Two horizontal forces P= 15 kN are applied to pin B of the assembly shown in the figure blew, Knowing that a pin of 16 m

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Two Horizontal Forces P 15 Kn Are Applied To Pin B Of The Assembly Shown In The Figure Blew Knowing That A Pin Of 16 M 1 (26.86 KiB) Viewed 56 times

Two Horizontal Forces P 15 Kn Are Applied To Pin B Of The Assembly Shown In The Figure Blew Knowing That A Pin Of 16 M 2 (46.24 KiB) Viewed 56 times

Two Horizontal Forces P 15 Kn Are Applied To Pin B Of The Assembly Shown In The Figure Blew Knowing That A Pin Of 16 M 3 (37.03 KiB) Viewed 56 times

Two Horizontal Forces P 15 Kn Are Applied To Pin B Of The Assembly Shown In The Figure Blew Knowing That A Pin Of 16 M 4 (37.03 KiB) Viewed 56 times

Two horizontal forces P= 15 kN are applied to pin B of the assembly shown in the figure blew, Knowing that a pin of 16 mm diameter is used at each connection. 10 mm B 50 mm PKN PKN 10 mm 50 mm 45° 70° C
a) Determine the member force. Please insert - sign if it indicates the compressive member. KN FAB 11 FBC = KN b) Determine the maximum value of the average normal stresses in link AB and in link BC. Please insert - sign if it indicates the compressive stress. AB = MPa Oc= MPa
) c) Determine the average shearing stresses in the pin at A, B, and C; Ta = MPa TA TB MPa Tc = MPа d) Determine the average bearing stresses at Ain member AB, at Bin member BC, and at Cin member BC. Bearing Stress A MPa Bearing Stress B MPa Bearing Stress C= MPa