = Y N m
60deg
9 m
Mo = 80 (9 sin 60) = 624 N m; CCW
EXAMPLE
9
60 deg

Resultants
To describe the resultant action of a group or system of forces
Resultant: simplest force combination which replace the original forces without altering the external effect on the body to whichthe forces are applied
R
R = F1+F2+F3+.. = F
Rx = Fx; Ry = Fy; R = ( Fx)2 + ( Fy)2
= tan -1 (Ry/Rx)

F1 F2
F3
F1 D1; F2 D2; F3 D3
F1 F2
F3
M1 = F1d1;
M2 = F2d2;
M3 = F3d3
R= F
Mo= Fd
NON-CONCURRENT FORCES
R
d
Mo=Rd
How to obtain resultant force ?

Principle of moments
Summarize the above process: R = F
Mo = M = (Fd)
Mo = Rd
First two equations: reduce the system of forces to a force-couple system at some point O
Third equation: distance d from point O to the line of action R
=> VARIGNONS THEOREM IS EXTENDED HERE FOR NON-CONCURENT FORCES
R= F
Mo= FdR
d
Mo=Rd

STATICS MID SEMESTER DYNAMICS
Tutorial: Monday 8 am to 8.55 am
1. Vector Mechanics for Engineers Statics & Dynamics, Beer & Johnston; 7th edition
2. Engineering Mechanics Statics & Dynamics, Shames; 4th edition
3. Engineering Mechanics Statics Vol. 1, Engineering Mechanics Dynamics Vol. 2,
Meriam & Kraige; 5th edition
4. Schaums solved problems series Vol. 1: Statics; Vol. 2: Dynamics, Joseph F. Shelley
Reference books

ENGINEERING MECHANICSTUTORIAL CLASS: Monday 8 AM TO 8.55 AM
07010605 (5 Students)07010601
Dr. Saravana Kumar120507010449 (36 Students)07010414TG5
07010413 (13 Students)07010401
Dr. M. Pandey120207010353 (28 Students)07010326TG4
07010325 (25 Students)07010301
R. Ganesh Narayanan1G207010249 (16 Students)07010234TG3
07010233 (33 Students)07010201
Dr. senthilvelan1G107010149 (8 Students)07010142TG2
Prof. R. TiwariL207010141 (41 Students)07010101TG1
ToFrom
TutorsClass RoomRoll NumbersTutorial Groups
LECTURE CLASSES: LT2 (one will be optional):Monday 3 pm to 3.55 pmTuesday 2 pm to 2.55 pmThursday 5 pm to 5.55 pmFriday 4 pm to 4.55 pm

Three dimensional force system
Rectangular componentsFx = F cos x; Fy = F cos y; Fz = F cos z
F = Fx i + Fy j + Fz k = F (i cos x + j cos y + k cos z) = F (l i + m j + n k)F = F nf
o
Fx i
Fy j
Fz kF
z
x
y
l, m, n are directional cosines of F

F
r
Mo
dA
A - a plane in 3D structure
Mo = F d (TEDIOUS to find d)
or Mo = r x F = (F x r) (BETTER)
Evaluating the cross product
Described in determinant form: i j krx rY rZ
FX FY FZ
Moment in 3D
Expanding

Mo = (ryFz - rzFy) i + (rzFx rxFz) j + (rxFy ryFx) k
Mx = ryFz rzFy; My = rzFx rxFz; Mz = rxFy ryFx
Moment about any arbitrary axis :
F
r
Mo n
o
Magnitude of the moment M of F about
= Mo . n (scalar reprn.)
Similarly, M = (r x F.n) n (vector reprn.)
Scalar triple product
rx ry rz
Fx FY FZ
, , DCs of n

Varignons theorem in 3D
o F1
F3F2
r
B
Mo = rxF1 + rxF2 + rx F3 += (r x F)
= r x (F1+F2+F3+)
= r x (F) = r x R
Couples in 3D
B
M
Ar
ra
rb
d
-F+F
M = ra x F + rb x F = (ra-rb) x F = rxF

Beer-Johnston; 2.3
F1 = 150N
30
F4 = 100N
15
F3 = 110N
F2 = 80N20
Evaluate components of F1, F2, F3, F4
Rx = Fx; Ry = Fy
R = Rx i + Ry j
= tan -1 (Ry/Rx)
Ry
Rx
R
R = 199i + 14.3j; = 4.1 deg
2D force system; equ. Force-couple; principle of moments

F1
F2 R =3000 N
30 DEG
45 DEG15 DEG
Find F1 and F2
3000 (cos15i sin 15j) = F1 (cos 30i Sin 30j)+ F2 (cos45i sin 45j)
EQUATING THE COMPONENTS OF VECTOR,
F1 = 2690 N; F2 = 804 N
R = F1 + F2
Boat

o
A
B
20 DEG
C
OC FLAG POLE
OAB LIGHT FRAME
D POWER WINCH
D
780 N
Find the moment Mo of 780 N about the hinge point
10m
10
10
T = -780 COS20 i 780 sin20 j
= -732.9 i 266.8 j
r = OA = 10 cos 60 i + 10 sin 60 j = 5 i + 8.6 j
Mo = r x F = 5014 k ; Mag = 5014 Nm
Meriam / kraige; 2/37

Replace couple 1 by eq. couple p, -p; find
M = 100 (0.1) = 10 Nm (CCW)
M = 400 (0.04) cos
10 = 400 (0.04) cos
=> = 51.3 deg
MP
-P
40100
100
100
100N 100N
60
1
2
1
2

80N
30 deg
60 N
40 N
50 N2m 5m45
2m
2m
1m
o
140Nm
Find the resultant of four forces and one couple which act on the plate
Rx = 40+80cos30-60cos45 = 66.9 N
Ry = 50+80sin 30+60cos45 = 132.4 N
R = 148.3 N; = tan-1 (132.4/66.9) = 63.2 deg
Mo = 140-50(5)+60cos45(4)-60sin45(7) = -237 Nm
o
R = 148.3N
63.2 deg237 Nm
o
R = 148.3N63.2 deg
148.3 d = 237; d = 1.6 mFinal LOA of R:
o
R = 148.3N
b
xy
(Xi + yj) x (66.9i+132.4j) = -237k
(132.4 x 66.9 y)k = -237k
132.4 x -66.9 y = -237
Y = 0 => x = b = -1.792 m
LOA of R with x-axis:

Couples in 3D
B
M
Ar
ra
rb
d
-F+F
M = ra x F + rb x F = (ra-rb) x F = rxF
F
AB
F
M = Fd
F
AB
F
-F
rB
Equivalent couples

How to find resultant ?
R = F = F1+F2+F3+Mo = M = M1+M2+M3+ = (rxF)
M = Mx2 + My2 + Mz2; R = Fx2 + Fy2 + Fz2
Mx = ; My = ; Mz =

EquilibriumBody in equilibrium - necessary & sufficient condition:
R = F = 0; M = M = 0
Equilibrium in 2D
Mechanical system: body or group of bodies which can be conceptually isolated from all other bodies
System: single body, combination of bodies; rigid or non-rigid; combination of fluids and solids
Free body diagram - FBD:
=> Body to be analyzed is isolated; Forces acting on the body are represented action of one body on other, gravity attraction, magnetic force etc.
=> After FBD, equilibrium equns. can be formed

Modeling the action of forces
Meriam/Kraige
Imp
Imp

FBD - Examples
Meriam/Kraige
Equilibrium equns. Can be solved,
Some forces can be zero
Assumed sign can be different

Types of 2D equilibrium
x
F1
F2
F3
Collinear: Fx = 0 F1 F2
F3F4
Parallel: Fx = 0; Mz = 0
F1
F2
F3
F4
X
Y
Concurrent at a point: Fx = 0; Fy = 0
X
Y
M
General: Fx = 0; Fy = 0; Mz = 0

General equilibrium conditions
Fx = 0; Fy = 0; Fz = 0
Mx = 0; My = 0; Mz = 0
These equations can be used to solve unknown forces, reactions applied to rigid body
For a rigid body in equilibrium, the system of external forces will impart no translational, rotational motion to the body
Necessary and sufficient equilibrium conditions

Written in three alternate ways,
AB
CD
PY
Px
QY
Qx
RY
Rx
W
BY
AX
AY
MB = 0 => will not provide new information; used to check the solution; To find only three unknowns
A B
C D
P Q R
RollerPin
Fx = 0; Fy = 0; MA = 0 I

Fx = 0; MA = 0; MB = 0 II
Point B can not lie on the line that passes through point A
First two equ. indicate that the ext. forces reduced to a single vertical force at A
Third eqn. (MB = 0) says this force must be zero
Rigid body in equilibrium =>
MA = 0; MB = 0; Mc = 0; III
Body is statically indeterminate: more unknown reactions than independent equilibrium equations

