並列部材の不静定（初期すきまあり）
剛体壁と剛体板の間に、長さの違う丸棒を並べて挟み込んだ不静定問題の計算です。短い側の部材にすきまがあるため、荷重が小さいうちは長い側だけが力を受けます。剛体板の変位・各部材の反力と応力・断面積を求めます。
Input
External force
Length of member 1
Clearance
部材1Quantity
部材1Diameter
部材1Young's modulus
部材2Quantity
部材2Diameter
部材2Young's modulus
Result
部材1Contraction
部材1Cross-sectional area
部材1Reaction force per bolt
部材1Stress
部材2Cross-sectional area
部材2Contraction
部材2Reaction force per bolt
部材2Stress
Notes
- Reactions and stresses are given per member. The totals are for member 1 and for member 2, and they add up to the applied force .
- The wall and the plate are taken as rigid, and the plate is assumed to move parallel to itself without tilting. This calculation does not cover a plate that bends, or one that tilts because the members are unevenly placed.
- Members are treated as round bars, with the area taken as . Buckling is not checked. Check it separately when a slender member is in compression.
- leaves member 2 with zero or negative length, so nothing can be calculated. Enter a gap smaller than the length of member 1.
- The applied force is assumed to push. Enter it pulling — a negative value — and the rigid plate lifts off the members, which this calculation does not describe.
Related pages
How to solve parallel members with an initial gap
Walks through adding the compatibility condition to the equilibrium equations, what to do while the shorter member is not yet in contact, and the sign convention.
Bar fixed at both ends (statically indeterminate, in series)
A statically indeterminate bar fixed to walls at both ends, loaded axially somewhere along its length. The members sit in series here, so the compatibility condition reads: the two extensions sum to zero.
Giãn nở nhiệt và ứng suất nhiệt
Stress from a temperature change in a restrained member. Working from the expansion the restraint cancels is the same compatibility argument.