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The beam is made of elastic-perfectly plastic material. Determine the maximum elastic moment and the plastic moment that can be applied to the cross section. Take { \sigma }_{ Y } = 36 ksi.

Step-by-step

Referring to Fig. a, the location of centroid of the cross section is

{ \overline { y } =\frac { \sum { \overline { y } A } }{ \sum { A } } =\frac { 7.5\left( 3 \right) \left( 6 \right) +3\left( 6 \right) \left( 3 \right) }{ 3\left( 6 \right) +6\left( 3 \right) } =5.25\quad in }

The moment of inertia of the cross-section about the neutral axis is

I = { \frac { 1 }{ 2 } } (3)(6^{3}) + 3(6)(5.25 – 3)^{2} + { \frac { 1 }{ 2 } } (6)(3^{3}) + 6(3)(7.5 – 5.25)^{2} = 249.75 \quad in^{4}

Here, { { \sigma }_{ max }=\sigma }_{ Y } = 36 ksi and { ¢=\overline { y } }= 5.25 in.Then

{ { \sigma }_{ max }=\frac { Mc }{ I } ;\quad \quad \quad \quad 36=\frac { { M }_{ Y }\left( 5.25 \right) }{ 249.75 } }

M_{Y} = 1712.57 kip\cdotin = 143 kip\cdotft

Referring to the stress block shown in Fig. b,

{ \int _{ A }^{ }{ \sigma dA=0; } }\quad \quad \quad \quad T – C_{1} – C_{2} = 0

d(3) (36) – (6 – d)(3)(36) – 3(6) (36) = 0

d = 6 in.

Since d = 6 in.,  C_{1} = 0 ,

Here,

T = C = 3(6)(36) = 648 kip

Thus,

M_{P} = T(4.5) = 648(4.5) = 2916 kip\cdotin = 243 kip\cdotft

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