Meriam / Kraige; 2/10
z
x
12 m
9
B
O
A
15 T = 10kN
Y
Find the moment Mz of T about the z-axis passing thro the base O
3D force system

F = T = ITI nAB = 10 [12i-15j+9k/21.21] = 10(0.566i-0.707j+0.424k) k N
Mo = rxF = 15j x 10(0.566i-0.707j+0.424k) = 150 (-0.566k+0.424i) k Nm
Mz = Mo.k= 150 (-0.566k+0.424i).k = -84.9 kN. m

Merial / Kraige; 2/117
Replace the 750N tensile force which the cable exerts on point B by a force-couple system at point O

F = f , where is unit vector along BC
= (750) BC/IBCI = 750 (-1.6i+1.1j+0.5k/2.005)
F = -599i+412j+188.5k
rob = OB = 1.6i-0.4j+0.8k
Mo = rob x F
= (1.6i-0.4j+0.7k) x (-599i+412j+188.5k)
Mo = - 363i-720j+419.2k

Meriem / Kraige; 3/4
MA = (T cos 25) (0.25) + (T sin 25) (5-0.12) 10(5-1.5-0.12) 4.66 (2.5-0.12) = 0
T = 19.6 kN T
25 deg
y
Ax
Ay
10 kN
0.5 m
4.66 kN
5m
1.5m0.12 m
Fx = Ax 19.6 cos 25 = 0
Ax = 17.7 kN
Fy = Ay+19.61 sin 25-4.66-10 = 0
Ay = 6.37 kN
A = Ax2 + Ay2 = 18.88kN
2D equilibrium
Find T and force at A; I-beam with mass of 95 kg/meter of length
95 kg/meter => 95(10-3)(5)(9.81) = 4.66kN

A
B
40 50 30 10
6060 a 80
mm, NBeer/Johnston; 4.5
Find reactions at A, B if (a) a = 100 mm; (b) a=70 mm
40 50 30 10
Bx
By
Ay
a = 100 mm
Ma = 0 => (-40x60)+(-50x120)+(-30x220)+
(-10x300)+(-Byx120) = 0
By = 150 N
Fy = 0 => By-Ay-40-50-30-10 = 0
= 150-Ay-130 = 0 => Ay = 20 N
a = 70 mm
By = 140 N Ay = 10 N

A B
20 20 20 20
C2.25
3.75
E
D
4.5
F
1.8
A B
20 20 20 20
C2.25
3.75
E
D
4.5
F1.8
150 kN
Ex
Ey
Find the reaction at the fixed end E
DF = 7.5 m
Fx = Ex + 150 (4.5/7.5) = 0 => Ex = - 90 kN (sign change)
Fy = Ey 4(20)-150 (6/7.5) = 0 => Ey = 200 kN
ME= 20 (7.2) + 20 (5.4) + 20 (3.6) +20 (1.8) (6/7.5) (150) (4.5) + ME= 0
ME= +180 kN.m => ccw
ME
Beer/Johnston; 4.4

Instructions for TUTORIAL
Bring pen, pencil, tagged A4 sheets, calculator, text books
Submitted in same tutorial class
Solve div II tutorial problems also
Solve more problems as home work
Tutorial : 10 % contribution in grading
Do not miss any tutorial class
QUIZ 1 FEB, 11TH, 2008

3D equilibrium
3D equilibrium equns. can be written in scalar and vector form
F = 0 (or) FX = 0; FY = 0; FZ = 0
M = 0 (or) MX = 0; MY = 0; MZ = 0
F = 0 => Only if the coefficients of i, j, k are zero; FX = 0
M = 0 => Only if the coefficients of i, j, k are zero; MX = 0

Modeling forces in 3D

Types of 3D equilibrium

Meriem / KraigeB
A
7 m
6 m
2 my
x
z
By
Bx
G
W=mg=200 x 9.81
W = 1962 N
Ay
AzAx
h
3.5
3.5
7 = 22 + 62 + h2 => h = 3 m
rAG = -1i-3j+1.5k m; rAB = -2i-6j+3k m
MA = 0 => rAB x (Bx+By) + rAG x W = 0
(-2i-6j+3k) x (Bx i + By j) + (-i-3j+1.5k) x (-1962k) = 0
(-3By+5886)i + (3Bx-1962)j + (-2By+6Bx)k = 0
=> By = 1962 N; Bx = 654 N
F = 0 => (654-Ax) i + (1962-Ay) j + (-1962+Az)k = 0
=> Ax = 654 N; Ay = 1963 N; Az = 1962 N; find A

Meriem / Kraige; 3/64

I.H. Shames
Find forces at A, B, D. Pin connection at C; E has welded connection

F.B.D. - 1
F.B.D. - 2
Mc = 0 => (Dy) (15) 200 (15) (15/2) (1/2)(15)(300)[2/3 (15)] = 0
Dy = 3000 N
F.B.D. - 2

F.B.D. - 1
MB = 0 => -Ay (13) +(3000) (21) 200 (34) (34/2-13) (300) (15) [6+2/3(15)] = 0
Ay = -15.4 N
Fy = 0 => Ay+By+3000-200(34)-(1/2)(300)(15) = 0
Sub. Ay here,
=> By = 6065 N

2D, 3D force system
Rectangular components
Moment
Varignons theorem
Couple
Force-couple system
Resultant
Principle of moment
Equilibrium equations
Fx = 0; Fy = 0; MA = 0
Fx = 0; MA = 0; MB = 0
MA = 0; MB = 0; Mc = 0
2D
F = 0 (or) FX = 0; FY = 0; FZ = 0
M = 0 (or) MX = 0; MY = 0; MZ = 03D

Structures
Truss: Framework composed of members joined at their ends to form a rigid structures
Plane truss: Members of truss lie in same planeBridge truss
Roof truss

Three bars joined with pins at end
Rigid bars and non-collapsible
Deformation due to induced internal strains is negligible
B D
CA
Non-rigid rigid
E
Non rigid body can be made rigid by adding BC, DE, CE elements
B
C
D
A
A
B
c

Instructions for TUTORIAL
Bring pen, pencil, tagged A4 sheets, calculator, text books
Submitted in same tutorial class
Solve div II tutorial problems also
Solve more problems as home work
Tutorial : 10 % contribution in grading
Do not miss any tutorial class
QUIZ 1 FEB, 11TH, 2008

StructuresTruss: Framework composed of members joined at their ends to form a rigid structures
Plane truss: Members of truss lie in same planeBridge truss
Roof truss

Three bars joined with pins at end
Rigid bars and non-collapsible
Deformation due to induced internal strains is negligible
Non rigid body can be made rigid by adding BC, DE, CE elementsB D
CA
Non-rigid rigid
E
B
C
D
A
A
B
c
Simple truss: structures built from basic triangle
More members are present to prevent collapsing => statically indeterminate truss; they can not be analyzed by equilibrium equations
Additional members not necessary for maintaining equilibrium - redundant

In designing simples truss or truss => assumptions are followed
1. Two force members equilibrium only in two forces; either tension or compression
2. Each member is a straight link joining two points of application of force
3. Two forces are applied at the end; they are equal, opposite and collinear for equilibrium
4. Newtons third law is followed for each joint
5. Weight can be included; effect of bending is not accepted
6. External forces are applied only in pin connections
7. Roller or rocker is also provided at joints to allow expansion and contraction due to temperature changes and deformation for applied loads
T
T c
c weight
TWO FORCE MEMBERS

Method of joints
This method consists of satisfying the conditions of equilibrium for the forces acting on the connecting pin of each joint
This method deals with equilibrium of concurrent forces and only two independent equilibrium equations are solved
Newtons third law is followed
Two methods to analyze force in simple truss

Example
AB
C
D
EF
L
Fy = 0; Fx = 0
Finally sign can be changed if not applied correctly

Internal and external redundancy
external redundancy: If a plane truss has more supports than are necessary to ensure a stable equilibrium, the extra supports constitute external redundancy
Internal redundancy: More internal members than are necessary to prevent collapse, the extra members constitute internal redundancy
Condition for statically determinate truss: m + 3 = 2j
- Equilibrium of each joint can be specified by two scalar force equations, then 2jequations are present for a truss with j joints
-The entire truss composed of m two force members and having the maximum of three unknown support reactions, there are (m + 3) unknowns
j no. of joints; m no. of members
m + 3 > 2 j =>more members than independent equations; statically indeterminate
m + 3 < 2 j => deficiency of internal members; truss is unstable

