Resistencia Dos Materiais

Problem 1-1

Determine the resultant internal normal force acting on the cross section through point A in each

column. In (a), segment BC weighs 300 kg/m and segment CD weighs 400kg/m. In (b), the column

has a mass of 200 kg/m.

(a) Given: g 9.81

m

s2

:= wBC 300

kg

m

:=

LBC := 3m wCA 400

kg

m

:=

FB := 5kN LCA := 1.2m

FC := 3kN

Solution:

+↑Σ Fy = 0; FA wBC g − ( ⋅ )⋅LBC − (wCA⋅g)⋅LCA − FB − 2FC = 0

FA := (wBC⋅g)⋅LBC + (wCA⋅g)⋅LCA + FB + 2FC

FA = 24.5 kN Ans

(b) Given: g 9.81

m

s2

:= w 200

kg

m

:=

L := 3m F1 := 6kN

FB := 8kN F2 := 4.5kN

Solution:

+↑Σ Fy = 0; FA w L − ( ⋅ )⋅g − FB − 2F1 − 2F2 = 0

FA := (w⋅L)⋅g + FB + 2F1 + 2F2

FA = 34.89 kN Ans

Problem 1-2

Determine the resultant internal torque acting on the cross sections through points C and D of the

shaft. The shaft is fixed at B.

Given: TA := 250N⋅m

TCD := 400N⋅m

TDB := 300N⋅m

Solution:

Equations of equilibrium:

+ TA − TC = 0

TC := TA

TC = 250N⋅m Ans

+ TA − TCD + TD = 0

TD := TCD − TA

TD = 150N⋅m Ans

Problem 1-3

Determine the resultant internal torque acting on the cross sections through points B and C.

Given: TD := 500N⋅m

TBC := 350N⋅m

TAB := 600N⋅m

Solution:

Equations of equilibrium:

Σ Mx = 0; TB + TBC − TD = 0

TB := −TBC + TD

TB = 150N⋅m Ans

Σ Mx = 0; TC −