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Shape factor, plastic hinge properties, T-beam

Short answer

Shape factor S = Zp/Ze = Mp/My. A plastic hinge forms at full yield, rotates at constant Mp, occurs at points of maximum moment, and reduces indeterminacy by one. For the T-section, Zp = 9.92×10^7 mm³, Ze = 5.48×10^7 mm³; S ≈ 1.81.

Define shape factor and write properties of plastic hinge. Find shape factor of the given T-beam section. [Figure: T-section; flange width 2000 mm and thickness 150 mm; web depth 950 mm and web width 175 mm (overall depth measured as printed).]

Shape factor

The shape factor is S=MpMy=ZpZeS = \dfrac{M_p}{M_y} = \dfrac{Z_p}{Z_e}, the ratio of the plastic moment of the section to the yield moment.

Properties of a plastic hinge

  1. It forms at a section where the moment reaches MpM_p and the whole depth of the section has yielded.
  2. It rotates at a constant moment MpM_p (like a hinge that carries a moment) and offers no extra resistance to rotation.
  3. It forms at the points of maximum moment: fixed supports, joints, under concentrated loads, and at the section of zero shear under a UDL.
  4. Rotation takes place only in the direction of the applied moment; on unloading or reversal the hinge locks and behaves elastically.
  5. It extends over a finite length of the member (the plastic hinge length), but is treated as a point in analysis.
  6. Each hinge reduces the degree of indeterminacy by one; when enough hinges form (degree + 1), the structure becomes a mechanism and collapses.

Shape factor of the T-section

Assumed dimensions: flange 2000 mm × 150 mm; web 175 mm wide, 950 mm deep below the flange; total depth D=150+950=1100D = 150 + 950 = 1100 mm. (If 950 mm is the overall depth, the web is 800 mm and the shape factor becomes 1.84.)

Area: A=2000×150+950×175=300000+166250=466250A = 2000\times150 + 950\times175 = 300000 + 166250 = 466250 mm².

1. Elastic neutral axis (distance from the top):

yˉ=300000×75+166250×(150+475)466250=271.11 mm\bar y = \frac{300000\times75 + 166250\times(150+475)}{466250} = 271.11\ \text{mm}

2. Moment of inertia about the elastic neutral axis:

I=2000×150312+300000(271.11−75)2+175×950312+166250(625−271.11)2=4.542×1010 mm4I = \frac{2000\times150^3}{12} + 300000(271.11-75)^2 + \frac{175\times950^3}{12} + 166250(625-271.11)^2 = 4.542\times10^{10}\ \text{mm}^4

3. Elastic section modulus (extreme fibre in the web at 1100−271.11=828.891100 - 271.11 = 828.89 mm):

Ze=4.542×1010828.89=5.480×107 mm3Z_e = \frac{4.542\times10^{10}}{828.89} = 5.480\times10^{7}\ \text{mm}^3

4. Plastic neutral axis divides the area into two equal halves, A/2=233125A/2 = 233125 mm². The flange area is 300000 mm² >233125> 233125, so the axis lies in the flange, at depth

yp=2331252000=116.56 mm below the topy_p = \frac{233125}{2000} = 116.56\ \text{mm below the top}

5. Plastic section modulus (moment of each half-area about the plastic axis):

Zp=2000×116.56×116.562+2000×33.44×33.442+166250×(33.44+475)=9.923×107 mm3Z_p = 2000\times116.56\times\frac{116.56}{2} + 2000\times33.44\times\frac{33.44}{2} + 166250\times\left(33.44 + 475\right) = 9.923\times10^{7}\ \text{mm}^3

6. Shape factor:

S=ZpZe=9.923×1075.480×107=1.81S = \frac{Z_p}{Z_e} = \frac{9.923\times10^{7}}{5.480\times10^{7}} = 1.81

Answer: S≈1.81S \approx 1.81 (elastic Ze=5.48×107Z_e = 5.48\times10^7 mm³, plastic Zp=9.92×107Z_p = 9.92\times10^7 mm³, plastic neutral axis in the flange 116.6 mm from the top).

✓ Answer checked by Claude Opus 5.5 (high effort)

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