AC E
F
DB
1000
10
1010 10
1000
I. H. Shames
Determine the force transmitted by each member;
A, F = 1000 N
Pin A
FAB
FAC
1000
AFx = 0 =>FAC 0.707FAB = 0
Fy = 0 => -0.707FAB+1000 = 0
FAB = 1414 N; FAC = 1000 N
Pin B
B
FBC
1414
FBD
Fx = 0 => -FBD + 1414COS45 = 0 => FBD = 1000 N
Fy = 0 => -FBC+1414 COS45 = 0 => FBC = 1000 N
1000
FAC
FAB
1414
FBC
FBD

Pin C
B
1000
1000FCE
FDC1000
10001000
1000
FDCFCE
Fx = 0 => -1000 + FCE + FDC COS 45 = 0 => FCE = 1000 N
Fy = 0 => -1000+1000+ FDC COS 45 = 0 => FDC = 0
SIMILARLY D, E, F pins are solved

B D
AC E
30 20
5 5
5
5 55 5
kN, m
Find the force in each member of the loaded cantilever truss by method of joints
Meriem / Kraige (similar pbm. 6.1 in Beer/Johnston)

ME = 0 => 5T-20(5)-30 (10) = 0; T = 80 kN
Fx = 0 => 80 cos 30 Ex = 0; Ex = 69.28 kN
Fy = 0 => Ey +80sin30-20-30 = 0 => Ey = 10kN
Fx = 0; Fy = 0
Find AB, AC forcesFx = 0; Fy = 0
Find BC, BD forces
Fx = 0; Fy = 0
Find CD, CE forces
Fy = 0
Find DE forces
Fx = 0 can be checked
FBD of entire truss
FBD of joints

Q = 100 N; smooth surfaces; Find reactions at A, B, C Q
Q
AB
30
c
roller
100
100
RA
Rc
roller
RB
F = 0 => (-RA cos 60 - RB cos 60 + Rc) i + (-2 x 100 + RBsin 60 + RA sin 60) j = 0
RC = (RA + RB)/2RB + RA = 230.94
RC = 115.5 N
RB
100
Rc
RAB30
F = 0 => (-RAB cos 30 - RB cos 60 + Rc) i + (RB Sin 60 100 - RABsin 30) j = 0
0.866 RAB + 0.5 RB = 115.5; -0.5 RAB + 0.866 RB = 100
RAB = 50 N (app.); RB = 144.4 N; RA = 230.94-144.4 = 86.5 N

Method of joints
This method consists of satisfying the conditions of equilibrium for the forces acting on the connecting pin of each joint
This method deals with equilibrium of concurrent forces and only two independent equilibrium equations are solved
Newtons third law is followed
Two methods to analyze force in plane truss
Method of sections

Methodology for method of joints
AB
C
D
EF
L
Fy = 0; Fx = 0
Finally sign can be changed if not applied correctly

5B D
AC E
30 20
5 5
5 55 5
kN, m

Method of sections
In method of joints, we need only two equilibrium equations, as we deal with concurrent force system
In method of sections, we will consider three equilibrium equations, including one moment equilibrium eqn.
force in almost any desired member can be obtained directly from an analysis of a section which has cut the member
Not necessary to proceed from joint to joint
Not more than three members whose forces are unknown should be cut. Only three independent equilibrium eqns. are present
Efficiently find limited information

AB
C
D
EF
LThe external forces are obtained initially from method of joints, by considering truss as a whole
Assume we need to find force in BE, then entire truss has to be sectioned across FE, BE, BC as shown in figure; we have only 3 equilibrium equns.
AA section across FE, BE, BC; Forces in these members are initially unknown
AB
C
D
EF
LR1 R2A
A
Methodology for method of sections

Now each section will apply opposite forces on each other
The LHS is in equilibrium with R1, L, three forces exerted on the cut members (EF, BE, BC) by the RHS which has been removed
IN this method the initial direction of forces is decided by moment about any point where known forces are present
For eg., take moment about point B for the LHS, this will give BE, BC to be zero; Then moment by EF should be opposite to moment by R1; Hence EF should be towards left hand side - compressive
Section 1 Section 2

Now take moment about F => BE should be opposite to R1 moment; Hence BE must be up and to the right; So BE is tensile
Now depending on the magnitudes of known forces, BC direction has to be decided, which in this case is outwards i.e., tensile
MB = 0 => FORCE IN EF; BE, BC = 0
Fy = 0 => FORCE IN BE; BC, EF = 0
ME = 0 => FORCE IN BC; EF, BE = 0Section 1 Section 2

Section AA and BB are possible
convenient

Important points
IN method of sections, an entire portion of the truss is considered a single body in equilibrium
Force in members internal to the section are not involved in the analysis of the section as a whole
The cutting section is preferably passed through members and not through joints
Either portion of the truss can be used, but the one with smaller number of forces will yield a simpler solution
Method sections and method of joints can be combined
Moment center can be selected through which many unknown forcespass through
Positive force value will sense the initial assumption of force direction

Meriem/Kraige
Find the forces included in members KL, CL, CB by the 20 ton load on the cantilever truss
Section 1 Section 2
x
Moment abt. L => CB is compressive => creates CW moment
Moment abt. C => KL is tensile => creates CW moment
CL is assumed to be compressive
yKL
20 T
C CB
CL
GP
L
K

yKL
20 T
C CB
CL
GP
L
K
x
Section 1 Section 2
ML = 0 => 20 (5) (12)- CB (21) = 0 => CB = 57.1 t (C)
Mc = 0 => 20 (4)(12) 12/13 (KL) (16) = 0; KL = 65 t (T)
Mp = 0 => find PC distance and find CL; CL = 5.76 t (C)
BL = 16 + (26-16)/2 = 12 ft
= tan -1 (5/12) => cos = 12/13

Meriem/Kraige
Find the force in member DJ of the truss shown. Neglect the horizontal force in supports
Section 2 cuts four members, but we have only 3 equi. Equns
Hence consider section 1 which cuts only 3 members CD, CJ, KJ
Consider FBD for whole truss and find reaction at A
MG = -Ay (24) +(10) (20) + 10(16) + 10 (8) = 0
Ay = 18. 3 kN => creates CW moment
Force direction
Moment abt. A => CD, JK Eliminated; CJ will be upwards creating CCW moment
Moment abt. C => JK must be towards right creating CCW moment
ASSUME CD TO HAVE TENSILE FORCE

MA = 0 => CJ (12) (0.707) 10 (4) -10( 8) =0; CJ = 14.14 Kn
MJ = 0 => 0.894 (CD) (6) +18.33 (12)-10(4)-10(8) = 0; CD = -18.7 kN
CD direction is changed
From section 1 FBD
From section 2 FBD
MG = 0 => 12 DJ +10(16)+10(20)-18.3 (24)-14.14 (0.707)(12) = 0
DJ = 16.7 kN

I.H. Shames FBD - 1
FBD - 2
From FBD-2
MB = 0 => -(10)(500)+30 (789)- FAC Sin 30 (30) = 0
FAC = 1244.67 N
From FBD -1
Fx = 0 => FDA Cos 30 (1244.67) cos 30 1000 sin 30 = 0 ;
FDA = 1822 N
Fy = 0 => (1822)Sin 30 + (1244.67) sin 30 +FAB 1000 Cos 30 = 0; FAB = -667 N

Frames and machines
Multi force members: Members on which three or more forces acting on it (or) one with two or more forces and one or more couples acting on it
Frame or machine: At least one of its member is multi force member
Frame: Structures which are designed to support applied loads and are fixed in position
Machine: Structure which contain moving parts and are designed to transmit input forces or couples to output forces or couples
Frames and machines contain multi force members, the forces in these members will not be in directions of members
Method of joints and sections are not applicable

Inter-connected rigid bodies with multi force members
Previously we have seen equilibrium of single rigid bodies
Now we have equilibrium of inter-connected members which involves multi force members
Isolate members with FBD and applying the equilibrium equations
Principle of action and reaction should be remembered
Statically determinate structures will be studied

Force representation and FBD
Representing force by rectangular components
Calculation of moment arms will be simplified
Proper sense of force is necessary; Some times arbitrary assignment is done; Final force answer will yield correct force direction
Force direction should be consistently followed

Frames and machines
Multi force members: Members on which three or more forces acting on it (or) one with two or more forces and one or more couples acting on it
Frame or machine: At least one of its member is multi force member
Frame: Structures which are designed to support applied loads and are fixed in position
Machine: Structure which contain moving parts and are designed to transmit input forces or couples to output forces or couples
Frames and machines contain multi force members, the forces in these members will not be in directions of members
Method of joints and sections are not applicable

Inter-connected rigid bodies with multi force members
Previously we have seen equilibrium of single rigid bodies
Now we have equilibrium of inter-connected members which involves multi force members
Isolate members with FBD and applying the equilibrium equations
Principle of action and reaction should be remembered
Statically determinate structures will be studied

