STATIC FORCE ANALYSIS
STATIC FORCE ANALYSISWEEK 1
COUPLEMOMENT OF COUPLECouple: Two equal and opposite forces (F & F’)Moment of Couple: A vector normal to the planeof the couple (M RBAX F)
Conditions of EquilibriumA system of bodies is in equilibrium, if, and only if:In 2-D (planar) systems:
Two-Force MembersNot in equilibrium F 0, M 0Not in equilibrium F 0, M 0In equilibrium F 0, M 0Condition of equilibrium, for any two-force member with no appliedtorque: The forces are equal, opposite and have the same line of action.
Three-Force MembersONot in equilibrium M 0In equilibrium F 0, M 0Condition of equilibrium, for any three-force member with no applied torque:Forces should be coplanarFour-Force Members: The problem is reduced to one of three-force member.Then the approach above is applied.
Example (Graphical Solution)Link 3 is a two-force member
b)1 st approach:c)2 nd approach: Since link 4 is is a three-force member; lines of action of forces P, sobtained.1414d) Force triangle is used.If 1st approach was used, this triangle gives direction and magnitude of F14If 2nd approach was used, this triangle gives magnitudes of F34,Since, graphicallye) Action and reaction forces are equal:f) F 0 andF14
(Analytical Solution) 5(cos68.4i sin68.4j) X 120(-cos40i-sin40j) 12(cos68.4i sin68.4j) X F34 (cos 22.4i sin 22.4j) 0F34 33.1 lb
Problem (Analytical Solution)(without friction)O2A 75 mm P 0.9 kNAB 350 mm M12 ? F 0F34 P F14 0F34(cos11,95i-sin11,95j)-900i F14j 0i: 0.978 F34 F34x 900 F34 920.25 Nj: 0.207 F34 F34y F14 F14 190.5 NFor the moment balance,all of the force vectorsshould pass throughpoint B.F34 920,25 /-11,95 N 900i-190.5j NF14 190.5j N
Two-force member F 0F23 -F43 F34 F 0F12 -F32 F23 F34 MO2 0M12 O2A x F32 0M12k 0.075(cos105i sin105j) x (-900i 190.5j) 0M12k – 3.69k 65.2k 0M12 -61.51 N.m M12 -61.51k N.m
Problem (Analytical Solution)(with friction) 0.2 (Between piston and cylinder)0.2N F 0 F34 P F14 0F34 P (0.2Ni Nj) 0F34(cos11.95i-sin11.95j)-900i 0.2Ni Nj 0i: 0.978 F34 -900 0.2N 0j: 0.207 F34 N 0 N 0.207 F340.978 F34 -900 0.2(0.207F34) 0 F34 882.61 NewtonN 182.7 Newton F14 36.54i 182.7j NewtonF34 882.61 /-11,95 Newton 863.48i-182.75j Newton F 0 F12 -F32 F23 F34 MO2 0 M12 O2A x F32 0M12k 0.075(cos105i sin105j) x (-863.48i 182.75j) 0M12k – 3.547k 62.55k 0M12 -59.003 N.m M12 -59.003k N.m CONCLUSION ?
STATIC FORCE ANALYSIS WEEK 1 . COUPLE MOMENT OF COUPLE Couple: Two equal and opposite forces (F & F’) Moment of Couple: A vector normal to the plane of the couple (M R BA X F) Conditions of Equilibrium A system of bodies is
6. Explain the procedure for static force analysis of slider-crank mechanism Steps in the static force analysisof slider-crank mechanism i) configuration diagram is drawn to scale. F is the input force on link 4 ii) L
Oasys GSA Contents Notation 8 Degrees of freedom 10 Active degrees of freedom 10 Degrees of Freedom with no Local Stiffness 11 Analysis Options 13 Static Analysis 13 Static P-delta Analysis 13 Modal Analysis 14 Modal P-delta Analysis 14 Ritz Analysis 15 Modal Buckling Analysis 16 Model Stability Analysis 17 Non-linear Static Analysis 18
Abstract: Static analysis relies on features extracted without executing code, while dynamic analysis extracts features based on execution (or emulation). In general, static analysis is more efficient, while dynamic analysis can be more informative, particularly in cases where the code is obfuscated. Static analysis of an Android application
Static force analysis makes direct use of static force equilibrium equations. An analysis that includes inertia effects is called a dynamic-force analysis and will be discussed in the next chapter. For an analytical solution formulation one must draw the 3 free-body diagram
3M ª Metal-in Static Shielding Bag SCC 1000, Open Top and Ziptop . Static Shielding Bag SCC 1300 3M . 3M ª Metal-Out Static Shielding Bag SCC 1500, Open Top and Ziptop 3M Metal-Out Cushioned Static Shielding Bag 2120R Metal-in Shield Bags are intended to provide a static safe environment for electronic devices. Metal-in Shield Bags
Static routes are manually configured and define an explicit . Configuring an IPv6 static route is very similar to IPv4 except that the command is now ipv6 route. The following must be configured before entering a static . IPv6 also has a default static route similar to the IPv4 quad zero (0.0.0.0) static default route. Instead, the IPv6 .
Configure IP Default Static Routes Default Static Route (Cont.) IPv4 Default Static Route: The command syntax for an IPv4 default static route is similar to any other IPv4 static route, except that the network address is0.0.0.0and the subnet mask is0.0.0.0. The 0.0.0.0 0.0.0.0 in the route will match any network address.
Module Objective: Troubleshoot static and default route configurations. Topic Title Topic Objective Packet Processing with Static Routes Explain how a router processes packets when a static route is configured. Troubleshoot IPv4 Static and Default Route Configuration Troubleshoot common static and default route configuration issues.