Force representation and FBD
Representing force by rectangular components
Calculation of moment arms will be simplified
Proper sense of force is necessary; Some times arbitrary assignment is done; Final force answer will yield correct force direction
Force direction should be consistently followed

AFAE
BD
Full truss
K, J are un-necessary here

Meriem/Kraige
A
B
C
D
E
F
30 lb
50 lb
12
12
30 ft
20 ft
20 ftFind the forces in all the frames; neglect weight of each member
AxAy
50 lb
30 lb
Cx
Cy
Mc = 0 => 50 (12) +30(40)-30 (Ay) = 0; Ay = 60 lb
Fy = 0 => Cy 50 (4/5) 60 = 0 => Cy = 100 lb
FBD of full frame

ED:
MD = 0 => 50(12)-12E = 0 => E = 50 lb
F = 0 => D-50-50 = 0 => D= 100 lb
(components will be eliminated)
EF: Two force member; E, F are compressive
EF: F = 50 lb (opposite and equal to E)
AB:
MA = 0 => 50(3/5)(20)-Bx (40) = 0 => Bx = 15 lb
Fx = 0 => Ax+15-50(3/5) = 0 => Ax = 15 lb
Fy = 0 => 50 (4/5)-60-By = 0 =>By = -20 lb
BC: Fx = 0 => 30 +100 (3/5)-15-Cx = 0 => Cx = 75 lb
FBD of individual members
D
Fx = -50 (cos 53.1)+15+15 = -30+15+15 = 0
F
Fx
Fy
53.1 deg
E

A
B
C
E
D
60100 150
480 N
160
60
80
Find the force in link DE and components of forces exerted at C on member BCD
A
B
C
E
D
100 150
480 N
160
Ax
Ay
Bx
80
Fy = 0 => Ay-480 = 0 =>Ay = 480 N
MA = 0 => Bx (160)-480 (100) = 0 => Bx = 300 N
Fx = 0 => 300+Ax = 0 => Ax = -300 N
FBD of full frame
= tan -1 (80/150) = 28.07 deg

FBD of BCD
B
CD
480 N
300 Cx
Cy
FDE
Mc = 0 => -FDE sin 28.07 (250) 300(80)-480 (100) = 0; FDE = -561 N
Fx = 0 => Cx (-561) cos 28.07 +300 = 0 => Cx = -795 N
Fy = 0 => Cy (-561) sin 28.07 480 = 0 => Cy = 216 N
E
D
FDE
FDE
DE: Two force member
D
FDE
A
E
Ax
Ay
FDECy
Cx
FBD of AE
FBD of DE

A
B
C
D
E F
400 kg
3m 2m
1.5m
0.5m
1.5m
1.5m
R =0.5 m
Find the horizontal and vertical components of all the forces; neglect weight of each member
Meriem/Kraige
FBD of full frame
AyAx
0.4 x 9.81 = 3.92
MA = 0 => 5.5 (-0.4) (9.81) + 5Dx = 0 => Dx = 4.32 kN
Fx = 0 => -Ax + 4.32 = 0 => Ax = 4.32 kN
Fy = 0 => Ay 3.92 = 0 => Ay = 3.92 kN
Dx

FBD of individual members
3.92
4.32
3.92
BxBy
4.32
CxCy
A
D
3.92
3.92
3.92
3.92
F
C
E
Cx
Cy
ExEy
E
EyEx
Bx
By
B
3.92
3.92
A
B
C
D
E F
400 kg
3m 2m
1.5m
0.5m
1.5m
1.5m
R =0.5 m
Apply equilibrium equn. And solve for forces

Machines Machines are structures designed to transmit and modify forces. Their main purpose
is to transform input forces into output forces.
Given the magnitude of P, determine the
magnitude of Q.
Taking moments about A,
Pb
aQbQaPM A === 0

Center of mass & center of gravity
A B
C
W
G
A
B
C
W
G
W
GAB
C
G
Body of mass m
Body at equilibrium w.r.t. forces in the cord and resultant of gravitational forces at all particles W
W is collinear with point A
Changing the point of hanging to B, C Same effect
All practical purposes, LOA coincides with G; G center of gravity
BODY

dw
G
w
z Y
X
Moment abt. Y axis = dw (x)
Sum of moments for small regions through out the body: x dw
Moment of w force with Y axis = w x
x dw = w x
Sum of moments Moment of the sum
W = mg
r
r
X = ( x dm) / m
X = ( x dw) / w Y = ( y dw) / w Z = ( z dw) / w
Y = ( y dm) / m Z = ( z dm) / m
1
2

r = ( r dm) / mIn vector form,
= m/V; dm = dv
X = ( x dv) / dv
Y = ( y dv) / dv
Z = ( z dv) / dv
= not constant through out body
3
4
Equns 2, 3, 4 are independent of g; They depend only on mass distribution;
This define a co-ordinate point center of mass
This is same as center of gravity as long as gravitational field is uniform and parallel

Centroids of lines, areas, volumes
X = ( xc dv) / v Y = ( yc dv) / v Z = ( zc dv) / v
Suppose if density is constant, then the expression define a purely geometrical property of the body; It is called as centroid
Centroid of volume
X = ( x dA) / A Y = ( y dA) / A Z = ( z dA) / A
Centroid of area
X = ( x dL) / L Y = ( y dL) / L Z = ( z dL) / L
Centroid of line

x
y
h
xydy
Find the y-coordinate of centroid of the triangular area
AY = y dA
b h (y) = y (x dy) = y [b (h-y) / h] dy = b h2 / 6
b
X / (h-y) = b/h
0
h
0
h
Y = h / 3

Beams
Structural members which offer resistance to bending due to applied loads
Reactions at beam supports are determinate if they involve only three
unknowns. Otherwise, they are statically indeterminate

External effects in beams
Reaction due to supports, distributed load, concentrated loads
Internal effects in beams
Shear, bending, torsion of beams
v
v
M M
SHEAR BENDING TORSION

Cx
W
D
E
B
CF
G A
SECTION - J
F
D
J
TV
M
V SHEAR FORCE
F AXIAL FORCE
M BENDING MOMENT AT J
T
D
CyFBE
AX
AY
J
AA
J
F
VM
Internal forces in beam
compression
Tension

Shear force and bending moment in beam
To determine bending moment and shearing
force at any point in a beam subjected to
concentrated and distributed loads.
1. Determine reactions at supports by
treating whole beam as free-body
FINDING REACTION FORCES AT A AND B

2. SECTION beam at C and draw free-body
diagrams for AC and CB. By definition,
positive sense for internal force-couple
systems are as shown.
DIRECTION OF V AND M
SECTION C
SECTION C SECTION C
+ VE SHEAR FORCE
+VE BENDING MOMENTV
M
V
M

EVALUATING V AND M
Apply vertical force equilibrium eqn. to AC, shear force at C, i.e., V can be determined
Apply moment equilibrium eqn. at C, bending moment at C, i.e., M can be determined; Couple if any should be included
+ ve value of V => assigned shear force direction is correct
+ ve value of M => assigned bending moment is correct

Evaluate the Variation of shear and bending
moment along beam
Beer/Johnston
MB= 0 =>RA (-L)+P (L/2) = 0; RA= +P/2
RB = +P/2
SECTION AT C
Between A & D
SECTION AT E
Between D & B

SECTION AT C; C is at x distance from A
Member AC: Fy = 0 => P/2-V = 0; V = +P/2
Mc = 0 => (- P/2) (X) + M = 0; M = +PX/2
Any section between A and D will yield same result
V = +P/2 is valid from A to D
V = +P/2 yields straight line from A to D (or beam length : 0 to L/2)
M = +PX/2 yield a linear straight line fit for beam length from 0 to L/2

SECTION AT E; E is at x distance from A
CONSIDER AE:
Fy = 0 => P/2-P-V = 0; V = -P/2
ME = 0 => (- P/2) (X) +P(X-L/2)+ M = 0; M = +P(L-X)/2
EB CAN ALSO BE CONSIDERED

V = V0 + (NEGATIVE OF THE AREA UNDER THE LOADING CURVE FROM X0 TO X) = V0 - w dx
M = M0 + (AREA UNDER SHEAR DIAGRAM FROM X0 TO X) = M0+ V dx
c1

c1 cclab9, 1/24/2008

Beer/Johnston
:0= AM
( ) ( )( ) ( )( ) 0cm22N400cm6N480cm32 =yB
N365=yB:0= BM
( )( ) ( )( ) ( ) 0cm32cm10N400cm26N480 =+ AN515=A
:0= xF 0=xB
The 400 N load at E may be replaced by a 400 N force and 1600 N-cm couple at
D.
Taking entire beam as free-body, calculate
reactions at A and B.
Determine equivalent internal force-couple
systems at sections cut within segments AC,
CD, and DB.

R. Ganesh Narayanan 119:02 =M ( ) 06480515 =++ Mxx
( ) cmN 352880 += xM
From C to D:
= :0yF 0480515 = VN 35=V
:01 =M ( ) 040515 21 =+ Mxxx220515 xxM =
From A to C:
= :0yF 040515 = VxxV 40515 =
V = 515 + (-40 X) = 515-40X = 515 - 40 dx
M = 515-40x dx = 515x-20 x2
0
x
0
x

Evaluate equivalent internal force-couple systems
at sections cut within segments AC, CD, and DB.
From D to B:
= :0yF 0400480515 = V
N 365=V:02 =M
( ) ( ) 01840016006480515 =+++ Mxxx
( ) cmN 365680,11 = xM

Shear force & Bending moment plot
AC: (35X12) + (1/2 x 12 x 480) = 3300
0 to 3300
CD: 3300 +(35X6) = 3510
3300 to 3510
DB: 365 x 14 = 5110
5110 to 0
AREA UNDER SHEAR FORCE DIAGRAM GIVES BM DIAGRAM

300 lb
4 ft 4 2 2
100 lb/ftFind the shear force and bending moment for the loaded beam

Machine

FrictionEarlier we assumed action and reaction forces at contacting surfaces are normal
Seen as smooth surface not practically true
Normal & tangential forces are important
Tangential forces generated near contacting surfaces are FRICTIONAL FORCES
Sliding of one contact surface to other friction occurs and it is opposite to the applied force
Reduce friction in bearings, power screws, gears, aircraft propulsion, missiles through the atmosphere, fluid flow etc.
Maximize friction in brakes, clutches, belt drives etc.
Friction dissipated as heat loss of energy, wear of parts etc.

Friction
Dry friction
(coulomb friction)
Fluid friction
Occurs when un-lubricated surfaces are in contact during sliding
friction force always oppose the sliding motion
Occurs when the adjacent layers in a fluid (liquid, gas) are moving at different velocities
This motion causes friction between fluid elements
Depends on the relative velocity between layers
No relative velocity no fluid friction
depends on the viscosity of fluid measure of resistance to shearing action between the fluid layers

Dry friction: Laws of dry frictionW
N
W weight; N Reaction of the surface
Only vertical component
P applied load
F static friction force : resultant of many forces acting over the entire contact area
Because of irregularities in surface & molecular attraction
A
W
P
N
F
A

P
W
N
F
A B
P is increased; F is also increased and continue to oppose P
This happens till maximum Fm is reached Body tend to move till Fm is reached
After this point, block is in motion
Block in motion: Fm reduced to Fk lower value kinetic friction force and it remains same related to irregularities interaction
N reaches B from A Then tipping occurs abt. B
Fm
Fk
F
p
Equilibrium Motion
More irregularities interaction
Less irregularities interaction

EXPERIMENTAL EVIDENCE:
Fm proportional to N
Fm = s N; s static friction co-efficient
Similarly, Fk = k N; k kinetic friction co-efficient
s and k depends on the nature of surface; not on contact area of surface
k = 0.75 s

Four situations can occur when a rigid body is in contact with a
horizontal surface:
We have horizontal and vertical force equilibrium equns. and
F = N
No motion,
(Px < Fm)
FmFk
F
p
Equilibrium Motion

No motion Motion No friction Motion impending
It is sometimes convenient to replace normal force N and friction force
F by their resultant R:
ss
sms
N
N
N
F
=
==
tan
tan
kk
kkk
N
N
N
F
=
==
tan
tan
s angle of static friction maximum angle (like Fm)
k angle of kinetic friction; k < s

Consider block of weight W resting on board with variable inclination
angle .
Angle of inclination =
angle of repose; = s
R Not vertical
ANGLE OF INCLINATION IS INCREASING

Three categories of problems
All applied forces are given, co-effts. of friction are known
Find whether the body will remain at rest or slide
Friction force F required to maintain equilibrium is unknown
(magnitude not equal to s N)
Determine F required for equilibrium, by solving equilibrium equns; Also find
N
Compare F obtained with maximum value Fm i.e., from Fm = s N
F is smaller or equal to Fm, then body is at rest
Otherwise body starts moving
Actual friction force magnitude = Fk = k N
Solution
First category: to know a body slips or not

A 100 N force acts as shown on a 300 N block
placed on an inclined plane. The coefficients of
friction between the block and plane are s = 0.25 and k = 0.20. Determine whether the block is in equilibrium and find the value of the friction force.
Beer/Johnston
:0= xF ( ) 0N 300 - N 100 53 = F
N 80=F:0= yF ( ) 0N 300 - 5
4 =N
N 240=N
The block will slide down the plane.
Fm < F
Fm = s N = 0.25 (240) = 60 N
= 36.9 DEG
= 36.9 DEG

If maximum friction force is less than friction force
required for equilibrium, block will slide. Calculate
kinetic-friction force.
( )N 240200N
.
FF kkactual
=
==
N 48=actualF
FmFk
F
p
Equilibrium Motion

Meriam/Kraige; 6/8
M
30
Cylinder weight: 30 kg; Dia: 400 mm
Static friction co-efft: 0.30 between cylinder and surface
Calculate the applied CW couple M which cause the cylinder to slip
30 x 9.81
NA
FA = 0.3 NANB
FB = 0.3 NB
M
C
Fx = 0 = -NA+0.3NB Cos 30-NB Sin 30 = 0
Fy = 0 =>-294.3+0.3NA+NBCos 30-0.3NB Sin 30 = 0
Find NA & NB by solving these two equns.
MC = 0 = > 0.3 NA (0.2)+0.3 NB (0.2) - M = 0
Put NA & NB; Find M
NA = 237 N & NB = 312 N; M = 33 Nm

Meriam/Kraige; 6/5
Wooden block: 1.2 kg; Paint: 9 kgDetermine the magnitude and direction of (1) the friction force exerted by roof surface on the wooden block, (2) total force exerted by roof surface on the wooden block
= tan-1 (4/12) = 18.43
Paint
Wooden block
12
4
Roof surface
(2) Total force = 10.2 x 9.81 = 100.06 N UP
N
F
10.2x 9.81
X
Y
(1)Fx = 0 => -F+100.06 sin 18.43 => F = 31.6 N
Fy = 0 => N = 95 N
Second category: Impending relative motion when two or three bodies in contact with each other

Beer/Johnston
20 x 9.81 = 196.2 N
N1
F1
T
30 x 9.81 = 294.3 N
F2
For 20 kg block For 30 kg block
F1P
N1
N2
(a)

(B)490.5 N
N
P

Beer/Johnston
A
B
6 m
2.5 m
A 6.5-m ladder AB of mass 10 kg leans against a wall as shown. Assuming that the coefficient of static friction on s is the same at both surfaces of contact, determine the smallest value of s for which equilibrium can be maintained.
A
B
FB
NB
FANA
W
1.25 1.25
O
Slip impends at both A and B, FA= sNA, FB= sNB
Fx=0=> FANB=0, NB=FA=sNA
Fy=0=> NAW+FB=0, NA+FB=W
NA+sNB=W; W = NA(1+s2)
Mo = 0 => (6) NB - (2.5) (NA) +(W) (1.25) = 0
6sNA - 2.5 NA + NA(1+s2) 1.25 = 0
s = -2.4 2.6 = > Min s = 0.2

Wedges
Wedges - simple machines used to raise heavy loads like wooden block, stone etc.
Loads can be raised by applying force P to wedge
Force required to lift block is significantly less than block weight
Friction at AC & CD prevents wedge from sliding out
Want to find minimum force P to raise block
A wooden block
C, D Wedges

0
:0
0
:0
21
21
=+
=
=+
=
NNW
F
NN
F
s
y
s
x
FBD of block
( )
( ) 06sin6cos:0
0
6sin6cos
:0
32
32
=+
=
=+
=
s
y
ss
x
NN
F
P
NN
F
FBD of wedge
N3
6
F36

Two 8 wedges of negligible weight are used to move and position a 530-N block. Knowing that the coefficient of static friction is 0.40 at all surfaces of contact, determine the magnitude of the force P for which motion of the block is impending
Beer/Johnston
s = tan1 s = tan1 (0.4) = 21.801
21.8
R1
FBD of block
20
21.8
530
R2530
R2
R141.8
91.8 46.4 (R2/Sin 41.8) = (530/sin 46.4)
R2 = 487.84 N
Using sine law,
slip impends at wedge/block wedge/wedge and block/incline

P = 440.6 N

Beer/Johnston
A 6 steel wedge is driven into the end of an ax handle to lock the handle to the ax head. The coefficient of static friction between the wedge and the handle is 0.35. Knowing that a force P of magnitude 60 N was required to insert the wedge to the equilibrium position shown, determine the magnitude of the forces exerted on the handle by the wedge after force P is removed.
P = 60 N s = tan 1 s= tan 1 (0.35 ) = 19.29
19.293
19.29
36
By symmetry R1= R2; in EQUILIBRIUM
Fy = 0: 2R1 sin 22.29 60 N =0
R1 = R2 = 79.094 N
WHAT WILL HAPPEN IF P IS REMOVED ?
R1R2

Vertical component of R1, R2 will be eliminated
Hence, H1 = H2 = 79.094 N cos22.29 = 73.184 N
Final force = 73.184 N
Since included angle is 3(< s) from the normal, the wedge is self-locking and will remain in place.
No motion

Screws
Used for fastening, transmitting power or motion, lifting body
Square threaded jack - screw jack V-thread is also possible
W- AXIAL LOAD
M APPLIED MOMENT ABOUT AXIS OF SCREW
M = P X r
L LEAD DISTANCE Advancement per revolution
HELIX ANGLE
M
Upward motion

2r
L
W
RP = M/r
One full thread of screw
To raise load
M
F
angle of friction
R
P
w +
tan (+) = P/W = M/rW
=> M = rW tan (+)
= tan-1 (L/2r)
To lower load unwinding condition
P = M/r
W
R
< Screw will remain in place self locking
=> M = rW tan (-)
= In verge of un-winding
Moment required to lower the screw

P = M/r
W
R
> Screw will unwind itself
=> M = rW tan (-)Moment required to prevent unwinding

Beer/Johnston A clamp is used to hold two pieces of wood together as shown. The clamp has a double square thread of
mean diameter equal to 10 mm with a pitch of 2 mm.
The coefficient of friction between threads is s = 0.30.
If a maximum torque of 40 Nm is applied in
tightening the clamp, determine (a) the force exerted
on the pieces of wood, and (b) the torque required to
loosen the clamp.
Lead distance = 2 x pitch = 2 x 2 = 4 mm
r = 5 mm
( )
30.0tan
1273.0mm 10
mm22
2tan
==
===
ss
r
L
= 3.7
= 7.16s
(double square thread)

a) Forces exerted on the wooded pieces
M/r tan (+) = WW = 40 / (0.005) tan (24) = 17.96 kN
b) the torque required to loosen the clamp
M = rW tan (-) = 0.005 (17.96) tan (9.4)M = 14.87 Nm

The position of the automobile jack shown is controlled by a screw ABC that is single-threaded at each end (right-handed thread at A, left-handed thread at C). Each thread has a pitch of 2 mm and a mean diameter of 7.5 mm. If the coefficient of static friction is 0.15, determine the magnitude of the couple M that must be applied to raise the automobile.
Beer/Johnston
FBD joint D:
Fy = 0 => 2FADsin254 kN=0
FAD = FCD = 4.73 kN
By symmetry:
4 kN
FAD FCD
25 25D

FBD joint A:
4.73 kN
FAC
FAE = 4.73
25
25A
Fx = 0 => FAC2(4.73) cos25=0
FAC = 8.57 kN
L = Pitch = 2 mm
W = FAC = 8.57
R
Joint A
P = M/r
(7.5)
Here is used instead of used earlier
MA = rW tan (+) = (7.5/2) (8.57) tan (13.38) = 7.63 Nm
Similarly, at C, Mc = 7.63 Nm (by symmetry); Total moment = 7.63 (2) = 15.27 Nm

Journal & Thrust bearing
Journal bearings provide lateral support to rotating shafts
Thrust bearings provide axial support
Journal bearing - Axle friction Thrust bearing - Disc friction
shaft
bearing
shaft
bearing

Friction between two ring shaped areas
Friction in full circular area
- DISK FRICTION (Eg., Disc clutch)
Consider Hollow shaft (R1, R2)
M Moment required for shaft rotation at constant speed
P axial force which maintains shaft in contact with bearing

Couple moment required to overcome friction resistance, M
Equilibrium conditions and moment equations are necessary to solve problems

A .178 m-diameter buffer weighs 10.1 N. The coefficient of kinetic friction between the buffing pad and the surface being polished is 0.60. Assuming that the normal force per unit area between the pad and the surface is uniformly distributed, determine the magnitude Q of the horizontal forces required to prevent motion of the buffer.
Beer/Johnston
O
M
Q - Q0.2 m
Mo = 0 => (0.2) Q M = 0; Q = M / 0.2
M = 2/3 (0.6) (10.1) (0.178/2) = 0.36 Nm
Q = M / 0.2 = 0.36/0.2 = 1.8 N

Belt frictionDraw free-body diagram for PP element of belt
( ) 02
cos2
cos:0 =
+= NTTTF sx
( ) 02
sin2
sin:0 =
+=
TTTNFy
dT / T = S d
dT / T = S dT1
T2
0
ln (T2/T1) = S ; T2/T1 = e S
Consider flat belt, cylindrical drum
angle of contact

V- Belt
T2/T1 = e S /sin (/2)
ln (T2/T1) = S ; T2/T1 = e S
Applicable to belts passing over fixed drums; ropes wrapped around a post; belt drives
T2 > T1
This formula can be used only if belt, rope are about to slip;
Angle of contact is radians; rope is wrapped n times - 2n rad
In belt drives, pulley with lesser value slips first, with S remaining same

Beer/Johnston
A flat belt connects pulley A to pulley B. The
coefficients of friction are s= 0.25 and k= 0.20 between both pulleys and the belt.
Knowing that the maximum allowable tension in the
belt is 600 N, determine the largest torque which can
be exerted by the belt on pulley A.
Since angle of contact is smaller, slippage will occur on pulley B first. Determine
belt tensions based on pulley B; = 120 deg = 2/3 rad
( )
N4.3551.688
N600
688.1N600
1
3225.0
11
2 s
==
===
T
eT
eT
T

( )( ) 0N600N4.355mc8:0 =+= AA MM
mcN8.1956 =AM
Check for belt not sliping at pulley A:
ln (600/355.4) = x 4/3 => = 0.125 < 0.25

A 120-kg block is supported by a rope which is wrapped 1.5 - times around a horizontal rod. Knowing that the coefficient of static friction between the rope and the rod is 0.15, determine the range of values of P for which equilibrium is maintained.
Beer/Johnston
PW = 9.81 X 120 = 1177.2 N
= 1.5 turns = 3 rad For impending motion of W up
P = W e s = (1177.2 N) e (0.15)3
= 4839.7 N
For impending motion of W downP = W es = (1177.2 N) e(0.15)3
= 286.3 N
For equilibrium: 286 N P 4.84 kN

In the pivoted motor mount shown, the weight W of the 175-N motor is used to maintain tension in the drive belt. Knowing that the coefficient of static friction between the flat belt and drums A and B is 0.40, and neglecting the weight of platform CD, determine the largest couple which can be transmitted to drum B when the drive drum A is rotating clockwise.
Beer/Johnston
For impending belt slip: CW rotation = radians
Obtain FBD of motor and mount; MD = 0 => find T1 and T2
Obtain FBD of drum at B; MB in CCW; MB = 0; Find MB
T1 = 54.5 N, T2 = 191.5 N
MB=10.27 N.m

Virtual work
We have analyzed equilibrium of a body by isolating it with a FBD and equilibrium equations
Class of problems where interconnected members move relative to each other; equilibrium equations are not the direct and conventional method
Concept of work done by force is more direct => Method of virtual work

Work of a force
U = +(F cos ) S (+ ve)
FA
A
S
FA
AS
U = +F (cos S)
Work done U by the force F on the body during displacement is the compt. Of force in the displacement direction times the displacement
Work is a scalar quantity as we get same result regardless of direction in which we resolve vectors
FA
A
S U = -(F cos ) S
U = 0 if S = 0 and = 90 deg

FA
A
drA1
A2Work done by force F during displacement dr is given by, dU = F.dr
dU = (Fx i + Fy j + Fz k).(dx i + dy j + dz k)
= Fx dx + Fy dy + Fz dz
U = F.dr = Fx dx + Fy dy + Fz dz
We should know relation between the force and their coordinates
Work of a couple
d
MdU = M d
U = M d
F
-FMoment can be taken instead of forces

Forces which do no work
ds = 0; cos = 0
reaction at a frictionless pin due to rotation of a body around the
pin
reaction at a frictionless surface due to motion of a body along the
surface
weight of a body with cg moving horizontally
friction force on a wheel moving without slipping
Only work done by applied forces, loads, friction forces need tobe considered

Sum of work done by several forces may be zero
bodies connected by a frictionless pin
=> W.D by F and F is opposite and will cancel
bodies connected by an inextensible cord
internal forces holding together particles of a rigid
body
Rigid bodyA, B particles
F, -F are acting as shown
Though dr, dr are different, components of these displacements along AB must be equal, otherwise distance between the particles will change and this is not a rigid body; so U done by F and F cancel each other, i.e, U of internal forces = 0

Principle of virtual work
Imagine the small virtual displacement of particle which is
acted upon by several forces F1, F2, .. Fn
Imagine the small displacement A to A
This is possible displacement, but will not occur
AA ---- VIRTUAL DISPLACEMENT, r (not dr)
Work done by these forces F1, F2, .Fn during virtual
displacement r is called VIRTUAL WORK, U
U = F1. r + F2. r + ..+ Fn. r = R . r
Total virtual work of the forces
Virtual work of the resultant

Principle of virtual work for particle
Principle of virtual work for rigid body
Principle of virtual work for system of interconnected rigid bodies
Work of internal forces is zero (proved earlier)

Applications of Principle of virtual work
Mainly applicable to the solutions of problems involving machines or mechanisms consisting of several interconnected rigid bodies
Wish to determine the force of the vice on the block for a given force
P assuming no friction
Virtual displacement is given; This results in xB and yc. Here no work is done by Ax, Ay at A and N at B
TOGGLE VISE

UQ = -Q xB ; UP = -P yc
In this problem, we have eliminated all un-known reactions, while MA = 0 would have eliminated only TWO unknowns
The same problem can be used to find for which the linkage is in equilibrium under two forces P and Q
Output work = Input work

Real machines
For an ideal machine without friction, the output work is equal to the input
work; 2Ql cos = Pl sin
In real machine, output work < input work => because of presence of
friction forces
( )
=
+=
==
tan
cossincos20
0
21 PQ
PlPlQl
xFyPxQU BCBOutput work = Input work friction force work
Q = 0 if tan = => = , angle of friction

Mechanical efficiency
Mechanical efficiency of m/c, = Output work / Input work
For toggle vise, = 2Ql cos / Pl sin
Substituting Q = P (tan ) here
= 1 cot
In the absence of friction forces, = 0 and hence = 1 => Ideal m/c
For real m/c, < 1

Beer/JohnstonDetermine the magnitude of the couple M
required to maintain the equilibrium of the
mechanism.
Virtual displacement = , xD, Work done by Ex, Ey, A is zero
By virtual work principle,
U = UM + Up = 0
M + P xD = 0
xD = -3l sin can be obtained from geometry
M = 3Pl sin

Beer/Johnston
A hydraulic lift table consisting of two identical linkages and hydraulic cylinders is used to raise a 1000-kg crate. Members EDB and CG are each of length 2a and member AD is pinned to the midpoint of EDB. Determine the force exerted by each cylinder in raising the crate for = 60o, a = 0.70 m, and L = 3.20 m.
Work done is zero for Ex, Ey, Fcg; Work done by W, FDH will be considered

W, y are in opposite direction, (-)sign will come
FDH, s are in same direction, (+) sign will come
1)
2) Express y, s in terms of
y = 2a cos ; s = (aL sin/s)
---- (1)
Substituting y & s in (1) gives,
-(1/2) W (2a cos ) + (FDH) (aL sin/s) = 0
FDH = W (s/L) cot
3) Apply numerical data
FDH = (1000 X 9.81) (2.91/3.2) cot 60 = 5.15 kN S is obtained from this triangle

The mechanism shown is acted upon by the force P. Derive an expression for the magnitude of the force Q required for equilibrium.
Beer/Johnston
YF
W.D. by Ay, Bx, By will be zero
U = 0 => +Q (XA) - P (YF)
Find XA and YF in terms of
(Check calculation of XA and YF)
U = Q(2l cos ) - P(3l sin ) = 0
Q Bx
By
P
Ay
Yf
XAXA
x
y

Work of force using finite displacement
Work of force F corresponding to infinitesimal displacement, dr = dU = F. dr
Work of F corresponding to a finite displacement of particle from A1 to A2 and covering distances S1, S2,
U1-2 = F . dr or U1-2 = (F cos) ds = F (S2-S1)A2
A1
S2
S1
S1, S2 distance along the path traveled by the particle
Area under curve = U1-2
Similarly, work of a couple of moment M, dU = M d
U1-2 = M d = M (2-1)2
1

Work of a weight
yW
WyWy
WdyU
WdydU
y
y
=
=
=
=
21
21
2
1
Work is equal to product of W and vertical displacement of CG of body; Body moves upwards; Body moving downwards will have +ve work done
( )
222
1212
1
21
2
1
kxkx
dxkxU
dxkxFdxdU
x
x
=
=
==
Work of a springF = k xk spring constant, N/m
+ve work done is expected if x2 < x1, i.e., when spring is returning to its un-deformed position
( ) xFFU 2121
21 +=

Potential Energy
Work of a weight: 2121 WyWyU =
The work is independent of path and depends only on
positions (A1, A2) or Wy
potential energy of the body with
respect to the force of gravity Wr== gVWy
( ) ( )2121 gg
VVU =
Vg1 < Vg2 => PE is increasing with displacement in this case, work done is negative
Work is positive, if PE decreases
Unit of PE Joule (J)

Work of a spring
( ) ( )=
=
=
e
ee
V
VV
kxkxU
21
222
1212
121
potential energy of the body with
respect to the elastic force Fr
Here PE increases, work done is (-ve)
Now in general, it is possible to find a function V, called potential energy, such that, dU = -dV
U1-2 = V1 V2 => A force which satisfies this eqn. is conservative force
Work is independent of path & negative of change in PE for the cases seen

Potential energy & equilibrium
Considering virtual displacement, U = -V = 0
=> (dV / d) = 0 => position of the variable defined by single independent variable,
In terms of potential energy, the virtual work principle states that if a system is in equilibrium, the derivative of its total potential energy is zero
(V/) = 0
Example:
Initial spring length = AD
Work is done only by W, F
For equilibrium, U = (Ve + Vg)
1
2

Total potential energy of the system, V = Vg + Ve
For W For F
dv/d = 4kl2sin cos Wl sin = 0
= 0 and = cos-1 (W/4kl)

POC
k
BA
b
b
b
b
Two uniform links of mass, m are connected as shown. As the angle increases with P applied in the direction shown, the light rod connected at A, passes thro pivoted collar B, compresses the spring (k). If the uncompressed position of the spring is at = 0, find the force which will produce equilibrium at the angle
Compression distance of spring, x = movement of A away from B; X = 2b sin /2
U = (Ve + Vg)
Find Ve, Ve; Vg, Vg; U (of P)
Vg = 0
Ve = k x2; Vg = mgh
U = P (4b sin /2)
2Pb cos /2 = 2kb2sin /2 cos /2 + mgb sin /2
P = kb sin /2 + mg tan /2
4bsin/2

Meriam/Kraige, 7/39
For the device shown the spring would be un-stretched in the position =0. Specify the stiffness k of the spring which will establish an equilibrium position in the vertical plane. The mass of links are negligible.
Spring stretch distance, x = 2b-2b cos
Ve = k [(2b)(1-cos )]2 = 2kb2 (1-cos )2
Vg = -mgy = -mg (2bsin) = -2mgbsin
V = 2kb2 (1-cos )2 - 2mgbsin
For equilibrium, dv / d = 4kb2(1-cos ) sin - 2mgb cos = 0
=> K = (mg/2b) (cot /1-cos )
kb b
b
m
y
Vg = 0

0=d
dV
0=d
dV
d2V / d2 < 0d2V / d2 > 0Must examine higher
order derivatives and are
zero
AB AB
Stability of equilibrium (one DOF )

Knowing that the spring BC is un-stretched when = 0,
determine the position or positions of equilibrium and state
whether the equilibrium is stable, unstable, or neutral.
Beer/Johnston
( ) ( ) cos221
2
21
bmgak
mgyks
VVV ge
+=
+=
+=
( )( )( )( )( )
8699.0
m3.0sm81.9kg10
m08.0mkN4sin
sin0
2
22
2
=
==
==
mgb
ka
mgbkad
dV
=== 7.51rad902.00

( )( ) ( )( )( )
cos43.296.25
cosm3.0sm81.9kg10m08.0mkN4
cos
22
2
2
2
=
=
= mgbkad
Vd
at = 0: 083.32
2
+=d
Vd stable

Spring AB of constant 2 kN/m is attached to two identical drums as shown.Knowing that the spring is un-stretched when = 0, determine (a) the range of values of the mass m of the block for which a position of equilibrium exists, (b) the range of values of for which the equilibrium is stable.
Beer/Johnston (10.81)
AB
AB

(Sin varies from 0 to 1)
(Cos varies from 1 to 0)

XV = ( xc dv) YV = ( yc dv) ZV = ( zc dv)Centroid of volume:
XA = ( xc dA) YA = ( yc dA) ZA = ( zc dA)
Centroid of area:
Moments of inertia : The moment of inertia of an object about a given axis describes how difficult it is to change its angular motion about that axis.
First moment of volume
w.r.t. yz plane
Symmetry plane
Centroid of volume xc dv = 0

Consider a beam subjected to pure bending.
Internal forces vary linearly with distance from
the neutral axis which passes through the section
centroid.
X-axis => neutral axis => centroid of section
passes
F = k y A vary linearly with distance y
moment second
momentfirst 022 ==
====
=
dAydAykM
QdAydAykR
AkyF
x
r
MX = y F = k y2 A;
Moment of inertia of beam section w.r.t x-axis, IX (+VE)

Second moments or moments of inertia of an area with
respect to the x and y axes,
== dAxIdAyI yx22
For a rectangular area,
331
0
22 bhbdyydAyIh
x ===
IY = x2 dA = x2 h dx = 1/3 b3h0
b
Rectangular moment of inertia

The polar moment of inertia is an important parameter in problems involving
torsion of cylindrical shafts and rotations of slabs.
= dArJ2
0
The polar moment of inertia is related to the rectangular moments of
inertia,
( )xy II
dAydAxdAyxdArJ
+=
+=+== 22222
0
Polar moment of inertia

Consider area A with moment of inertia Ix. Imagine that
the area is concentrated in a thin strip parallel to the x axis
with equivalent Ix.
A
IkAkI xxxx ==
2
kx = radius of gyration with respect to the x axis
A
JkAkJ
A
IkAkI
OOOO
yyyy
==
==
2
2
222yxO kkk +=
Radius of gyration

Beer/Johnston
( )h
hh
x
yyh
h
b
dyyhyh
bdy
h
yhbydAyI
0
43
0
32
0
22
43
=
=
==
12
3bhI x=
Determine the moment of inertia of a
triangle with respect to its base.
For similar triangles,
dyh
yhbdA
h
yhbl
h
yh
b
l =
=
=
dA = l dy
Determination of MI by area of integration

yMI of rectangular area:
dA = bdy
xb
h
y
dy
Ix = y2 dA = y2 bdy = 1/3 bh3; Iy = 1/3 hb3
0
h
MI - Ix and Iy for elemental strip:
y dIx = 1/3 dx (y3) = 1/3 y3 dx
dIy = x2dA = x2y dx or 1/3 x3dy
x
YX
dA = Ydx From this, MI of whole area can be calculated by integration
dx
dy
About centroidal axis (X, Y): Ix = 1/12 bh3; Iy = 1/12 b3hX
Y

y
x
a
by = k x5/2
Find MI w.r.t Y axis
Beer/Johnston (9.1)

Triangle: bh3/12 (about base)
Circular area: /4 r4 (about dia)
Rectangular area: bh3/3 (about base)

Parallel axis theorem
Consider moment of inertia I of an area A with respect
to the axis AA
= dAyI2
The axis BB passes through the area centroid and
is called a centroidal axis.
( )
++=
+==
dAddAyddAy
dAdydAyI
22
22
2
2AdII +=
dA
A A
y
CB B
d
y
C Centroid
BB Centroidal axis
MI of area with centroidal axis
0
Parallel axis theorem
Jo = Jc + Ad2First moment of area

Moments of Inertia of Composite Areas
The moment of inertia of a composite area A about a given axis is
obtained by adding the moments of inertia of the component areas A1,
A2, A3, ... , with respect to the same axis.
x
y
It should be noted that the radius of gyration of a composite area is not
equal to sum of radii of gyration of the component areas

MI of some common geometric shapes

Moment of inertia IT of a circular area with respect to a
tangent to the circle T,
( )4
45
224412
r
rrrAdIIT
=
+=+=
Application 1:
Application 2: Moment of inertia of a triangle with respect to a centroidal axis,
( )3
361
2
31
213
1212
2
bh
hbhbhAdII
AdII
AABB
BBAA
=
==
+=
IDD = IBB + ad2 = 1/36 bh3 + 1/2bh (2/3h)2 = bh3

Find the centroid of the area of the un-equal Z section. Find the moment of inertia of area about the centroidal axes
shames
Ai xi yi Aixi Aiyi
2x1=2 1 7.5 2 15
8x1=8 2.5 4 20 32
4x1=4 5 0.5 20 2
Ai = 14 Aixi = 42 Aiyi = 49
1
2
3
Xc = 42/14 = 3 in.; Yc = 49/14 = 3.5in
y
x
1
6
2 1 4
1
1
2
3
Xc, Yc
Ixcxc = [(1/12)(2)(13)+(2)(42)] + [(1/12)(1)(83)+(8)(1/2)2] + [(1/12)(4)(13)+(4)(32)] = 113.16 in4
Similarly, Iycyc = 32.67 in4

Beer/Johnston:
Determine the moment of inertia of the shaded area
with respect to the x axis.
Rectangle:
( )( ) 46313
31 mm102.138120240 === bhIx
3
Half-circle:
moment of inertia with respect to AA,
( ) 464814
81 mm1076.2590 === rI AA

( )( )
( )23
2
212
21
mm1072.12
90
mm 81.8a-120b
mm 2.383
904
3
4
=
==
==
===
rA
ra
moment of inertia with respect to x,
( )( )46
362
mm1020.7
1072.121076.25
=
== AaII AAx =25.76x106 (12.72x103) (38.2)2
moment of inertia with respect to x,
( )( )46
2362
mm103.92
8.811072.121020.7
=
+=+= AbII xx
46mm109.45 =xI
xI = 46mm102.138 46mm103.92

Product of inertia, Ixy
= dAxyI xy[Similar to Ixx (or Ix), Iyy (or Iy)]
When the x axis, the y axis, or both are an axis of symmetry,
the product of inertia is zero.
The contributions to Ixy of dA and dA will cancel out
Parallel axis theorem for products of inertia:
AyxII xyxy +=
Centroid C is defined by x, y

Moment of inertia, Product of inertia about rotated axes
Given
=
==
dAxyI
dAxIdAyI
xy
yx22
we wish to determine moments and product of
inertia with respect to new axes x and y
x, y rotated to x, y
2cos2sin2
2sin2cos22
2sin2cos22
xyyx
yx
xyyxyx
y
xyyxyx
x
III
I
IIIII
I
IIIII
I
+
=
+
+
=
++
=
The change of axes yields
Ix+Iy = Ix+Iy

y
xa
ImaxImin
Assume Ixx, Iyy, Ixy are known for the reference axes x, y
At what angle of , we have maximum and minimum I
Minimum angle will be at right angles to maximum angle
These axes are called Principal axes & MI are Principal MI
Principal axes & Principal MI
Imax, min = (Ix+Iy/2) (Ix-Iy/2)2 + Ixy2
tan 2 = 2Ixy / (Iy-Ix)

For the section shown, the moments of inertia with
respect to the x and y axes are Ix = 10.38 cm4 and Iy =
6.97 cm4.
Determine (a) the orientation of the principal axes of the
section about O, and (b) the values of the principal
moments of inertia about O.
Apply the parallel axis theorem to each rectangle,
( ) += AyxII yxxyNote that the product of inertia with respect to centroidal axes parallel to the xy
axes is zero for each rectangle.
56.6
28.375.125.15.1
0005.1
28.375.125.15.1
cm,cm,cm,cmArea,Rectangle42
=
+
+
Ayx
III
II
I
Ayxyx

( )
=
+=
=
=
255.4and4.752
85.397.638.10
56.6222tan
m
yx
xym
II
I
== 7.127and7.37 mm
( )22
22
minmax,
56.62
97.638.10
2
97.638.10
22
+
+=
+
+= xy
yxyxI
IIIII
4min
4max
cm897.1
cm45.15
==
==
II
II
b
a

treliça statica.pdf

Documents

LOBO CARNEIRO TRELIÇA DE MÖRSCH

1º Condição de Treliça Isostática

RMC - Treliça (1)

Mecânica Aplicada I Trabalho Laboratorial Treliça ·...

Tabuleiros de Pontes em Treliça Mista

Capítulo 2 - Método dos Elementos Finitos para treliça 2D...

RELATORIO TRELIÇA MADEIRA

Dimensionamento de treliça-UNIUBE+PROFESSOR DOGMAR

Capítulo 3 Método dos Elementos Finitos para treliça 3D.....

Análise Matricial da Treliça de Howe

Modulação Em Treliça

Análise de Treliça Espacial Constituída de Material ...

Excel Treliça

Statica -...

PROJETO DE TRELIÇA METÁLICA PLANA DE GALPÃO INDUSTRIAL

CÁLCULO E DIMENSIONAMENTO DOS ELEMENTOS ESTRUTURAIS DE...